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English Pages [332] Year 2014
Problems Plus in
T Matiematics A Das Gupta
29881909
Part 2
A-150
Problems
a,
Let
and
B
be real and
i
a
i
Pls
=1and a be an arbitrary complex number
= 1.} B
f(a-106-1)- (B+1a-1)}a a(a
1
MB 1)-
(a
-
1MB
where the sets A
tAnB=o,
in lIT Mathematies 1
such that
(a-1KB-1)=0
-1)=0.
,p,
=
2,
then find the set of
possible values of a.
.
Find the quadratic equation whose roots are nth powers of the roots of r* - 12 x+1=0. t the point representing the complex number z, is a point on or inside the circle whose
l.
centre is the complex numberr+ i -0 and the radius is r then prove that the greatest value + of +3+ lisn(n +1).
n
Answers 1. (i) A,D
5.
(-b--2e
6.
1
9.
3
12.
(ü)C
(i) A.E
2.(i) 0 (i)
3z (ii) 0, 1, respectively
+20)= (C+ (6-b}, C0-
c-C
5 12
-1,1+ i, -i} 1
11. 4S
13.x-2r cos+1 =0
5. Permutation and Combination Recap of Facts and Formulae Counting of number of ways to do some work
3. Counting
consists of two parts W, IW, of which one aurt can be done in m ways and the other part in n ways
fa work
W
The number ofcombinations (selections)
"C,=n_ "C,!n-T) The total number of selections of one or more objects from n different objects
2-1="C
the work W can
be done in mn ways, if both the parts are to be done one after the other to do the work W. (Multiplication latw of counting)
number of ways to do the work under the restriction = (the number of ways to do the work without restriction) (the number of ways to do the work under opposite restriction).
formulae for permutation
Thenumber of permutations (arrangements) of ditferent things taking rat a time
"P,
n-r)!
wheren!=
..+"C,).
(when selection of 0 things is allowed) (when at least one thing is to be selected).
n+1
.The total number of selections from
the
. Counting
+ "C, +"C,+
.The total number of selections of any number of things fromn identical things
Similar is the law for works that have 3 or more parts.
fa work is to be done under some restriction then
of n diferent
things taking r at a time
ehen
the work W can be done in m+n ways, if by doing any of the parts the work W is done. (Addition lawofcounting)
formulae for combination
q
p like things, like things of another type and r distinct things
=(p+10q+1)2"-1
if at least one thing is
tobe selected) (P+1Mg+ 1)2'-2 (if none or all cannot be selected) The total number of selections of r things from n different things when each thing can be repeated unlimited number of times ="c,
n
For example If 8 balls are to be selected from balls of 3
1.2.3. ...
colours, balls of each colour being available unlimited number of times, then the number of selections
1.
ne number of permutations of n things taking all at a
38-1Cs
e
of which p things are identical, q things are Iaentical of another type and the rest are different
10!
10C%12! 4.
he
number of arrangements of n dijfërent thinigs round a closed curve -1)! if clockwise and anticlockwise arrangements are considered different,
herer
=8, n
=
3)
10 x9
245
Number of distributions The number of ways to distribute n dijiirent things between two persons, one receiving p things
other qthings,wherep+q=n
"Cx-C
2-1)! if clockwise and anticlockwise
n
(-p)! p!(n-p)! q!n-p-q)!
arrangements are considered identical. (Circular permutation) A-151
and the
A-152
Problems Plus in IIT Mathematics
= coefficient of
n=p+q
p!q!
r" in
Similarly for 3 persons, the number ways of
!p!q!r'
where p +q+r=n.
The number of ways to distribute
m
among n persons equally = mn)!
.
(m
r(r
xn
n
The number of ways to divide n different things into three bundles of p, q and r
things!air p
3!
different things
m * n
(m 1)
The total number of ways to divide among r persons
n identical things
=nr-'Cr-1 For example The number of different ways to divide 20 identical apples among 4 persons
=204-C4-, (* here n =20,r= 4)
23C23 *22 x21 = 23 x 11x7
n!
T(r+ 10r+ 2)... (r+n-1)
!"
The number of ways to divide into n equal bundles= n)
+1)r+2).. (r+n-1)
different things
(r+n-1)_n+-'C,n!(r-1) ! The number of integral solutions of
x+*2+X +
+I=n where x,21, x22 1,... ,x.>
is the same as the number of ways to distribute identical things amongr peršons each getting at least1, This is also equal to the coefficient of x" in the
expansion of
(r+x*+rt.)
=coeficient of r" =
in|
coefficient of x " in x'(1
-
x)
coefficient of r" in
r1++Tr+1),2,
6
1771.
rr+1)r+2).(+n-1),»,
.The toal number of ways to divide n identical things amongr persons so that each gets at least one
-Cr-
For example The number of different ways to distribute 20 identical balls in 4 different boxes so that no box remains empty =
20-1C-1
=
C
=
969.
l'
coefficient of "- in
1+
224.. rtr
here n =20, r=4)
-18 17
+ 1Xr+2). (r+n1
_r(r+ 1)(7+2)... (7+ n-r-1)
6
r(+
.
(n r)! 1(7 +2) (-1)D (-r)
5. Use of solution of linear
equations and coefficient of a power in expansions to find the number of ways of
distribution
The number of integral solutions of X +X2 +s+.. +X,=n where x 20, 20, ., x,20 is the same as the number of ways to distribute n identical things among r persons. This is also equal to the coefficient of x" in the
expansion of (x°+x'+x2+x*+.)
in 1-
=
coefficient of
=
coefficient of r" in
(1
-
x)
(n-1)! (n-r)!(r-1)!
n-1C-1
Note The number of solutions of x +X2 4+X3 + 4 wherex2 0, x21,x 23, x25 is equal to the coefficient of x in
2
(x+x+x2 r+x2+x3+ ..) x(r+x*+x5.. )(x+r+x+..) 6. Some useful results
C,="C-r .C,+"C-1="*'C
A-153 Permutation and Combination
1
"= 1-1)
-Co+"C,x +"*'C3x+"*C,x'+
+1 +2), n(+1)2, Y,2 (nn +10n
+ nx +
+"-Cxt
3!
2!
nn +10n+2). ( +r-),,
where
n is a
positive integer.
Selected Solved Exanples is divisible by ( )*. Showthat(kn)! 1. ... (kn) (kn)!= 1/2:3 Here u)l(n + 1)n + 2).. (n + = (1-2-3.
-'C+("* m -'C+" n)}
="*"C +""Ck-1
+ I(2n+ 1(2n + 2)... (2n n)..
(rn
Now,
+ 1)
[(k 1)n
x
+
1)(rn
1)n
|k-
N 2)... (rn + n), wherere (rH + n)
(rn +n)
(rn+1(rn +2)... by (1),
+)
Tm
and 2.
!
= (n !
M,
M,,
M)... (u! Mg-1 integer = (n M... M- !*x
divisible by )* (n*) ! is divisible by Putting k= u, we get
(n
j-0
'C+"Ck-1) +("*'Ck-1+""C-1t
Ck-1
+'Cg-1+"**Cg-t. n+m-C-1+""Ck-1
=("C+"*'C-1)
Ck1t+n-Ck-1+
n+Ck-1)
++-1C+"*"Ck-1)
="C+("*2Ck-1 +..+
r- 1) r)!
r-2)!
(r+2)! (-
1
or or
+(r*+3r+ 2) (n-r)-(2r+5)(n-r)
replacing r by Similarly, from (2)
(n-r-
1)*-(2r+7)(n
+I(r+1)+3(r+1)+21
or
(nr)- 2(n-r)+1
or
(n-r)-(2r
-
0
.
r+1 in (3), we get
r -1)
-
0
(2r+ 7)(m- r) =
+3 +2) 0 +(2r +7 +r+2r+1+3r = (4) 14) 0 +9(n )+(r* +7r+ -
identical (3) and (4) must be
+
+
(2)
(r+2)(r+ 1) 2 (n- r- 1)(n- r) (r+ 1)(-r- 1) 1)(0n- r) + 1)+(n-r2(r +2)(- r) = (r+2)(r . (3) =
'Ck-1+"**Ck-1t.
nnCk-1)
-
r!(n-
!.
2. Evaluate "C+ E"*lCk-1
"'C+("
.
"Cr+3
"C,2= "C,+
where M, is an integer
="C+ ("C+"u*
(1)
"C, + "C,+2
From(1),2r+1)!(1
!
(tn
is
in AP
"C,, = integer
is divisible byn
+n) =n
(rn
*
"Cr,2"C,,a are
2."C=
(kn)!=n! (n! M) (n!
Value
"Cr, l
"C
(rn)!
n!
( (rn +1)(ru +2)...
Note
found n, r cannot be
integers 3. Show that two positive "C,,2, "C,.s are in AP such that "C,,"C,,1,
+n)! +2).. (rn + )(rn (rm)!n! -
(kr)!
Ck
Ik- 1)n +n] (1)
+ 1)(rn+2).. 1-2-3. rn1-2-3 .. (rn)
(1 +1)(rn
+ 21
mCk. 1)
+
quadratic in (n- ),
127+5r'+3r+2
2r492+7r+14 2r 9 5
+9
2r +5
and
r+7r+ 14 =r+3r +2
= -3. or 4r=-12, i.e., r or are absurd. Both 9 5 and r =-3 hold at the same time. (1) and (2) cannot + 3 cannot be in AP.
"C"C
"C.2. "C
A-154 4.
Problems Plus in
IT Mathematics
How many numbers ofn digits can be made with the nonzero digits in which no two consecutive digits are
the same?
There are nine nonmzero digits, namely 1,2,3,... and 9.
Now, a number is divisible by 6 if it is even evor as well as divisible by 3. So, the number of 4-digit numbers divisiblot
by
6 that can be made with 1,2, 4, 5 =2 x 3! ( the numbers should have an even digit in the units place)
8
In order to make an n-digit number we have to fill n places by using the nine digits. As no two consecutive digits are to be the same, a digit used in a place cannot be used in the next place but it can be used again in the place
3P3
coming after the next place. So the first place can be filled in 9 ways; the second place can be filled in 8 ways
The number of numbers of 4 digits, divisible by can be made with 0, 3, 4,5
the required number of
=9 x8
5.
x
numbers Sx 8x... to n factors = 9 x 8"-.
How many numbers of 4 digits can be made with the digits 0, 1, 2, 3, 4, 5 which are being unrepeated in the same divisible by 3, digits number? How many of these wil1 be divisible by 6?
We know that a number is divisible by 3 if the sum of the digits in the number is divisible by 3.
Here 0+ 1 2 3+4 +5 omitted whose sum is 3 or
15; so two digits are to be 6 or 9.
Hence, the number of four digits can be made by either 1,2, 4,5 or 0,3,4,5 (omitting two digits whose sum is 3) 0,1,3,5 or 0,2, 3,4 (omitting two digits whose sum is 6) 0, 1,2,3 (omitting two digits whose sum is 9) The number of 4-digit numbers that can be made with
1,2, 4,5 =
*P, = 4 !.
The number of 4-digit numbers that can be made by the digits in any one of remaining four groups (each
containing =
the
0)
4!-3!
(':
the number of numbers beginning with 0=*P^ = 3 !)
required number of 4-digit numbers divisible by 3
= 4!+4(4! 24
+4(24
3) 6)
=
96.
that
= (3!-2!)+3! numbers should have 4 o 0 in units place and should not come in thousands place).
(rejecting the digit used in the first placc)
the third place can be filled in 7+1, i.e., 8 ways (rejecting the digit used in the second place but including the digit used in the first place) and so on.
6 .
(the
(
0-O|0)-Tn 2P2
3Pa
3P3
Similarly, the number of numbers of 4 digits, divisiblo by 6, that can be made with 0, 1,2,3
=(3!-2)+3!
The number of 4-digit numbers divisible by 6 that can be made with the digits 0, 1,3, 5 = 3 !. The number of numbers of 4 digits, divisible by 6, that canbe made with 0, 2,3, 4
= (3!-2!) +(3! -2!)
+3!
(the
numbers should have 4 or 2 or 0 in units place and 0 should not come in thousands place). the required 4-digit numbers divisible by 6 =2
x3! +(3!-2!)+3!+(3!-2!) +3!+3! +(3!-2
12+4
+6+4+6+6+4
6. How many words can be INTERMEDIATE if
+ (3!-2) +3! +4 +6 52. )
made with letters of the word
(i) the
words neither begin with I nor end with E (ii) the vowels and consonants alternate in the words (iii) the vowels are always consecutive
no vowel is between two consonants consonants does not change (vi) the order of vowels does not change? (i) The required number of words = (the number of words without restriction) the number of words beginning with 1) (the number of words ending with E) +(the number of words beginning with I and ending with E) (iv)
(v) the relative order of vowels and
A-155
Permutation and Combination
vords beginnin (because word
with I as well as words contain me words beginning with ding with Ewith E).
consonants if all the Consonants become consecutive the required number of words when all the the number of arrangements consonants are consecutive
uv) No vowel will be between two
ending number of words without restriction The there are 12 letters in which there 12-(
Iand
213:2!
number of words beginning with
The
312!
extreme lett place we are left to with I in the letters NTERMEDIATE in which there are ange 11 three Es). two Ts and number of words ending with E212121 The
the extreme right place we are left to MEDIAT in which there are arrange 11 letters Ts). Es and two two Is, two I number of words beginning with and ending
with
7!6! 2137
are two Is, three Es and two "Ts).
151200. and consonants will V The relative order of vowels of letters the not change if in the arrangement 1st, 4th, 7th, VOwels occupy places of vowels, i.e., occupy their 9th, 10th, 12th places and consonants 11th places. places, i.e., 2nd, 3rd, 5th, 6th, 8th, the required number of words
.
E in
The
6!6! 2!3!*2T
if no two order of vowels will not change arrangement vowels interchange places, i.e., in the all the vowels are treated as identical. have the same For example LATE, ATLE, TLAE, etc., ETLA, TLEA, order of vowels A, E. But LETA, E. So, etc., have changed order of vowels A, LATE is counted but LETA is not. V then LVTV If A, E, are taken as identical, say arrangement by does not give a new V.) V, interchanging The required number of words in the number of arrangements of 12 letters identical which 6 vowels are treated as
extreme in the extreme left and E in the 10 letters to arrange are left right places we Ts and two Es). NTERMEDIAT in which there are two
(with
I
of words the required number 11! 10! 12! 11! 2!2! 2!2!2! 2!3!2! 3!2! 10! 10 11 - 11 2-11.3+6)-85 24 213121(12 Gi)
There are 6 vowels and 6 consonants. So the number of words in which vowels and consonants
number of words in which vowels occupy odd places and consonants occupy even
(the
7 In how many ways can 5 identical black balls, identical red balls and 6 identical green balls be arranged in a row so that at least one ball is separated from balls of the same colour?
places) +(the number of words in which consonants occupy odd places and vowels occupy even places)
7.
6!
The required number of ways
2131 22
2131 43200.
219 (1)
are two Ts also)
there
alternate =
21600.
(vi) The
10 22!
with E
(as above)
Considering the 6 vowels IEEIAE as one thing, the consonants number of arrangements of this with 6 (
there are two Ts in the consonants).
=
without restriction) (the number of ways when balls of each colours
(the number of ways -
are consecutive)
18!-3!( 5!7!6!
6 consecutive in themselves VOwels can be arranged among 6! Ways.
For each of these arrangements, the
2!3 the
required number of words
2x
151200.
18!
5!6!76.
there are altogether 18 balls in which 5, 7 and 6 are identical; and considering balls of the same colour as one thing there are 3 things, there being no arrangement between balls of the same colour)
A1 Prews
12rs in
Ten
guests are to be seated adies 1he ladies insist on in a or of which tlhvee are sitting together while Iwo ot the gentlemen eiuse to take consecutive seats. in how many ways can the guests be seated?
Considering 3 ladius as one objort, the mumber ot arrangennents =*F
gentlemen
to be arrangad). and i But the 3 ladies can
goup of ladies, ie.,S things ar
be arrngei among themselves in
3! ways
the number of arrangenments in which 3 ladies ane together= *Psx3 Next, considering 3 ladies as
one object and 2 gentlemen efused to sit together) as another object, the number of arrangements (who
="P-
x3x2!.
the required number =
-
of arrangementss (the number of arrangements in which 3 ladies are together) {the
number of arrangements in which 3 ladies are together and 2 gentlemen (who refused to sit together) are together}
= $Ps
3!
-P,
3! -2!
=8!3!-7!.3! 2!
=7 !(Sx3!-3!2!)=5040(48 -12)
IT NMatlheratis After printing all these words, the wond cp RICKET
printed. the nJuirecd place ot the word CRICKET ( x4!+2!)+1 =531
5!+2
the wond will be printed in the 531th plae 10.
,
In a partieular progranming language. variable name can consist of a sequence of one lid alphanumerie characters A, B, C, tosix beginning with a letter. Find the Z, 0, 1,2 ,9 total valid variable names. ver
The number of valid variable names of one charaa
26
of
racter
(
there are 26 letters and the variable name must begin with a letter). The number of vali variablenames of two characters 26 x36 ( the first place shou have the second place can have a letter and any one of the26+10 alphanumeric characters,
repetition being allowed). The number of valid variable names of three characters ters 26 x 36 x 36 and so on. the total number of valid variable names of one to six characters
26+26 26(1+36
5040 x 36 181440.
Wilbe
x36+.. +36+.. +36 S)
x 36 +26
=26 36-1-2
+ 26 x 365
36-1).
9. A
dictionary is printed consisting of 7-lettered words that can be made with the letters of the word CRICKET. If the words are printed in the alphabetic order, as in an ordinary dictionary, find the position of the word CRICKET in that dictionary.
In alphabetic order the letters are C, C, E, I, K, R, T. Words beginning with C will be printed at begins with C. The number of wordsfirst. Our word also beginning with C and having C or E or I or K in the second place =
many numbers of five digits can be made with at least one repeated digit? Total number of numbers of five digits when digits can be repeated = 9x 10 ( there are ten digits 0, 1,2, ..., 9 and 0 cannot go in the ten thousands place) 11. How
4x5!.
E
x
x
K: x CRICx
CRIC
x
Exx
the number of numbers which may or may not have repeated digits = 9 x 10 But the number of numbers which do not have repeated
digits
1
T
The number of words beginning with CR and having C or E in the third place = 2 x 4!. The number of words beginning with CRIC and having E in the fifth place = 2!
Ps-P
( 10P
9PA
the required number of numbers having at least one repeated digit
A157 Permulation and Combination
4 in units place The number of numbers with
digits) of numbers of live digits having
nber of numbers
(total
of five
dhe nber (tle digit) no repeated P)
P-P the
9x10-"P 10-9.8.7.6(10- 1) 9x 27216 62784
=0x6+ 4x 4 =
.
The or
-
LL
10
10
10
used). 1,3,5, 7,9 will be number of numbers (20,000 being .the required excluded)
=4x 10 x 10 x 10 x 5-1= 19999. of four digits can be made How many even numbers numbers. 4? Find the sum of the with the digits 0,3, 5, 0 or 4 in the units place. The even numbers have
the
3
the 0 of four digits with in The number of even numbers .. (1)
units place = "Pa 4 in the of four digits with The number of even numbers (2)
units place="P,-?P2
4
= 2! in tens place
=
2.
= 32. =0x2+3x3 +4 x2 +5x3 hundreds place = 32.
Similarly, sum of the digits in place with 3 in thousands The number of numbers = 2x*P2 =4.
place with 5 in thousands The number of numbers
=2xP2 =4. The number of numbers 2! 2.
the
place with 4 in thousands
thousands place sum of the digits in
3x4+5 x4+4x2even40.numbers
.the
P=3!=6
=2.
with 5 in terns place The number of numbers caseof 3). 2!+1=3 (as in sum of the digits in tens place
13.
required sum of x 40x 1000 +32 x 100+32
10 + 16
43536.
of identical balls of four There are unlimited number arrangements of at most different colours. How many made by using them? 8 balls in a row can be arrangements of one ball = 4, The number of different balls. because there are only four balls of arrangements of two
14.
The number
2
numbers total number of even 10. = P3+ PP3 -2P2) = 6 +6-2 4 in units place. The even numbers have 0 or with 0 in units place 1he number of numbers
2
with 3 in tens place
with The number of numbers
55
=
=2!+1 =3.
filled in 4 ways. thousands place can be the ten thousands, ndreds and tens places can be Each of ways. filled in 10 10 x 10x 10 places can be filled in 4x four first the So ways. used is either these the sum of the digits After filling even or odd. if the sum can be filled in 5 ways (' 8 will be 4, 6, the last place 2, 0, digits even, one of the of the digits is digits the of one is odd, of the digits used and if the sum
P
16.
The number of numbers
Co
3
for all the in units place
0 in tens place The number of numbers' with
natural numbers are there lying between How many60,000, the sum of digits being eyen? 20.000 and five digits beginning with 2,3, 4 mbers will be of numt
12.
the digits
fron (2))
numbers
=
90000
sum of
=4
4, etc. arrangements the required number of = 4x 4 =
=4+43+4++4= 4(4"-) 41
from (1)1
-(4-1)=x
65535=4 x 21845 87380.
A-158 Problems
hen
rom
a
given
5
well-shuftled pack of
Pus
a
cards, player is cards. If the cands have consecutive values then it is said that the player has a run and if they are also trom the samne suit it is said that the player nasa runing tiash. In how player hold a (i) run ii) many different ways can a running flash?
are
4 suits ot having 13 cands, thecards ina pack
1( In a K.
A), 2, 3,
..
game of cards
values being
52
in
tT Mathematics The total nuniber of points of intersectinn on = the number of selections of 2 lines fro
om n
"C
lines
of 52 cards, cach suit
10, 11(= ). 12(= Q), 13(= K). A is also given
the next value above
our problem is to select 5 consecutive cards to have a unning flash when the cards of a suit are arranged as follows:
Clearly, each line will cut the
n-1 points.
out of "C points, n-
1
remaining n
points are on each
The number of lines obtained by joining any
ines
at
line. ithe
"C points n{n-1) fn(n -1)
"caC"
The nunmber of same suit from selections of 5 consecutive cards of the
the above arrangenment = 10, (1,2,3, 4, 5), (2,3,4, 5, 6), ..
mt-10n-n-2) 8
because they are like (10, 11, 12, 13, 14).
The number of selections of 5 cards of consecutive values from the pack 10 x 4, because in each place we have 4 choices of suits. the total number of running flashes 10
10
10
But from each of the n sets of (n
-1C lincs have been counted
=
10
x
4
=
the required number of new lines
n(n-10--2)1x"-'C
10240.
8
nn-10--2),1
16. There
are m bags which are consecutive positive integers numbered by m number k. Each bag contains starting with the flowers as the number levelled as many different against the bag. A boy has to pick up k flowers from any one of the bags. In how many different ways can he do the work? The total number of ways +***C C
=*C+'C
points
which are not new li nes. So nx"-C, lines are not new lines.
10 =40
and the total nunmber of runs
- 1) collinear.
x-D-2)
8
n(
+..
8
2-n -2-40n -2)}
=no-1%-
51+6)
=n-1)-2)01
-3).
'C+**C.1- ** 'Ck.1)
Ck1**Ck-1)+. +(
C. 1-** m-'Ck.1)
18. There are two sets of parallel lines, being xcos a + ysin a = p; p=1,,
3,..., m and ycos a-Xsin a = ; q = 1, 2, 3,..., 1(n> m) where a is a given constant. Show that the lines form
because C-1= "Cn +"Cm- 1 i.e., "Cm="* "Cm-1"Cm 1 "*
k 17. There
m(m -1)(3n-m-1) squares.
Ck-1
are n coplanar straight lines, no two being parallel and no three are concurrent. How many different new straight lines will be formed by joining the intersection points of the given lines?
their equations
Clearly xcos a + ysin a = p; p = 1,2, 3,., m represent m parallel lines, the distance between consecutive lines being 1. = q; q = 1,2,3,.. ycos a xsina , n represent " parallel lines, the distance between consecutive lines being 1.
Also
A-159 Permutation and Combination 3 are ladies and 4 of them are relatives, them 19. A man has relatives, 3 of different ways 8entlemen; his wife has 7 how many 3 In gentlemen. 4 adies and 3 ladies and of party can they invite a dinner man's relatives 3 of the are there that so gentlemen relatives? and 3 of the wife's man's relatives by HL, lady man's relatives Let us denote the wife's gentleman gentleman relatives by HG, by WL. by WG and wife's lady relatives = 3. n(WG) = 4, n(WL) = 3, n(HG) = Here n(HL) 4,
m
7
n-1
each line of the first set
cot a;
the slope of Again, line of the second set = tan a. slopeofeach the (tan a) =- the two sets cut orthogonally (-cot a)
it, at
=-
3HL, 3WG
90o
consisti of two linessof the first set and two lines second set will form a square, if they are the distances. lines of equal taken at sides of length 1 unit number of squares of 2 The nunmber of selections of lines from m lines
theat
2 lines from unit distance and
distance
The
the
lines from
lines at
a
distance distance of 2 units ... (1,3), (2, 4), will be selections (°: ... = (M -2)(n-2) (m-2, m) and (1, 3), (2, 4), a
=
(1
number of squares 1Mn 1) + (m-2)(n -
-
2) + (m
1
-
21+ {mn -3(m +mn-(m
1)mn
-
1)mn-
(
=m(m 6
-
men if
-
1)(m +n)
The number and 4 men:
(m
1)(3n
=3x4x 4 x 3= 144 *C C C; xCx 3 6x 6x3 324
324
=
485.
Combinations
=7056
6w, 4m
+(m-1)9.
+n)
-
1)*)
(77-)m(21711)
2 + n)
m -1).
+21-1}
8:
CxC4!5!6!2! 9824 x*3528
5w, 5m
+ (m
x *C2
work with MrY insist to work together? least 4 women of committees of 10 with at
m-1)
(»1 +n)(1+2+3... +m-1)
6n-3(m
6
-
+ 1n) +31
+(1+2+3... (1
(n
16
(i) Ms X refuses to (ii) Ms X and Mr Y
+ n) +1)
(m
+{mm 2(n1 + n) +
(1
8
-3)n -3)
m-1)
=
of 10 be selected many ways can a committee women and 4 men from 9 with at least 4 women and
4w, 6m
+.. +(m = {mn
x
9!
and so on.
total
Cx'C,
x4
20. In how
Possibilities
n)
(-2,
4
=1 CC=1x1 'C,
1HG,2HL, 2WG, 1WL
16+1+144
sides of length 2 units number of selections of 2 lines from m lines at of 2 units and 2
2WL 2HG, 1HL, 1WG,
CC3
ways Therefore, required number of
umber of squares of =
3HG, 3WL
lines at unit
2), (2, 3), ... f: elections will be (1, (m 1, ni) from the first set and . n-1, n) from (1,2), (2, 3), the second set).
1(n-1)
(nt
n
Combinations
Possibilities
C C=5 24 CxCS,876-5-5880
at least 4 number of committees of 10 with women and 4 men 3528 7056 + 58S0 = 16464. in which Ms X and The number of such committees MrY are present: of committees of In this case we have to find the number 8 women and 8with at least 3 women and 3 men from committees in which 7 men. As above, the number of Ms X and Mr Y work together
total
="C^x
Cs+
"C,
x
"C+"Cs x 'C3
don uny ways can tvwO distinct subsets 3) elements be selectei so that the Aot k ney set etty owO nmon eleutents? n
of
hare
wv sutes
tt
e
allai P and Q Tha lements foe
ederettsut ot elenruts torboth th
r tte sudset tnielen
eNns
nents tr
nts
tnm theem and tien: can vary tum --eneas ir, te nunter ot slrtions s auss themunter of FRr a
ot any olrtins things trom r number
things
is
urber oi wlartions
the tal
--1 eadin
ut
te cas when both the subsets ant equal er having oy the two mon dlements, are every rair ot Qis apoaring twie like
.
Hene the myuirni number ot ways
end
osngC="C
C-2i.
*C.)-1}
D.2-1-1}4
hai of ie series of incmial coeficients
-2 Ifis
side, 4 on the row side and one to steer. There are 11 crew of which 2 can stroke only, 1 can row only while 3 can steer only. In how many ways the crew can be aranged for the boat?
2-
odd. ie.r
=
2
-
1
thenr m =
1.
the required number of subsets
m-C1
C
z-Co-1 2m-C
=
23. A boat is to be manned by 9 crew with 4 on the stroke
The number of ways to select the man to steer Of the remaining 8 persons, 2 (who
-=22=2-1
"C.
can stroke only) are already seletted for the stroke side and 1 is selected for the row side (who can row only). So, of the remaining 5 crew we have to select 2 for the stroke side and 3 for the row side.
ToW
half of the series of binomial coefficients
22
=
steer stroke
A-161
P'errutstion arid Cembinatin
t
identical), 13 are cards is (i.e.. two where each card ne same denomination irom 52 cards
nunber otselections- 'c, x 'C x'C uitd tho stroke side as well as nw tthe peisons on 4 ways tor every selection of the natranged in thenuirnd
wo in nunmber. selections of ne number of
'C4!s4!= 17280. «
elements.
A
aining contain isasetThe set A is nconstructed
subset
P of A is
many
In how mangoes and 4 apples. made if 27. There are 5 of fruits be selection a can different ways different samme kind are (i) fruits of the kind are identical? (ii) fruits of the same number of to select any ways of number () The mangoes
r
,
"Cs =25. =C+C+C+. number of ways to select any
The nunmber of
suisesand Prove from
apples
objects ol ways to select n that the number rest the which n are identical and 3n objects of
are difterent,is
=2x2-1
2**2r
n-r
0 to n. Herer varies from of selections the nequired number
2C-,
=2"C,+2"C,-+2"Cn-
=2-1. (i) The required
=(514
number 37800. Also number of factors of the proper divisors of the find the sum of the odd
2C%
number.
Here
3"C+"C,"C+... C
37SO0 = 378 x 100
coefficients+
C
23- 2 26.
shuffled together. Two different packs of cards are among 4 players, each getting
Cards are dealt equally a player get his cards 13 cards. In how many ways can samne same suit with the if no two cards are from the
denomination? each card being 2 in Here, there are 52 distinct cards, the same suit with number. As no two cards are to be of
2x5
2
5C,}+5.*C of the series of binomial
x 126 x
3
x5 =33x 42 x23x5 =3 x3x3x7 =233x5 x7factors
writing in reverse order
sum of the first
number of ways = x5-1 =29. 1)-1 6
28. Find the
r0
-C-3CChalf
C=2
fruits of ways to select number required the way in which 0 lencluding the selected} mangoes and 0 apples are
objects from u identical of selections of r number objects The objects from 2n different and objarts
=
2
C2"13!39
by eplacing the
A is chosen. Find the A subsc Q of lemen nts ot : of selecting and Q so that P and Q l' of ways nunnoninterseeting. ar elenents is ovailable for selection in making conmon t the But and Q must not have any vell as Q. element can be taken in 3 ways ch mt. EQ or e P, e Q) or Q which is can be taken in 3"-1 rays lementsumber ot ways to select nonintersarting are al to the case when P,Q ). Q (eacliuding the
o
distinct cards tron 52
13
cards C selected in 2 iways be can cards But each of the 13 packs). (belonging to either of the two ways the required nunmber of
number ot ways
C
A A
Caris ae to be seleted
the nequired number of from the total nunmber of selections =
(2, 2,2,), (G, 3, 3). (5, 5),. 7)
=3+13 1211
+
1)-2
(excluding 2°.30.5.70, i.e., 1 and 23 33 5.7, i.e., 37s00 as factors)
=4x4x 3 x2 The required sum
(333°
2 96-2
94.
-3%5°+ 5' 2
-5 x7° +7')-1
as a factor and 1 as a divisor are to be excluded)
A-162 Problems Plus in lIT Mathematics
3-
14141 12! =91
12!. However, the first r nethod metho is advised because it covers all situations.] When 3 girs seat together in the back row of an I: Possibilities Combinations Permutations Van I Van lI
8-1
(3-15-18 = (3
2x 4
-
1(5
1)-
-
Note
1
=
80
x
124
1
9919.
(i) If we have to find the number of odd proper
3g. 4b
5b
divisors then the required number = (3 +102+ 101+ 1)
3g, 3b
6b
we
3g. 2b
7b
-1 have toexclude 2s and the
number30.50.7)
(ii) If we have to find the number of even proper divisors
3(3
then the required number +102 + 101 +1)-1
weat
the
havetoexclude 23.33.52.7 and
+°Cx1 x2 9!
= 9!3!x7x6+9!x 14 =9!3!(42
5
Cs
CxCsx P;
x"Ps
12!
7!5! 77!
6
6
12CxC
C,xCx
1
x3!xx7!
x2x3!4!7!
x4!+9!x 3x24
56 +12) = 660
x9
the required number of ways
2x 660x9!= 1320x9!. 30.
1C7
girlsin
Similarly, the number of sitting arrangenments with3 girls in Van II = 660 x9!
Permutations
Van II
3
+1712x3x4!7!
persons = 3+9 I
9!
3!6!
29. In how many ways can 3 girls and 9 boys be seated in two vans, each having numbered seats, 3 in the front and 4 at the back? How many
Van
number of sitting arangements with
=Cx1x2x314!xC,*1*2x3!x
least one 2 is to be selected). find the number of proper divisors divisible by 10 then the required number = 3 x (3+1)*2x(1 +1)-1 at least one 2 and one 5 are to be selected).
(ii) If we have to
There are 7 seats in each of the vans. The total number of = 12. Now: Possibilities Combinations
CCs CxCx (2x3!)4!xp, CxC,* 2x3)°P^ x7P, C CC CxCx (2 x 319 Pxp
because the 3 girls in the back row can take place i seatsremaining together in 2 x3! ways. in 4 Van I
sitting arrangements are possible if 3 girls should sit together in the back row on adjacent seats?
x
In the given figure you have the road plan of a city. A man standing at X wants to reach the cinema hall at Y by the shortest path. What is the number of different paths that he can take? UA
'P, x "P
L
12
6!67!.7!
5
total
12CxC
7
- 21 (12
!) +
12!.7.7.7.6 49 (12
!) +21 (12
2
L4
number of sitting arrangements
_121 7:6 [Note
CsxC x 'P; x"Pz 12 517727
!) = 91
(12
!)
We could have directly arranged 12 persons in 14 seats. The number of ways for this is "P12, i.e.,
L
As the man wants to travel by one of the many possible shortest paths, he will never turn to the right or turn downwards. So a travel by one of the shortest paths is to take 4 horizontal pieces and 4 vertical pieces of roads. As he cannot take a right turn, he will use only one or the five horizonal pieces in the same vertical column. Similarly, he wili use only one of the five vertical pieces
Permutation and Combination
row sanme horizonta L the in hortest path is an arrangement of eight things sho a so that the order of Ls and Us do U,, U,, U,U, L, L. clearly L cannot be taken without taking change (.'c not cannot be taken without taking U, etc.)
L Uy
Hence, the
=84x36 9 x (3360
Us do not change the order of of arrangements treating Ls as the number identical and Us as identical
1
1
five digits that can be The number of nunmbers of different digits where nmade with 1, 1,0 and two other
24
1,1
made numbers of five digits can be of How many identical digits?
How many two having exactly the repeated digits in consecutive these wil1 have
places?
digits be 0, 0 then three other digits "C ways. 1, 2,3, .,9 in can be made of numbers of five digits that The number digits other different with 0,0 and three can
5-4 the
without restriction number of arrangements 0 is 4 number of numbers beginning with
and the
is !)
that can be
number of numbers of five digits different digits made with 0, 0 and thrce other
the
-'C,
are consecutiv
aretreated as one object) two of numbers with Hence, the required number (4! -3!)
C2 x
1
(* 1,
identical digits consecutive
= "C,(4!-3!) +9/'C x 4!+ "C2 x (4!-3 ! =
84
x 18
+9(56
x 24
+
28
x
18)
18144. = 1512 + 16632 =
digits can be made with at most each of whih can be used
nunmbers of five 32. How many
the digits 1, 2, 3 thrice in a number?
make numbers digits 1, 1, 1, 2,2, 2, 3,3, 3 to will be as follows of five digits. The digits (constructionwise): (i) three identical, one pair We have the
(forexample: 1,
1,
1,2, 2, etc.)
different (i) three identical, two
4
The number of numbers with 1,1 and three other
made of five digits that can be 0 otlher than different digits
(forexample: 3,3, 3, 1,2, etc.)
(iii) two
5
pairs, one different
(for example: 2, 2, 1, 1, 3, etc.) possilbilities start with In order to cover all the the identical digits and go on reducing all
that can be made The number of numbers of five digits with 1,1,0 and two other different digits
5!
1
=*Cx4
8 7:6-5-2.7.5-70.
identical f the from selected be
45360.
one object). (.0,0 are treated ascan be made ="C44! -3) five digits that different The number of numbers of 9) and three other with 1, (or 2, 2 or ... or 9, 1, are consecutive digits other than 0, where as one object). C 1, are treated
U,
31.
=
1344)
of five number of numbers other Next, similarly as before, the with 0, 0 and three digits that can be made consecutive different digits when 0,0 are
of arrangements of L, L, Ly, L the number U U, where the order of Ls as well as
8!
42336
3024
number of shortest paths
4!4!
24)+9x {56 x 60 28x (60 12
84(60
4
digits that can be number of numbers of five made with 1,1 and three other different digits
the
ultimately number of identical digists, reaching all different digits. identical digits, The number of selections of three one pair = °C, xC. of Corresponding to each selection, the number
TIP G)
numbers that can be made
=
5!
3!2!
total number of numbers of three identical digits and one pair
the
etc. Similarly for numbers having 2, 2 or 3,3, having exactly two the required number of numbers
repeated digits
5! CC,3121 3x2x
60
ii) The number of selections of three identical
and two different digits - 'C,
x *C,
(1)
digits
A-i6
PobiemsPusim UT Mathemttis total numer of digits and two ditterent numlers ot thre digits the
-'c, (i)
idetntical
Pc-3
digit 'C 'C
'cC;
=
'c,
5
30=90.
the aquind nunmber of numbers 0 60
0=
x
3
identical,
1
CC
2 1
pairs, different
1
pair,
3 different
.the
CC CC
CC
CC2121 5!
CC
'C, +*C,
=
1
+2x
1+6
=
coefficient of
C
+ *C,
3+3x3+3 + 9+9+12
The required number
r
in
+r"x1-x3 -x* +x)
(1-x-
4 5.6 7445:6:7-8 ,5 4!
5
r
in
1+
4x
coefficient of
-xS)
+
10x+ 20x* +35+56x1
neglecting powers higher than -4 1
=41.
34. In how many different elements be partitioned
xC, +C,
« 3C
CC+C,°C3 =
(1-x1-xX1 -x*1 -x 1 -*)
= 56 10
required number of selections =
coetticient of r* in
x
3identical.
different
1- 1- x 1-x
(1--
C,C
+xr+r+x)
coefficient of xin
=
pair
2
=
'C
4 identical
+x"Mr+x'+x*+r+a
x(++x
Find the total number of selections of 5 letters from five As, four Bs, three Cs and two Ds. Also find the number of 5-letter words that can be made. There is one letter Ato take 5 identical letters, two ietters A, B to take 4 identical letters, three letters A. B, C to take 3identical ietters and four letters A, B. C, D to takepairs. Therc ane only four different letters. Possible structures, selections and arrangements are given in a tabuiar form as below: Possibilities Combinations Permutations 5 identical C
different
CCx
+30 + 90 + 180+ 360 + 240 =901. Note The number of selections is also equal to the coefficient of x* in 1
=
210.
33.
1
CC
ot two pairs, one dieent
the total nunmber of nunbers of two pairs and
onedifferent digit
CC
(2
Thenumber of soloctions
3!2
x
4
3 +6x2+4x
1
= 41.
of words (i.e., permutations)
5!
ways can a set A of 3n into 3 subsets of equal
number of elements? (The subsets partition if PuQUR =A, P nR =¢, QnR
P,
Q, R form a
=
¢, RnP=¢.) The number of ways to partition the set - the number of ways in which 3n different things can be divided in 3 equal groups = x (say). Then 3n different things can be distributed among
persons equally in xx (3
!)
ways.
But the number of ways to distribute 3n different things
equally among 3 persons
= 3"C
31
n!2n!
C,,x "C,
2! n!n!n
3n
!0!
(n )3
A-165 Permutation and Combination
a box giving 1 ball to equal after Similarly, identical. are two be divided in the remaining 4 are to groups. of ways the required number
3n
6(1!3
(
x3
boxes. Each can hold are to be placed in three 1h Fiveb slls balls. In ihow many different ways can we the five remains empty, if balls so that no box
the
place are all different balls and boxes i balls are identical but boxes are different (ii) but boxes are identical balls are fferent are identical boxes (ii) as as well ) balls identical but boxes are are boxes as as well (v)ballsin a row? kept can have balls in to remain empty, boxes box hox is numbers: no Asfollowing 1,2, 2. the or Possibilities 1, 1,3 bute the balls in number of ways to a) The
goups of1, 1,3 -
=CxCx C
2!
distribute Similarly, the total number of ways to 1,2,2 balls to the boxes
3!
required number of ways
. ..
36. Let
=
whole answer in tabular form, permutations possibilities combinations
Note Writing the
CxCxC
C x'C, x°C,x = 5
1,2,2
C x'Cx*C
,),
x4x3
= 60
C,x'C,x*C;x
x1
+2+*+...+Xk= =
y1+y2+ Y3t
or
.
k.
yk+ Let I = yi +1, x = y2 +2,.X Putting these in the equation, 1+1) + (2 + 2) + (y3 +3) +. +(
+k)
=
n
+
=n-(1+2+3...
+k)
+
,
=
+ n-kk21)
*2t1x23+3 (1) When balls are different and boxes are identical, atter giving 3 balls to a box, the remaining 2 are to
divided in two equal groups because the boxes
(1)
= m
equal to The number of solutions will be things identical m number of ways in which distributed the number 1) can be thing persons (here k variables) and this is
the total (here the
among k equal to
m+k-Ck-1 (1) number of non-negative integral solution of and hence that of the given equation
the
mi+K-"Ck-1
(m+k-1) (k-1) ! m where
= required number of ways 60 +90 150.
1) When balls are identical but boxes are different the number of combinations will be 1 in each case. the required number of ways
be
x21,
satisfying the condition
=5x6x3=90 the
-25
integers such that and k be positive of solutions Find the number integers *22,.. ,xk2 k, all
1+ 2+ V3t.
= 5x4x3+5 x6 x3 60 + 90 150.
3
n
n
x*C, x°C,xi+°Cx'CxC,x
-'C
10 + 15
0. m (say) where m is an integer2 non-negative Now, we have to find the number of integral solutions of the equation
-'C,x*C, x*C,xi the
-
G,c
the are identical, boxes as well as balls (V) When arrangements will be and combinations number of 1 each in both cases. ways the required number of =1x1+1x1 =2. will be treated in a row, they kept are boxes When ways (V the number of as different. So, in this case will be the same as in (ii).
CxC x'C. can interchange their content, no
the boxes a new way when boxes exchange giving numbers interchange. equal containing balls in ways to distribute 1, 1, 3 number the totalboxes of balls to the But
=Gx
m =
n- kk+1). 2
Note (1) The
number of solutions of
1+ 2+ 3 t.
+Uk= m where 1, y2 etc., and m are non-negative integers, can also be obtained by finding coefficient in a suitable expansion. Consider the expansion of (1 + x+**+...)*.
A-166 Problems Plus
(+x
+**+...) x{+x
n IT Mathematics
where n20, +a*t
.)X.
Then
(+a+a*+...)
there being k factors. The coefticient of a power of r, say 1 " in the RHS = 1number of ways in which m can be split ink parts (non-negative integers), sum of the parts being
+112 + 13 t
=
coefficient of x " in
=
coefficient of
x" in (1
coefficient of "
++r+...)a++*+...)
x(+r+x*+)(r+x'
.)
=
-
+k-1)!
! (k-1)! (2) The number of non-negative integral ofyi + y2 +.. +Y S m can be obtained bysolutions finding the number of non-negative integral solutions of V*y2+... +Y+Yx-1 =m (where yk. 1 is also non-negative integer). Their non-negative integral solutions will number of be equal. 37. Find the number of non-negative integral solutions of 2r+y + = 21.
Clearly a =0, 1,2, 3, ..., 10. Let = x k; then 0 sks When r = k, y +z = 21 2k. The number of non-negative integral solutions
10.
(103-) 45-6.. (100 k)! (103-)
the number of ways to distribute (21 -2k) identical things (each thing is the number 1)
among 2 persons
21-2k2-1C.-, = 22-*C
=22
required number of solutions
=
2(1 +2
=
=
-C,
k0
10
Cy+ 10
C.
+*C
coefficient of r* in I(1
)"
+ (1 + x)
coefficient of x in (1+x)3, =
coefficient of r* in
(1 +
x)*
= coefficient of x* in (1+x)
(1
..
-1
+x)
((1 + x)
+(1 +x)1
-1 11
104
C
39. How many triangles with sides of lengths in integral cm can we get if the length of the greatest side is given to be 2m cm? Also find how many of the
xsy
ni).
possible to make an unsuccessful attempt to open the lock.
A-170
Problems Plus in
IT Mathematics
is made of the words that can be ma 26 dictionary Atelegraph of the word PARKAR by arranging the letters N capable of taking 4 distinct positions, including the wvord "PARKAR" in that diaatis the position of the ionary position of rest. Ii 1023 diiferent signals can be sent in the:same order as that if words are printed in all then find the number of arms. of an ordinary dictionary? many 10-digit numbers can be made with odd How can 6 boy's and 4 girls sitin. 27, In how many ways digits so that no two consecutive digits are the row two girls? between is so that no boy same? each of wo books ana 28. There are three copies two many odd numbers of five digits can be How of three volumes. consisting works cach In now formed with the digits 1,2,3, 4,5 if the digits cannot many ways can a booksel arrange the 12 books in be repeated in the same number? a shelf of one row so that neither the copies of the same n E n N and 300 of Un where u, rove 7.
in
17. Find the coefficient of
-2)(n -3) n1-17(212
independent of 10
termn
o V
x in
the expansion of
A-198 Frubles Plus
(iv (1r)
-
(v23x 28. 29.
4
30. If the
P
4)
,,
coefficient
prove that
c-bd 3c consecutive terms in a binomial expansion with positive 280 and 560 integral index be respectively 14, 84, then find the expansion. Determine
o1o')
36.
38.
+
39.
40.
expansion of
1)th and
the
(r +
(1
+x)"
the coefficient of the 2)th terms are equal, findr.
Prove that in the expansion of (1 x)", cocfficient of x" is double the coefficient of x" expansion
of
t
the in the
2x+ x*"
If in the expansion of (1 +a)", 5th,
the coefficients of the the 6th and the 7th terms are in AP, findn.
Prove that three consecutive binomial coefficients in the
.
expansion of (1 + x)", t e N cannot be in GP for Is there any n such that they are in AP or HP? 41. Givcn positive integers coefficients of 3rth term and r> 1,n > 2 and the (r binomial expansion of (1 x)" +2)th term in the are equal, find r. Ff any
in the
tha
are
expansion of (1+x)", the three consecutive Cuefficients are 165, 330 and 462 then find n.
.e
H
in
b, c, d
c)
xb-
2(a3-
46. Let (1 )(1 in AP, find
bd).
=
If a, az and
#.
are
n
47. If n be a
positive integer then prove that at the integral part P of (5-2v6)* is an od integer. If f be the fractional part of (5+2v6)". prove that = P
45. If (9
45)"
integers and
that(1 Bxp 49. If
+
(2r
)
in
4
(1
37. In the
a,
(
the value of r in the expression if the third term in the expansion is
Prove that the coefficient of the middle term of a)" is equal to the sum middle terms of (1 x)2'-1 of the coefficients of
the
be any tour onsecutive cocfticients ticients in a binomial expansion, show that
45. If
34. If four
10,00,000.
a is
2
+
binomial expansion then
tthat
Also show that
.
a
three consecu then prove
the expansion. 44. If a. b, c. d be four consecutive cocfficients binomial expansion of (1 - r)" then prove
4th term in the binomial expansion o
in
of (1 x)", a, b,c
baC. b-ac
1
is
(x
be
2ac
independent of x, find the value of Also calculate p if the fourth term is 31. Prove that there (independent of x) cannot be a constant term unless na is an integral multiple ain (xx" of b. 32. Find the term which does not contain irrational expression in the expansion of (3 33. If a, b, c 2. and d be the 6th, the 7th, the 8th and the terms respectively 9th
35.
expansion
coefficients
10
x
Find the term independent of x in the expansion or 4
tT Mathemuztics 43. If in the
1
Deternune the constant term in the expansion ot
(
in
50.
=p B
B
where r and
is a positive
B)
=
1
p
are positive proper fraction, prove
represents the least integer greater than then prove that ((fv3+ 1")), n e N, is divisible by 21 Find the greatest coefficient in the expansion ((x)
of
(1z) 51. If x
=find
the greatest term in the expansion of
(14x 52.
(a)
Find the nunerically greatest term in the expansiom of (3 2x) " when x 1.
b)
Find the value of the greatest term in the
expansion of v3 |1 53. In the
expansion of (r + a), if the eleventh term is the geometric mean of the eighth and the twelfth terms, which term in the expansion is the greatest?
54. In the expansion of
when
r it -
isknown
that the 6th term is the greatest term. Find the
possible positive integral values
of n.
At
Binomial 7heerem for Postex Integral Index
geatest cocfficient
the
in
expansion at the greatest the coetficient double in the 1)*"is
s
h
( of enpansion of(1 x) sum Find the . C,
C
Sum the
b6. Find the
series
65. Sum the series
1
57.
31(n-3)1 5 (-5)1
1!(n-)!
58.
Find the
"C; 2
."C"C,+2 P'rove that
59.
(1
.
Provethat "Co "C,-"C;"C, 61.
are
.!, If to
terms
the
s
in
C 1
the epansion of
x("C "C+
'C,+ "**C,++ C
'C:
Prove that
72.
+ If (1 x) " Co Cr+
"*
k
C
2
71.
Ca+.
+
+ (n
+
)"C
(1)a, +
++ (1 I*)=d
(iv)a4s4
(iii)a, +a,
4,X
+
then tind
76. P'rove
that "C,
1-(1+) 77. If
(it) 4d and
()4 b4.
ats-
2-
pove that
a
a,(3"+a,)
coefficients in the The sum of the binomial coefficient of is 1024. Find the expansion
ofa
the binomial expansion. exceeds that of another The eaponent of a binomial coefficients in
in 65.
binomial by 3. The sum of the taken together is 144. expansions of both binomials twoexponents. Find the smaller of the
..
"C-2"C,+3. "C s-
+
"C, +2.
,
then find (IV)a
+
(2n+
(1
+2x)
"C, + 3.
+
= (n+212"
= (u+
1) "C,
75. Prove that (i1)
+
+
n- 1d)C
+ C+{a +d)C +(a 2d)C,++(a+
"C+3.C, +5. "C,+
1f
C,"
series. find the sum of the following
74. Prove that
63.
""C
C+
+
C,
C,+
Co+2-"C, +3. "C,+
then find
"C
"C,
73. Prove that
prove that
(a)then
"C
C+
"C,
-t"C
"C)? *C, 12C,"C
"C,+ "C;+"C,
1001
Prove that
70.
sum:
"C,
3 1
'C
2
c,C,
(n- 1)!1 were
the exprst*
ain
Evaluate
69.
ever positive integer. is an
expansion o
the
+21(1 1)*
x)
(1
in
(1
Find the coefficient of
2nC
"
x
(1x
(1) 67.
C; 'c-
coetficient of
"C -(1 "C t
"C, =0
1
(-1) "(o 31)
12
"C,+= 0
+"C
Also prove that
("C,)"s "C, ("C) "C)'... 78. Find the sum:
1"Cot2. "C2 +3. "C+4 79. Show that
"C
M-
80. Sum the
"C, (m
-
"C,+
.
-2)+(-1) "C, {-
)"C2
(
)0
series
3.C-8.
"C +13. "C
18 "C,+
to(n1) term.
A-200 Problems Pius in irT Mathematices
81.
Evaluate x
82.
first r natural numbers. Prove by binomial expansion that 2
**. "C =
."C,
wherep, denotes the sum of the
214+1
nta
94.
1)(b-
-
1)+
S4. Prove that 1+3n+ 3
(a
2)-
95.
=
2 r=0
86.
96.
2 +8)
ne N.
97.
(r+ 1)2."C, r
then prove that
-*+
Ca+33
+.
+C"
..
10C10
90.
(b)
100.
"CT
Prove that 2n 1CR+21+1C?+2 =
101.
"Cak-
k=
C
(+ 1)2
r=0
91. Find
2
02
2
C,x'. prove that
3"+'-2"1
0
C
(b) (6)
2k
(-1)
2 T
Find the sum:
=
+2C .
c.. 21
21C?+2C3+
+2
C
ic3 +...
+ 27
1c? 1
Find the sum
C1015C +20C 5C
+1
+Cg 1C +20C%.
Ci0
k
(-3),
3nC2r-1 =0
r=1
(k+ 1)k + 2)
"
n!n!
102. Prove that 2
Ck
If (1 +x)
+2"CŽ= (-1)".2n)!
the sum of the products of binomial coefficients "Co, "C, "C "C taken two at a time.
11
Find the sum:
k1
n10.
99. Find
3"+-11 (a)
-
2)! (-)" 2) 2[I2n)2+(-1)". (n !)3
4-1 +2-
even.
... 2n*1c?
-
Prove that
98.
n
+3.
*..
2."C+23+2
(a)
C?+27 +1C
2 C 2c+ 2C3-..
(-1)"
+3
(ii) Prove that
89.
21
4n)+
87. (i) Prove that
k-0
according as n is odd od or
C+2C? + 2"C? +...
(Gi)Co-C
88.
"C
Prove that
Co
3 .
or (-1)2.
21CR
2" -s,
"C7 +2.
Prove that
1
If (1+x) "=Co + Cr+Cr+Czx+... )
0
2
)n +2)
(n+7n
=
-"Cf+"Cs-. +(-1)" ."c
5 n(n-1 M-2)
t Evaluate
Prove that "C =
3!
85.
n!n!
2
Prove that 2 "C "Ck,2
n)(b- 1) =0, n e N.
-
"C;+ "CZ+... + "C}=2n)!
k 0
a-2)(
+(-1)"
+
93. Prove that "C
83. Prove that
ab
+... +(-1) " Cu .
-.
n(n + 1)2
92. Evaluate
k +2
C Cr
where
k=and n is an even integer.
103. Prove that 1
2."C-2
"C-1+3.4."C,-2
+(-1)'(r+ 104. Show
that
C,(1-)-1-
(n+1)2 "1
"C,x ."c,x".
3.
1)(r+2) =2."-cC
-
"C3
+-1)
,1-,
-1. (1-x)"
A-201
Binomial Theorem for Positive Integral Index
Prove
that
i)
-npq+np 2."C,p'q"
105.
2."C?+3. "C} + ... +n.
106.
Prove that 107.
3
C2
+"C2-*C,-+. +"C2.2C,+ .). + 11) a,=2"("C, "C.2-C,
if p+q=1.
Provethat
"C=2n-1)! I(n-1)
Objective Questions
C
C
Fill in the blanks.
+T"C-1
The coefficient of a
118.
"Co
Prove that 10S. "C("CG "Co+
"CC
+1)."C"C2."C
x-when expanded in
..."C 120.
s.e1.1 n1C+'C2 S1 +"* 'C3
s2+ ... +H*'C
.1S,
2
Prove that
+(-1)".
S
"C (n
+1)
is
in the polynomial sum of the coefficients 2163 is_ expansion of (1 +r-3x2) in the coefficients (6) The sum of the numerical expansion of (2x + 3y) is of (a 2 +a)" does 122. If the 5th term in the expansion not contain a then the value ofn is. the expansion of 123. The sum of the rational terms in
(V2+35 10is
p1 2"*
m=P
C+2"*'C+2 1c?+.. 4 2n +Ic? C"C; where 0sis10,0sjs10.
112.
Evaluate
113.
Evaluate 2
114.
If (1+x+x*)"a +a1+ a2x+.
ij
aaa-ay43 +a2.. (ii) a= a2n -i* (ii)
126. If
x ocurs in the expansion
of
then
its
coefficient is
the sum of the binomial coefficients in greatest expansion of (a + b)" is 4096 then the binomial coefficient in the expansion is expansion of 128. The number of terms in the (1+x5) 5 which are free from radicals is . x the expansion 129. If n is even then the coefficient of in
+za +@21-2an 4+1
(1+x+x*+ x*) " = ao+a,x + azx+... +n " #3n
-r
= ag t
y +ay
t.
+2n
then prove that a, = a2n -r Also prove that + + 30 + 5a2 +..+ (4n +1)42 (2n +1)(2 p).
+2)"=ag + a,x + at nEN then prove that
117. If (1 +2x
of the positive integer. If the coefficients of (1 + x)" 2nd, 3rd and 4th terms in the expansion are in AP then the value of n is
-
a3-a+a2-a3 +
If(1 +py+ y"
x+is
127. If the
+421
then prove that
prove that a,=
middle term in the expansion of
125. Let n be a
"Cm"Cp|
111. Evaluate
116.
descending powers of
xis 31. Then n is equal to , 9950 100 S0 and 101 The larger among
124. The
115. If
expansion of
121. (a) The
Prove that
)
of (a + b) in the expansion
the The coefficient in the third term of
119.
snl+9+9**..+9",
110.
b
1s
+"C3)... ("C-1+"C,)
Given
109.
a, = "Co 2C,+"C,.2-*C-2
t
.
2n
of (1+)" |1-is 130. The sum of 2Co+2C +2C,+.. +2C0 is equal 131. The sum of 132. Take the
Co-2C
+2C
to
...+2C1o is equal to
expansions
(1+)"= "C, + "C,x+"Cx+ ... + "Cx" +"C (y+1)"="Coy"+"Cy"-l+"C2y-2
A-2202
P'nblems Plus
+)=
"C"+
"C"y+
The coetficient of
a1+y(r+ Choose
thhe
where
""
the expansion
in
(1+52a)+(1 (a) 5 135. In
(c) 2 in
coctficicnts of r" and enpansion of (l + a)"' " are (a) equal (b) equal bul opposite in sign (c) reciprvcal to each other (d) none of these The number of dissimilar terms
1
142.
(c)9
the expansion ofja*
(a + 2h+3c)
(d) 10
independent of r
(c)-C
(d) none of these The largest cocfficient in the expansion of
156.
is
(a) 137.
C
(Cu
(b) 6
(d)C
(c) 5
The sum of the series
"C C, +C+.
(a)2
(b) 2
(c)2+
"C
of
(c)
(-1) ",
m
(d) none of these
integral part of (8 +3V7)
is even.
there is
independent of x and a expansion
lernm
a term
independent of y but
of x and y both.
no term independent
of (1
+ 27+*°there
term whose coctficient coefficient of any other term. one
coefficients in the expansion
or false.
the expansion of
146. In the
the tern
is
is
is
exactly
not equal
to
1
(b) 2
(c) 2 1-.16 2 (8 =
4
147. In
the expansion
"C,, the sum of the series
148.
If fx)
of
every term
-
function of x.
(d) none of these
)
If C,
C
145. In
19
(d)2 19
139. The sum of the last eight (1 +) is equal to
(a) 2
+"C0
10
(b) 2C
(a)
144. The
(d) 2
2"-
State whether the statements are true
(d) 3
+
is
140.
)
35 when divided by 8 leaves the remainder 1
158.
(b)*C
(1 +
in th
in the expansi.. siono
(c) 45
expansion of (1+)
143. In the
(b) 0
p,9N)
is
(b) 24
(a) 9
the constant term is
-
C
(a)
(d) 2+ the expansion ot
-5v21)° is
(b) 7
(-1) " n
141. The
of x, is
number of terms
(cd)
2)
(e) none of these
number of terms in the exPansion ot (2a+*)" wrhen expanded in descending
(a)
()(-1)2 (n+1)
(c)-1)" (+
v))" is the series.
1
even integer, is
n is
(a) 0 ot
133. The
134. The
T Mathematus
"C"*v*t +"C"
correet option(s).
powes
in
x+
=
then
fa)
is a
is
polynomial function which is an even function.
IC6-2C
+3C
-
... +(-1) "(n + 1)C
149.
6Co-°C+bC2-... + "Cg =0.
Answers 2. 198
3.
84 9.1255 2
2.0020001 11. (i) -20
4. 2
(ii) (-1)
",
(21)
8.
*
10.
672, 5376
(iv)
(2
+ 1)
n! (n +1)!*
"
(iii) -560xs, 280x
21+1)! +1). n!
(
*
a
A-203
Binomal Theorem for P'ositiue Integrl Inulex
(v)(-1)
.
il
nis even,
3
63, (6i)
(i)
(iv) 3"-1 2
(ii)
1
(n+1) 2
(-1)
and
Do-1*
(-1)2
x
if n is odd
65. 4
1002! 67. 501952
68."'"Cm
12"n(1 (2"- 1 Ja (n 2"-
72.
12!8 2. 20!
13.
14. (i) 56 (ii) 0
77. (n-
45
80. 0
16.30! 155!
no
15.
(21)! 17 -1) ! (n +
18.714 1)
(2"- - 1) "C,
24.
4x+
25. 14
10+
5
27. (i)
16x+.. ; 19
(ii)
+)
n
34.
14
14 +84 + 280 + 560
i.e.,
(1
35. 10, 10 40.
+
2
"
39. 7 or 14
124.
possible to be in APbut not in HP
41.n 46.
2,3,4
52. (a)
53.
provided is even and n>2 n
54.
8th term
()(i)
57 (ii) (iv) 0
()
+2)x
1 (1+)
1+ x)"
(n +1)x
2)x"
(271)
99.
111.3"-2"
1132
(b) 5
2
42! !2
(21
120. 101 122. 10
0
10! (5 12
123. 41
125. 7
11
57344
243
127. 924
128. 12
20
132. "C
131.
20!
(10
27 x7! 13!
n= 49, 50, 51,
58.3
+
126. 42.
51. 6th term, 1.e.,
4th and 5th, i.e., 489888
56. 2 2
62.
50."C
(1
119. 32
121. (a) -1
37.14
-
(4 + 2) 1(21+1)!
118. 126
+2)
3
+3
35 10!25!
112.
672 + 448+128,
(b) n
2)
1(1 4 )"
+
2)x
(21+1)!
101.
32.22
32.
n= 6,p=
4
2 (u !2
(v)-336
28. 11851
+
(n + 1)(n
92,
29. 168
1)(1
(4)"*2
o1 1.
25215.3 10! 15!
81.(n +2)2"-7-5
22+ 1
78. (n + 4)2 "-
+1
2"''+1 +1(u +2) ( (a)+1 b)
89.
(211)(iv)(m 30.
88. (a)
20.0 23. Bn
1
al=
1)2 "
3)!4
2" +1)d
-
(n+21+ 1)x+ (3n
85.
19. 924
21.
4
-
I|
1)--
'C
"
69.
(
"
64. 120
..,
59
!
133. (d) 137. (d) 141. (a)
true 149. false
145.
129. 0 +
"C+ "C2+.
130. 2 2 + "CA
134. (a) 138. (c) 142. (c)
135. (c) 139. (c)
136. (c) 140. (e)
143. (c)
144. false
146. true
147. true
148. true
Encl
Ter
for Pesae integr i-dz
Chapter Test Time:75 mirztes
folowing, eech statement is incormplete. Fll in the blank so that ezcr sultin statemant becomes corect. () E the sum of binomial coeicients in the expansion of (a y"is 1024 then tne grea binomial coeficient occurs in the th term. () The coeficient of expansion in the of (2-z-3x) is. * (ii) If < (2-1)
1+a+ + a+... + a +
(1)
1X+b-1+ i.e.,
+
3 '+am
37amx
relations a, = a-1+bn-1 etc.)
bt +
Cmo
+
+1
(a +a24+...
Hence, the problem.
x-1
a+...
..
(3)
2
+a")+a
+2
+am- ') +a2+ ... +a ") + am-2
(a 1+x=x, given)
8M2|a+a*+ m1 +
Pm+1) is true.
.+a ,*2, a" +
+a *4+a
(1)
for alln e
(2)
Also from (3),
N.
or
for all ne
a1-a
-2 a+a+... +a ") Ta 1
N.
or
2>0
m(m + 1)
Prove using induction that
1+a+ a
a+..
a+a... (
.
+a?+... +a m-)21+a+a2+... +a"-2
P(m+1) is true; thus, P(1) is true by PMI, P7) is true
+a"1
If (2) is not true then
i.e.,
a
m
aa?+...
b, +
=x31+x x3"*, using =x3 (1x) =x 3" -x3
2.
+
bnX +Cm
+
=x3"xja,,x
a2
we
From (1),
using (1)
x=x+ 1
Let
(1)
true. For this
m+3
I
1+a +a*+ ... +a "**>
T+a
13.
+a"")
+b)x +ambn+Cm+ mCm
(
(from given
.
+a2
Now, we want to prove that P(m +1) is have to prove that
l
Now, a =
becausea22
P(1) is true.
n = 1,
When
+1)+3a
3>3a
x" =a,x b, + n
Let P):
n= 1,
1, co=F0.
=
1+2
+ a"
+a +...
a
1+a+a(a-24
b-1+ Ca -1 Cn-1
a
1+a+at... +a"
Let P():
-2(a+
+
1-a "2>
1
a+... +a") + m{m
+
3)a "*
m(m + 1)a " *2> 00
0
Principle of Mathematical
12+... +a")+ mna" +a ") +ma" or +a 2a possible for a 22 or
"l(m
+
1)a
"l(m
+
1a
1)-21 n
for all n e N.
58. Prove
(n
.
).
+.
that
46.
Prove thatfor all n
47.
49.
50.
E
1+V4a+1
,
b=perfect square and and
a"- b" Vr
ab is an integer
a"+b" =c" +d" where n e = 0, prove by 5x+1 xiof roots B the be If a, not divisible by induction that o"+B" is an integer N.
63.
for all
where ne N. '" Prove that 10" -(5+ V17) by 2"*for all n e N.
4 64.
N.
65. If
x+is
N.
y" Prove, by mathematical induction, that n. integer odd is divisible by x + y for any positive prove that then If n is any odd positive integer nn-1) is divisible by 24. Prove, by induction, that x(r"-- na"-1) + (n -1)a" is divisible by (x a) for all n e N, n>1.
"+
Prove that (v2 - 1)" can be expressed as Nt- Nt-1, positive where t is a positive integer for all odd integral values of n.
= 1,
66. If
-
(5
-
V17)" is divisible
induction, that an integer, prove by N. an integer for all n e
"+is
-
51.
+abi +bs +...+b)
where r is not a integer. Use induction. then prove that a,, is an a2+b?=c2+d2 for the real that 62. If a +b=c+d and prove by, induction, numbers a, b, c, d then
4 Prove, by induction, that "> "C
for all n 48.
E
+.. +a,b,) < (a +a +..
1.
61. If
are positive integers. where a 2, etc.,
s
+ a2b2
a
N.
2 31+
at least
1
+X2+ (a,b
Prove, by induction, that
=
1.
1
..+ 45.
all >
where each x>0 and x *z 32 If x, then prove, by induction, that one all n 2 2. X3 +. +Xn > n for ... x,
60. Prove
Va + Na + Va +... to n times 44. where a> 0.
13
-> vn for
that-
forn E
1+ nr
n e N.
n
+2)"23"
+x)">
-
induction or otherwise that
for all n
" > 1)! for n 2 2, n e N and that for n > 1, 57. Prove by induction
all "
Prove that 2"x (n
56.
inductiory that|
Prove, by
54. Prove that 55.
nduction, that n!Sn'" forall
Prove, by
>
3"
induction, that 3
.by by Prove,
for
-
induction, that
Prove, by
andb
53.
24-1+l,b,=a-120-1
and
integral values of all for Prove that 10"-2> 81n n25. ! n 2 7. Prove that 3"< n for ( !)2>n" for n 23.
U2 =
1
and u +2= lUn1
then
for n21
+ 1l
prove by induction,
-
=
l2 5 and -1=
67. Let
Show,by induction, that 68. If
for all 21.
,
+61l- , 122. =
3"
= 3un-7t-1 +5t-2 -3 0,
18,
112
forn20.
-
(-2)" if
23 and = 17 then prove that 2u,, 20n 7 n
n
21.
=
1,
+32-"
A-230
.
Problems Plus in IT Mathematics
a
If
# = 3 and a, = 34,,-1-2n-2 for 1 22 then prove by induction that a,, =2"+1 for all n 2 0. 70. Use induction to prove that =2,
s
forall
sin(ax 72. Prove, by
+ b)}
=a
1,
(a)
"sin|ax + b+n
induction, that for
m
=
J
mm,
prove that
m 0, 1, 2,..
Prove, by mathematical induction, that
75. Let
1)(2cos 20 1)
-1)... (2cos 2"- "0-1) +sin nx
u
=
J sin
4
Ccos x sin"-'r+
Sin. 77.
2
2
2 2.
Sin
4
sin n*| *
is
s not
=
!log
| sin x|,
an
+
(1
cos
X
forall ne
integral
(-1)"-1,
Sin 2110
2cos 0
n times
2cos
2
220..
cos
2-1e=Sin 29
2 "sin 6
that
2.cos+ 4
isin
4
cot11
+ n(u + 1)}
=
tan(
+
1)-neN. 4
92. Prove, by induction, that
multiple
to prove + cos 2x + cos 3x +.+cos
cos
20 cos
+i)"=2
=
where n is any integer. 91. Prove, by induction, that
N.
n e N.
mathematical induction
cos
1)0
prove that
90. Prove, by induction,
for all even positive integral values
n
lpx +(1 -p)yl
prove that
89. Using induction,
cos
of n.
( "log x)
ys sin
N2+V2 +V2+.. to +(n + 1)
27|2 dx =0
p) sin
88. Use induction to
x
where n e N.
0)" = cos n0+ isin 0 for all positive as well as negative integral values of n. 87. Use induction to prove that sin sin 30+sin 50 ..
u.
e*sin
A
(cos 6+ isin
for n EN =
Use induction
1.
and 0 Sx, y sT.] 86. Prove, by induction, that
where 0 sps1
cos "x sin nx dx
79.
1
= 1,2, 3, ...,
+(-1)"- sin(21
dx =
e"(sin x + cos x)}
78.
r
i
sin A, Su sin
p sin x+ (1
"x dx. Prove, by induction, that
sin(21-1) 1)x
for
cos(1+1)x1 Sin a
[Given the fact
(-1)"-1.-1)!
for all positive integral values of n Prove, by inductio1, that
82. If
that
x + sin *2x + sin *3x+..
dx. -cos*
dxoa)=
|
-
(2cos
1
85. Let 0
X cos
23. Solve: cos
+ cos 2
.
0Sx sT. 1.
+ cOs 3x > 0, xe [0, n|
24. Solve: cos 3x +V3 sin 3x 25. Solve:
tan x +3cot
x>
+v2 0, cos I
1, cos x*1
x*nn
Sin 2x +lo81o cos x=
t
x+y*7+2**y**.y2**
sinx
x=10= 100.
so
orlo810 5in 2x+ log10
AM2 GM 3. Given
log
7, Solve for
Again, take the positive numbers x+*z,
43) = logn (43) = RHS.
are real if sin But here the logarithms
ie, 5-3 = 3-2 or ie, y=2 2=0, yThis is possibleif
log +log
z=0 log(r y+3.y3**.z*)=0,
log
43
Logarithm
x=log2a
4, y =
log,a 24 and z = lo843a
then prove that xyz + 1=2yz.
=loga
(ab), (bc), y = log, (ca) and z=loge
x+1 y+1 4.
Prove that loga(bc) logb(ca) log.(ab) =2+ log,(bc) +logs(ca) +log (ab).
-61 B-60
5. If
Problems Plus in
2+2-)_y2**-)ZT *Y2, log log y log
prove that
2
y-zV=z*.x*=xV.y .
If
O8= O8 V_Og E, y-2
-y
Z-X
8. Solve: logio sin x + logi00 (Bcos x)
-
logy =1.
Logarithm
1213 then Iflog 20.30103, log3=0.4771213 then th. thenumbey of digits in the number equal to 3 12 x 2*s 17. If log 3 0.4771213 then the number 16.
between the decimal point and the first digit in the value of 9-16 is_
prove that*+yy+zi23.
7. Prove thatx °%-logz.ylog-log .zlog
9. Solve: logz tan x +
IT Mathematics
Iflog2 sin
18.
-1 then the set of general aluesof vali
19. 1f 2loga+2log
Objective Questions
cos.log coslogsinlog(
+21og a"
- 55loga
(a) 10
(b)
(c) 20
(d)5
81 =
20. If
12. If logx + log y = log(x + y)
o-logy -C
then y as a function of x is
(a)
given by y=_
log z
C-a
The sum of the series
loge'
1
a?+b2=23ab then log
(a) b
is the AM of
22. If a, b, c
then
but
log(1+**)= log(.
log, *(log, a +log,
211 +
9 11.
3
2n7+cos 10.
(a) 1
(b) 2
(c) 3
(d) none of these
; neZ
272
2nm+12neE
16.9 S
18.
4nn,
1
17.15
+112+5,3T
4n
19.
13. log.9
14. log a, log b
20.(6
21.
()
22. (b)
+2,
a >0
c) is equal
Answers 8.
4.
is equal to
are positive numbers (#1) in GP and
1, the value of
nEZ
A + log
to
tan B+ log tan C.
triangle ABC?
log, sin+log sin +log 3. Prove that 4+
the value of xyz is
(b) c (d) none of these
()d
AABC, Prove that in an acute-angled log tan C)= tan log(tan A + tan B+
Can this hold for any
b)3
21.loga bx logbCX iO8
and 15. If0 0.
11. logs log2
Time: 30 minutes
gnifican
is
=logo3
Chapter Test
sin
3x > 0,0sxs 21. Solve: 1+log2 sinx+ log2 sin
+log2 sin =log25.
Problems Plus in IT
B-62
Mathematics
Answers
1cos0
)sin
2.
no
6.Properties of Triangle Recap of Facts and Formulae Angles ofa triangle triangle ABC, three angles are A,
1.
Ina
A+B+C=
sin(B +C)= sin(t-A)
and
C.
24 etc.
Vs(s-a)(s - b(s -c)
.A
= sin A
B cos(C+A) = cos(t- B) = -cos
in4sin
formula BC in the ratio m:n at D and In the AABC, AD divides = B. LBAC in two parts ZBAD = a, LCAD
4. Ratio
c
conco .COS 2 2.
=
.sin A B
in
Trigonometrical relations between sides and angles
For any AABC we have
m
2RR
where R = circumradius (Sine-rule)
b+c2-a2etc. .cos A =
(Cosineformulae)
2hc
If LADB
= 0
D
then
(m
+ n)cot 6
.(m
+ m)cot
=n cot p- m cot a = m cot C- n cot B
acos B+bcos A = c, etc.
in-
etc.,
bc
5. Some
where 2s =a+b+c
etc.
bC
tan-6-bMs-0_ s(s--a)
sin
-)4sts-a)
(s-bXs
tan 3.
2A + sin 2B + sin 2C
=
4sin A sin
1+cos2A+ cos 2B+ cos 2C= 4cos
etc,
sin A+ sin B + nC=4cos
where A= area of AABC
cot ASS-a)
important identities for angles of a triangle
In a AABC, we have
etc.
otetc.
B
sinC
A cos
B
cos C
A cos B cos C
+cos B + cos C=1+4sin
sin
sin
cos
A
cos
A + cos B + cos"C+2cos A cos B cos C = 1
A+ tan tan C= tan A tan B tan C, ie., cot B cot C+cot C cot A + cot A cot B 1
tan
B+
tantan-1,
Area of a triangle
the area of a AABC is denoted by A then sin A, etc.
Abe
B-63
B-64
Problems Plus in iIT
Mathematics
B-65 Properties of Triangle
an
6. Circumradius, inradius and exradii
stan
In the AABC, let the circumradius = R, inradius = r ana the three exradii corresponding to the vertices A, B and Cbe r and r respectively. Then
AAabc
R-2sin
-4Rsin
7.
A
6-tan-(6-tan=6-tan
=32R 28
coo
2. Prove
Regular polygon
i
the 20AA
ZA,0A2=2
that in a
= k(cos
b
-
LHS LHS
-+8-b)2
la
Also,tan Each
3. In the AABC,
2-2290.
cos
o
B + COS C)
BC.cot2
tan2
A)
os
COSAA
HSsin
cot
-2ksinsin
+b-c)
Dlcosin
+b+c-2c) =
ccot
in
2ksin
sin B).
C-cos
=k(cos
A
Similarly for others.
y+tancot
prove that
CotA +cot B+cot
interior angle of the regular polygon
cos B)+
+b-(acos B +bcos A))
+b+C-a-b)(a
exradius
Ckgiven)
A
2s-2)AD
cos Bcos
cosA
sin C.
sin
learly, in
coc
sin
B
AABC,
(2--b(a
tan-4R
3sin A sin
c
because AOAA, will *
one of the n equal triangles with a vertex at 0.
inradius
2-
+ z)tan
Aregular polygon of n sides will have s vertic a circle. If O be the centre and rbe the radius o circle, and a be the length of each side then
4Rsinsin ID
-4Rsincos
c-4+6°+c
=2 k(cos
4
cos B cos C sin B sin C
=
C-cos
B)
-cos
kl(cos C- cos B) + (cos A
C)
+(Cos
Each exterior angle of the regular polygon=2x
=kx0
=
B-cos A)
0.
-+
2R
Selected Solved Examples 1. In a AABC, prove that
sin sin A
A cos(B-)
3sin A sin B sin
C.
A sin
=
sin A
=
sin
sin
sin(B + C) cos(B-C)
sin
A (sin 2B + sin 2C).
R l
sin +
(bcos B+ccos
*A(sin 28+ sin 2C)
sin "B(sin 2C+ sin 24) +
c
cos A
22
A 2sin(B + C) cos(B C)
Similarly for others
LHS=
cos C+22
2422
A cos(B-C)
A sin
B
dbc2-a2+c?+a'-b+a?+b-c abc
A
2R sin C
acos A + bcos B)
8R3 1aDacos B+ bcos A) + ac(acos C+CcOs +bc(bcos C+cos
sin C(sin 2A + sin 28)
8R3 2bc
acos B+ bcos A =c, etc
atc Na-ac
2c0s Zcos-C
+b2+C2 a2bt2 be 42R be 45 sin A
ab+-RHS. 44 and z are proportional to cosines of the angles A, B and Cofa AABC respectively, prove that
y
8))
2+)tan
co0.
+c
= Here A, B, C are in AP; so B A -a,
4+C_180 Also B
abe la6+c3-a+b+ Rbe
+21
C)+b(ccos C+acos ) +c
In a AABC, the angles A, B, Care in AP. Show that
(a+b-c3 2R
cos(B C)
sin
5.
2a
B
90°
B2
2
90
B=60
-a
=
and C=A
60°, i.e.,
C=A -2a.
B
A = 60° +a
- 2a = 60° +a-2a
=
60°- a
C =2a.
A-
sin(60°+a)
sin 60°
sin(60°-a)
sin(60°+a)
a+CL 3 3 cosa 2cos
2
+C +
.
sin(60°
a
-
a)
a+C
v3 cos a
B-66
Problems Plus in IT
B67
Matkematics Properties of riangle
Now,
1-cosa-cos b- cos'c+2cos a c
a+
a +C
2-ac+C*
sin b sin c
VE)-4-E
1-cos a- cos "b
2tan -
cos
sin
= 2cos a (60° +a)sin(60° +a) 60° sin 60°
sin
sin
sin60
sin
60°
bcos
sina sin b sin'c
sin'a
sinB sin
(s-a)s-b%6=C)_ 5=1-4 1s6-aNs-bs-c)
sin
a -cos
Thus,
a-
sin
b+cisrational,becausea is
But
2cos
Vacos'asin'a-cos eos
sin A
sin
= sin C
B
-
=
cos b
4
6
sin
c-
-
-
rational rational
B + Sin
rational. But
a is
rational. So sinA is
Clearly
aa=
A;
Similarly (ii).
45in D= rational,
(i) a,
c
tantan
are rational
:Now, tan
-),
tan
and and
.
C(s-aXs-b
s
cos c)-cos 'a - cos b cos
+2cos a cos b cos c
Thus(Gi) 8.
=
s
a,b, c;A, are rational.
rational.
Thus ()
such that LOAC= 20CB= 2OBA = a. A Prove that cot a = cot + cot B+ cot C.
10. Let O be a point in the AABC
sin A = rational. ).
B)=
a
find the area of the triangle
Here, cos(A-B)-: By componendo and
oc
14AOC,-AC AAOC sin From AOCsin a
1- tan
A-B5
But LAOC=x-(2OCA + LOAC)
=T-(C-a +a)=r-C AC OC
dividendo,
sin(t-
).
Let(i)be true,
ie, a, tan
or
tanbe rational. Or
Now, sin B=
tan2
zC sin
tan
2Brational,
ceuse
tanis rational
But
tan
sina
C)
OC
(1)
sin
Similarly, from AOBC we get
2tan
tan
+cot
If in a triangle ABC, two sides are a = 6, b=3 and cos(A
nand tan C are rational because sum, differene So tanand
B
rational
2
Now, )
(cot
A + cot
festablish as in solved example 3).
4
sin A
Abe A
and
Sn
44
A
AsininC
AABC ) The sides a, b, c and the area A are rational
Let ) be true, ie., a, b, c and A be rational numbers.
bcos c
e
sin "b sin 'c
sin
rational.
a, sin A, sin B, sin C are rational. Prove that i) (i) (i)
+2cos a cos b cos c
b1
(cot
rational.
b
b+C
product and quotient of nonzero rational number
1 cos
A
from the altitudes of the AABC If a, B and y are the that show then respectively C vertices A, B and A+ cot B+ cot
sin sinC
isSln
()
cos a-cos "b cos
sinb-sin
sin
7. Consider the following statements concerning
Sinb Sin c sinA =1 - cos A _Cos a+cos b cos C-2cos a cos b cos c sin'b sinc
sin b
gin A
C63 sin 9
B, sin C be rational. Let (ii) be true, i.e., a, sin A, sin
cos c
sin b sinc =OS +Os cos C-2cos a cos
But
Thus(i)
thatsinasini
Cos
in AABC,
...\\
2cos a =2cos
cos a= cos b cos c + sin b sinc cos A cos b=cos c- cos a + sinc sin a cosB cos c= cos a cos b +sin a sin b cos C then prove
24A
sinb sin c
aina
6. If in a AABC,
Here, cos A
sin Bsin c
SinA
sina
9.
s is rational
e
area of the triangle
-zabsin
S
1-cos'a-cos b-cos C+ 2cos a cosb Cos
C=
cot1;
the
(s-cXs-a) (5-a\s-)
sin 'a sin "b sinc
cos a
tanis rational
Now, tan 2
C Similarlyinbsin'c
2cos
or
because
+2cos a cos
sinA
sin(60-a), (sin(60-a)
2 =rational,
1+tan
t:
OC
A> Bl.
sin
2)
Bsin(B-a)
(1)+(2)
bsin Bsin(B-a) asin C Sin a
B-69
Problems Plus in lIT Mathematics
B-68
Properties of Triangle
sinA sin C
sin
a
inA
sin B sinc cot a -cot B sin(A+C)= acot B sin
OinA sin
Cot
A sinC
cos C+ cos A sin A sin C
sin A or
cot C+ cot A cot a
=
=
sin Ccot
ar(AABC)yz n-(A +C)=B))
a
-
ABDM,
:In
L BC. So BD =DC.!
BDtan tan A
D
cot B
cot A + cot B+ cot C.
or
=
tan. A, ie,
-; LBMC=
2y
tan
A + tan B +
and tan
A
tan B, tan C
tanA + tan B+ tan C=
sin
sina cos C+ cos a sin
a,
cos
cos a sinC
or
tan A tan B. tan C
c-
or
cos C= sin "C
12. If
C= a"b sin C.
the distances of the sides of a AABC from its x,
circumcentre be
ay
c
y, z then prove that
abc 4xyz
a
AC
From the ACAD, cos C
CD
9BC-2a+c)-b3
=b
2b
2
BG=2a+2-b). 26-c Similarly CGor
... (2)
a
BG2CG2-BC2
13. 1f x, y, z are
the distances of the vertices of the AABC respectively from the orthocentre then prove that
BD From the AABDsin(A 90°)
cos A Let Hbe the orthocentre.
-oBG*+bi
.(1) 2bc
a+
Now, cos a =
2BG
CC
2a+22-b)+2a2b3-c)-a2
AB
sin ADD
2BG CG
-5a2
b
sin (90° + C)
2BG CG Then
ACos C
2cos
acosCOSC -2c
-,
a
-2c
c
LBHC=180° 2HBC-LHCB
180°
(90-C)-(90°-B) =B+C=t-A. ar(ABHC)= BH-CH sin LBHC
zyz sin(r-A) =;yzsinA.
b2+c2-a2-2b2
or
c-a2-3b2
Now,cos
A.cos
2ca2 3ca
ar(AABC)
2b_b2+c2-a2
cOs Sin
from (3)
= 3 ar(ABGC)}
a'+c-5a a
2bc
3b2+3(2-4)a-c+3¢82, 3ca
a
ar(AABC) .(3)
C+c-4
sin a
ar(ABGC)
c2-5jsin
(1)b+c-a-b 2bc
or
sin a
bc2-5a)sin a
from (2)
from
C1-0
ap+b*q-2alqcos
or
C=4C42
cos A
cos 3ca In a AABC prove that A and AD L AC. where AD is the median through
Cos
or
2 Now, BC2+ BA =2(BE 2+CE)
xyz
C
Cc
C+cos a sin C
BG= BE, CC-CF.
ie, 4 =
R
.abe4xyz
sin(a+)
AD
nd AG
From the AABC, cos A
But in a triangle ABC,
AC sin a,
geometry
AB2+AC2=2(AD?+ BD ), etc.,
or
anC =
b
We know from
a
abc
yz
tanC 14.
Let 2ACE =a. Clearly, from the figure, we get
+5zxsin B+xy sinC
(1 gives, 4R 4R
tan A.
tan B,
vertex C then prove that
sinC=a'p*+b'q*-2abpgcos C.
Also, we kmow that
C
Similarly,
sin A
... (1)
fromgeometry, ZBMD
cot a- cot B
11. If p, q are perpendiculars from the angular points A and B of the AABC drawn to any line through the
ab
,
ar(aAH)2ysinC Let Mbe the circumcenre. MD
sin A
(:
15.
ar(aCHA)=zxsin B Similarly
cot a sin B- cos B or
ABC make angles a The medians of a triangle that Prove other. each with coty+cot A + cot B+ cot C=0. cot a + cot ß+ etc. AABC and LBGC =a, Here, G is the centroid of the
BT
sin B cos a-cos B sin a
sin B
r
cota
124
bc2-52 12
3ca
Similarly, cot p-
5,
12A
coty2+b2-52 124
B-70
Problems Pius in
Properties of Triangle
.
cot a +cot ß + cot y=-
ab+c 44
Again gain cot A +cot + cot B+ B + cot C=OS 4
COS
(b+)sin B,
Cgin Asin B
R.
cos C sin C
Similarly
2R C
-(+c?-a2+c+a2-b?+a2+b?-c) h
+b2+c)
(a + 8) sin
Now,
R
sin(PIC
+
2QIC) 17.
44
0
(6
The internal bisectors of the angles of the AABC meet the sides BC, CA and AB at P, Q and R respectively. Show that the area of the APQR is equal to
+ c) sin
206 + eMe
2abc A (a+b(b +c)(C+a)
21BA +
2IAB-+
-BIP-
LBID
sin
+oysin
sin
o
bcr coDs 2c+aa
+b) sin
sin
2a + bX0 +) sin
sin2
Now, ar(APQR) = ar(APIQ)+ar(4QIR) + ar(ARi) abcr2 2(a + b)(b+c)c+a)
sin
sin
sin
cos
=
in 2(a+ b)(b+ cMc +0)
or
in
asinB-+in2vde SNa
r4Rsin
(7-cos a)r*=1-cos a
cOS
7-Cos
-cos
rTe x= where
.:
1, 1+x are in AP and from the question Clearly the sides a, b, c of the triangle are in AP. Hence
sinsin
a a
19. Prove that, fora scalene AABC,
cos a
a, c, b are in AP
cos A +cos B = 4sin
1-x,
abCk 1-r11+x
(suppose)
We have to prove that x
c
cos A +cos
B = 4sin
2cos A+B cos
=4sin2
=
2sinco-2sin0
N7-cos a
k{1+)_3k
cos
2
C is the is the greatest side. So +a. is the least angle. Hence C=A
cos
sin A+ sin
oninin
a+b
r: A+B=r-c
A--2sin 2
2sin
a=CA
n Cb n Ain
a 1-7x =
its greatest 18. If the sides of a triangle are in AP and the sides are angle exceeds the least by a, show that
aKing X Positive,
4R
cos
(1-x)
a+b+c-24 b+C-a
8earestangle and A
abcr
-1-
-2
+3a
S+b*K1-x)+k+ 2 2
COs 6
a -2cos-1
s(S-a)s-bS-c)S-a
Now,a=k(1-3), b=k, c=k(1 +X) Now, from ADIP,
1-X
V-2)1+2)1-x
A
inthe ratio (1-):1:(1+)
carcos2
ar(ARIP)
cot2
44 2. 34-a2a
sin
(3k-k) k(1-x) k(1 + X)
2.
9_s(6-b(s -c)
2a
Similarly, ar(AQIR) =-
-90P.
V
2S
+b+C
2(6+ cX+ a) sin
Let ZDIP =0, PI= x, Q! = y and RI =z. Then =
6-.6+s-b) b
are sides of a AABC and 3a = b+c,
that cot
LHS
abr cos
Let the bisectors of the angles meet at I. Then I is the incentre. Let ID L BC. Then ID =inradius = r. =
(c+a) sin
abr
Ifa, band c show
br
a+b+c
44
Clearly, 2BIP
abcr
a+bb +cXc +) 2abc A (a+bb+ ¢ +a)
ar(aPIQ)=PI1Q sinLPIQ
*y
-bX2.6-bX6-2
ab
abc r 4+bNb+Mc+a)tb+c)
and z=-
cot + cot y + cot A + cot B+ cot C
a+bC 16.
y=
+b+c)
bc
cot +
abcRr (a+b(b+C\C+a)
br
(c+a) sin
a+b 2ab
-
B-71
IT Mathematics
=
2c
B
=
4sin
4cos
sin=
2sin C
a,c,b are in AP.
8-7 Properties of Triangle
20. In a AABC, if atanA + btan B =(a
A+B
+bjtan2
sinA sinB sin?c
then prove that the triangle is isosceles.
Here, a tan A
"cos A
tan A
cos
=
X
B, cos Care also in AP. prove that cos A, cos
B
1
-coA1.1
sinA
nn
Sin
sin A
or
anun
i
tan
-
sin2B
1-cos "B=1-(2cos A A
1-4cos
A4+
a sin
bsin
A+B cos A cos
r
But cos
-B,
sin sin
cos
-14
-bcos A} =
{acos B
A cos
sin(A
-
B- sin
B)
=
B
x+6x+9
(1+ 9
cos A) =0
sin C
+x)
2+3)2
1+x 9 +x)
1
from (2)
1+cot C
from
2
tangents of the angles of a triangle are in AP prove that the squares of the sides are in the ratio
+9):(x+3):91+r) the greatest tangent.
Here tan A, tan B, tan 2tan
B
or
C are in AP;
tan A + tan
22.
A+C=T-B}
A
2cos A cos
sin B. cot B C=cos B =-cos(A +
or
2cos A cos C=-cos A cos C+ sin A sinC 3cos A cos
C= sin A sin
sin C
x9+
2+3)
... (1)
cos A+ Cos
9+x
c
+c2-a2,
2bccos A, 2cacos B, 2abcos C are in AF b
Also, s =
are in AP {dividing
2nbo)
COS COS eia are in AP Sin A 2R sin B2R tA, cot B, cot C are in AP (multiplying by
n+5
and
a+b+br b+br-a +br-b +b-br 2
a'h
x(2ab
But cos 20 2cos0-1;
2
(a+b2- b',}
4ah2 =4a2-(a2+b2-b?r32
or
=4ab- ta2+(1 r*)b2 (1-)5*-(4a-2(1-*ab+a'+ 4a h?=0
or
(1-O)*-2(1
+ra*b2+a*a2+ 4h3) =0o.
4(1+a-4(1-)a
2
2n(n+1)
so 21
n+5-1
1
n+2)=nl(n + 5)-2(n +2)9
(2)
(2ab+ (a2+ b2-br*))
Now, the values of b* are real, so
2n(n +1)
(+5)
.
-
21(n+1)
220+
br-(a-b
or
(n+1)(n-3)-3
(-3
(1)
(a+b+br
2(n+2)
+2)2 cos 20-+(n+1)-(n
n-2n-3
or
..
2)
n+1(n+5) 2n+1(n+2)
c+a?62, a2+b2-care in A
cosA, cOsB, cos C
Cos A
B
)+(n +2)-n n+6n+5 2(n+1)(n +2)
Now, cos e-
laddinga2+ b*+c
. (2)
B
From (1) and (2),
Imultiplying by -2
Let the altitude be h.
Now, (area of the A) =|
.
and
b, br.
A, cos B, cos C are in AP.
2n+1M+
are in AP
of the two sides is In a triangle of base a, the ratio triangle is less r1). Show that the altitude of the
Let the other sides be
2cos
C=2cos
=4; hence sides are 4,5, 6.
(a +b)-b
nis
9(1+x)
cot A, cot B, cot C are in AR
-2a2,-26,-2
n
neN
Ifa,b,care in AP then show that
C
tan A tan C =3 Let tan A be the greatest or the least tangent. Then tan A =x.
b Now,sinA sin B
(1+r9+2)
or
4
i.e,-and-3 which are not natural numbers
than or equal to
Cos B
side From the question, the largest angle opposite to the side n+2 is 20 while the smallest angle opposite to the
a,b,carein AP
C=
or
9
Hence, the problem.
sin(A +C)2 cos A cos C
nB cos cos C 2cos A cos
1+
where x is the least or
+3)2
-7t V4924
4
Cos
2cos
= =4 or 2n+7n +3 0.
Roots of 2n+7n +3 =0 are
are three 24. The lengths of sides of a triangle consecutive natural numbers and its largest angle is of the sides the twice the smallest one. Determine triangle Let the lengths of the'sides be n, n +1,n +2 where
3), (4), (5) and (6) we get
21. If the
(n-4)(2n+7n + 3) = 0
25.
A-C
or
COs
0
Therefore, the triangle is isosceles.
nn B Sin Cos
or
or
n
*-90-
cossi
-3)n+4n
-
or
cos + coscosco6
in2n
or
from (2))
(n
:or
C
2Sn
5
COS
or
0
B
2cos
17) +4) = n(-n + 2n + 17n -n*+2n+ 12= n*+n*-8n 2n-n*-25n- 12 = 0
or
C
r or
=0, ie., A =B.
A-B
or
cos C
lfroe
4x2 1- (1+ 9+x) (1+X9 +x)-4 (1+x9+x
sin A- B
cos C)
B
0.
sin
or
2
2sin
11+tanA 1+1tan'C
A +B
.cos(acos B- bcos A) =0.
sin
or
cos
2 tan+tan
Here, 2tan
X
2
or
iftantantan9are in AP,
AABC, 23. In a
(a2 + 4h3)20
or
(1+ra*-(1-75-(a2+4h3)20
or
i(l+r-1--5la-
or or
ra2-1-r3h*20
(1-
4h220
8-74
Problems Plus in iIT Mathematics
B-75 Properties of Triangle
ra
1-h.
120°
y sin
ar(AADC)=
Hence, the altitude is less than or
we get (from the question),
3 =xy
1
equal to
ar(ABCD)=4V3 =* CL
... (1)
xy =6 26. If in a triangle ABC
C
cos A cos B+sin A sin B sin "C=1,
In
AAC'B, LABC'= ZB'OC+ LACB
n eN then prove that the sides are in the ratio 1:1:v2.
Weknow that in AABC, 12 0
0
S cos A cos
B
In
from the question, 1 S cos(A
But cos(A
=
B
and
Ab
sin(C-0)sin Bsin
B).
and thenC=
C-
2AR
8=
or
from similar
C
0,
y> 0)
sin2cossin= sinsin sin sin n sin
-t:
(3)
triangles
sin(0 + C) and BC' = 21R sin(C-0) BC'=2ARisin(C +0) +sin(C-8)
4
7T T;
or
*3
n=7
4R sin C
41R sin C
regular polygon of n sides has the circumradius R and inradius r then prove that each side of the
-yt1 using (3), we get x3 y=2 or x= 2, y=3.
15
29, Let ArAy Ay .,
30. If a
polygon is equal to 2(R +) tan
An be the vertices of an n-sided
regularpolygon such that AA
b
Let AAz be a side and 0 be the centre. Let OB LA,A2.
1
AA AA4
Find the value of n.
ninn
cos
fOis the centre of the n-sided regular polygon, then
A,O42A,OA,
r
28. The
theratio of thesides =1:1:v2.
two adjacent sides of a cyclic quadrilateral are and 5 and the angle between them is 60°. If the area or the quadrilateral is 4v3, ind the renaining sides. Let the cyclic quadrilateral be ABCD in whc artABCD) =43, AB= 2, BC =5 and ZB = 60° Then D = 120°.
Let
OAOA==
LBAC
BCA =C.
=
A, ZA'BC' = LABC= B,
2r sin
Also, ZA,OA2
An
Similarly
Let AD =x, CD =y.
Now, ar(aABC)=2-5
AA=2r sin
and A1A=2r sin sin 60°=
=and
and
AB
Otanie,
AA
AABC and AA'B'C' are similar where
B'A'C'=
Clearly, OA =R and OB =r. =
LA,OB
InAA,OB,cos LA,OB=cos
cos =5
ZBC'A=
... =* OAn =r.
learly we get AA =24,M
27. In the AABC, a similar A A'B'C' is inscribed so that B'C BC. If B'C is inclined at an angle 0 with BC,
prove that
Or
13
sin
Also (r-y)=(r+y)-4xy =5-4.6=1
=2AR 2sin C cos 8 COS
or
(r+y)=13+2xy = 13 +2.6=25
cAB=AC+
.
AB
(+-2xy
or
Thus, we get
AC 2R A-B =0
(2)
+y=5
A'C
CAC B) =1
2+y+y=19
C -0
sin
sinsin-sin sin
From(1)and (2), *+y2=13
BC
B) s 1. -
=
sinsin
120°
y-2iyay
BC A'C sin(C-0) sin B
B'C
Now, cos(A - B) =1 A
-
23R. 2R.
ABAC, ZBA'C"= LA'C'B'-LA'OC
+sin A sin B = cos(A -B),
equality holding when C=;:
So, we get cos(A
sin A
sin(0+C)sin A
sinsinsin T
AC2-AD+DC-2AD DC cos
ABC
or
cos 60°
-4+25-2-2.5.-19
B'C
AC
+sin A sin B sin "CC
0+C
sin(0+C) sin A
sin "Cs1
COSA
=
BC Also, AC=AB+BC-2AB
2rsin
2rsin
2rsin
(1)
AB =r tan n
24,3 =2r tan
Now, 2(R +) tan
2
.
anfrom
(1)
(2)
B-77
Problems Plus in IIT Mathea ics
B-76
2r
1+ cos
1-cos
cos
sin
ung1-cos
sin 0
and y= rcot
a
Ina AABC, we have the identity
a Prove that the harmonic mean of the exradii of inradius. triangleis three times the
Let the
exradii be r, 2 3 and inradius=r.
equilateral 37. Show that the AABCis Here R 2
2r COS 7
in
CoS
tan=A^Az
oraB-a
Thus,
Ts are the inradius and the respectively of a AABC then prove that
OC = R, BC =a.
3
2.4sin
(6-)+(6-(6-
A sin B sin C,
36.
=4s-25ta +b +c) +a2+b*+c}
28+ sin 20)
-2
2+a+b+c
p,y:
then prove thatr=
a+B4y
wnere
r= inradius,
Let the incircle touches the side AB at P where AP = a. Let I be the incentre. Then Al bisects the BAC.
Now, r+ 34. If the exradii r, 7,
of a AABC are in H,
its sides a, b, care in AP
r
We know that
23
A
a
h
are in HP
are
in HP
arein AP from the right-angled APA,
= tan
a=rcot
s-carein AP -c are in AP
S-,
s-b,
b, a, b,
care in AP.
s-c
Now,
show
-sineo
T2
-co
and inradius =,
circumradius R. Then we have to prove that r,
bc)-RHS. 32. If the distances of the vertices of a triangle from the points of contact of the incircle with the sides be a,
.
sinsinsin
Prove that in a triangle, the sum of exradii exceeds the inradius by twice the diameter of the circumcircle. (or, Prove that 7 +2+7=r+4R.)
Let the exradii be r,
(using identity)
sinini R-
r-Brsinsin sin
Hence, the problem.
2ks= sin 2C
2A + sin
Ca
3r.
or
26sin
Vs-a)Ns-b)
ab
R
(a+b+c
LHS
and
V-6 -a)
16-as-b
S-a+Sb+5-C 3s-(a +b+c)
exradi
We know thatr=
sin 28
aXs-b\s-)
sA
-bN9.e-
24 lusing sine rule in ABOC)
Similarly, 2R
s(s
A4
1,1 Now
33. If r,ry
R =
=
s-bs-c)
13r
1
atB+7
circumradius and R, Ry R, are the circumradii of the triangles OBC, OCA and OAB respectively then
Clearly, in the AOBC, 2BOC =24, OB
we have to prove that
ay
31. If in the AABC, O is the circumcentre and R is the
2in
(given).
6-X6-bX6-)
sin
from (2).
rove that
if its circumradius
is double of the inradius
We know thatr
Sin
2
2r
Similarly, P-rcot
:
1-cos
Properties of Triangle
+72+r=r+
sinin
4R.
equality holding when
But
+3r
co-
max
B = C.
gan-ni)-ng-sin]
than
*
stls6 ) f(s-b)(s-c)
for 25=a +b+c
+c)+bc +s-as sS-a)(-b(s - c)
A.SS{b
Aa2-s(a+b+)+be
(25-2 +a=r+4R.
be)=4R:R-A
equality holding when sin A
2-
sin
sin equality holding when
From (1), equality must hold.
and
sin
(1)
Problems Plus in IIT
B-78
B
C
and sin
ie.,=30°
A
So
=
B+C
=60°
Mathematics
B-79
or
ccos A=+a> 0
or
a-csin?A >0
or
a>csin A.
a, B and y be three angles given by cos 16. If
a > Csin A
From the question cos A)b
180° - A = 120°;
b2-(2
Properties of Triangle
Pc+a
+(c-a)=0
So the triangle is equilateral.
According to the question, the values of baare b,, ;
38. In a triangle ABC, the sides
17.
=2c cos A and b, 2b, = e?-a? cos A and 2b2=c2-a2
In a AABc, if +C
18.
a+b+c
C+a
or
=
orc-
2h0
b-(2c cos A)b+(c2-a)=0.
or
AABC,
cos B+ y cos C= 4R () a cos A+B R, the the altitudes of a triangle and are p Ps If Pz radius of its circumcircle, prove that
if c*-2(a+bJe +ad+ab2+ b=0
A=VRP
C-or
thenprovethat
32.
followingholds:
2c08
Prove that cosec
A
proportional to the (b) cosines of two angles are
33. In a APQR, QR
opposite sides.
1.
for each of the admissible values of
4'sin(B-C)
that Ina AABC, prove tnatsin =0. B+sin C
2 In a AABC, show that
11.
4.
ab
Ina AABC. prove that a sin
(a
+b) sin=ccos
cot A
en prove
Ifin the AABC,
=
sin
12.
=
7. In a AABC, prove that
(b+
7
13.
In a AABC, if
the
24.
C
25
tan= 5 and tan B2then 23
p
15.
=
+B) sin(C-A) = -sin(B+2) Band Determine the measures of the angles A, cosine the that tha Can a triangle be constructed such s
ofis
angles are
B
cot
C
=v3 then prove
f in a
tan,
AABC,
then show that the
In a AABC, prove that
and
cosec
7
reason Give
26. 1n
B
2
cosec
2C in are
cot
+cot
C = cot
o+ cot
C) = csin(A
-
B),
prove that
B.
angle A of AABC meets BC atD then prove that
38. The bisector of the = m
bc
39. If
(i) m-
2bc cos b
the median AD of AABC be perpendicular to AB
thenprove that tan A +2tan +2cos B+ cos C=2, prove that
abcare
length of the side BC. on In a 4ABC, 2C =60°, LA = 75°. If D is a point AC such that the area of ABAD is v3 times the area of ABCD, find LABD. 37. AD is a median of AABC and LBAD = 0, 2CAD= ¢ then prove that
If AD
unun 27 in a AABC, if cos A a,b, care in AP. 28. In a AABC, if asin(B -
and it
the median AD=
divides the angle A into angles 30° and 45°. Find the
AP.
aAABC, if sin A, sin B, sin C are in AP, show that
in AP
12,
36.
a",b,c*are in AP. cotA,cot B, cot C are in AP in HP, prove that c a b, are a, AABC, sides if the 25. In cosec
=
and point on PQ such that MP : MQ =^:1(a>1) Find the line RM divides the 0
In a
19.
a
tan
tan
a+b+c
31.
This is a quadratic equation in b. So, bwill have two values if D
tan
sides a, b, c are in AP are in HP hen prove that the the a AABC from of altitudes 30. If a, B,y are the prove that (where respectively, C B vertices A, and Ris the circumradius)
en prove that
a=+8sinA. We have, cos A
triangle ABC, the quantities
triangle are in AP and its area is th The sides of a triangle of the same perimeter. an equilateral of are in the ratio 3:5:7. Prove that the sides
2c
of which given. If the side b has two possible values one is double the other then prove that
tntn
tantantanž=tan
has two real values.
b+2b,
b+c
and cosy= +b where a, b, c are the then prove that AABC a sides of
sB CoS
because
but B = C; so B = C= 60°. Hence A= B=C= 60°.
are a, c and the angle A
29. If in a
B = 0.
40. In a AABC, the lengths of the bisectors of the anglees A, B and C are x, y andz respectively. Show that
B-80
Problems Plus in
IT Mathematics B-8T
41.
Properties of Triangle
In a AABC, the median AD and the altitude AM divides the angle A in three equal parts. Show that
cossin2
R
53.
where A = area of the triangle. 42. Acyclic quadrilateral ABCD of
3N3
areais inscribed
(4-V10+2V5)42.
quadrilateral ABCD, if AB = 10, 2DAB =75°; 2CBA = 90°, 2CAB=45° and ZDBA =30° then findCD. 46. Prove that the sum of the radii of the circles, inscribed in and circumscribed about a regular polygon of n sides, is a is the length
cotwhere
of a side of the polygon. 47. The ratio of the area of the regular polygon ofn sides circumscribed about a circle to the area of the
regular polygon of equal number of sides inscribed in the circle is 4:3. Find the value ofn.
L441=1,1,1 55.
1B-IC=abctan LA
sides are equal.
a of the circle Prove that area of the polygon tan49. Three points A', B' and C' are taken on the plane
nn
of a AABC such that AA'BC, AABC and AABC are equilateral
with their circumradii R,
r
T,r
Rg, R; inradii respectively.Prove that
R): (rrr)=1:8:27.
', are exradii and r Is the inradius of a triangle then prove that
50. If
tan
60. If in a
AABC1
two angles of a triangle are 30° and 45° and the included side is (V3 + 1) cm then the area of the
82.
triangle of side Acircle is inscribed in an equilateral the circle is The area of any square inscribed in
4.
83. In a AABC, a
:b:c=4:5:6. The ratio of the radius
incircle is of the circumcircle to that of the
then
A =
A, B
and
C
)3
B =
are in AP and
-B)then
85. Ina AABC,
(b-c)*+ 4bc sin
++)tan2A=4a2
m4 m 1PrOve
72.
If in a AABC; 2tan
73.
In any AABC,
that there are
values ofc, one of which is (m-1) times tneother
Ifin AABC,a:b:c=8:10:12 a ihen the greatest
and the lea angle in terms offa is
a=0 then the value of
c
= 87. In a AABC, B
AD =h. Then
and
(a) 1:2
and the altitude
(b) 2:1 (d) none of these
(c)1 4 a cos 8. If in &ABC,a
cos2=bthenthesides
(a)AP
(b) GP
(c)HP
(d) none of these
B
tan
89. If in a AABC, e(a +b) 76.
C=
h:a is equal to
trigonometrical ratios of the angles, is
If in the AABC, cos A
=and
cos C-
then
4--then e:a=,
In a AABC, if (a +b +c(b+ c-a)= 3bc then 90.
B c-b na AABC,if tan A-tan A + tan B c
tan
tnen
a
AABC,
4oS
COS B
ZCOS
a
A=
a AABC, AD is the altitude from A. Given b> c
cC-23
and AD
-then
2B-.
then the
In a AABC, if a* +b°+c*= 2e"(a+b) then the measure of C is
(b) ()3
A= in
cos=be+a) cos
triangle is (a) isosceles (b) right angled (c) isosceles or right angled (d) none of these
a:b:c
On angle s
abx2-cx + ab =0 (b) abx3-(a2+bx+ ab =0 (c)c*-abx+ c2=0 (d) ax bx+a=0
a, b, c are in
If in a AABC, c= 2a then tan
are the
(a)
sin A +sin B+ sin'C in terms of products of
then
Fill in the blanks. 64.
75.
a AABC, |b
B
roots of the equation
sin C
oL
Objective Questions
In a AABC, C=90°. Then tan A and tan
S
A= two
86.
cotthen b: c=_
a'sin(BC b sin(C-A) sin B sinA (A-B) PGc2sin +
78.
that
circumircum-
The smallest angle of the triangle, whose sides are 7,413and V13, is
7. If in the ABC,
le-o=2. Na-bsin A :63. In a AABC;a , prove that
=
B
or
centre. The angle of elevation of the aeroplane whet nearest to the man, is a and when farthest it isp"
(1)
(3)
, i.e,sin
from 0-
sin(a
o.
or
p
Again, from APOC, if LPOC
3. Aman standing at a distance d from a tower of height h finds that an aeroplane is describing a horizonlal circle of radius r with the top of the tower at 1s
The corner of the top of the tower nearest toPis E and F is one of the outer corners.
22BOR =2(180°20) 2T-40. 40 (formula) we get 2n-40 =
.(3)
0
cross-multiplication we get from (3) and (4),
sina
-212a + V404
sin e=aS
0
sin From -
x2a
- V2a +V2. V5a =av2(V5-1).
The base of a vertical tower 1s a square. A man on a diagonal (produced) of the square is at a distance 2a from the tower. He observes that the angle of elevation of each of the two outer corners of the top of the tower is 30° while that of the nearest corner is 45. Prove that the breadth of the tower
Let the breadth of the tower = BC =x.
(2)
24. A
Vhcota -d
av2(vs-1).
=
(d + r)sin ß=hcos
+22ax-8a2=0
-22a+
-2Vhcota-d-d tan B-h) =
2
(d-r)sina
(2)
or
135°+(20)-(2av3)
=-h
POB=20 radian. Then, clearly hv3
... (1)
=hcos a
-22at N8a-4:-84 )
B
-
180-45 135
-824ar
r
tan
Also, in AAEP,
==
B=
rsin a-dsin a +hcos a
or
BC=x, CP=24, BP =2av3
and BCP
"a-d2
BP =2av3.
3
BP
hcot a
have, tan a
2a= FB.
:From the right-angled AFBP,
the right-angled ABFP PE tan a
:in
Now, from the right-angled AECP
Let
cos=t;
then v3(2t- 1) =t
2v3-t-v3=0 1+43,4N3 2131+5
Problems Plus in IIT
8--100
Mathematics Heights and Distances
flagstaff is mounted in the middle a Squ. based vertical tower.A man standino on the midway in front of one side of the t ower, groun ound finds u elevation of the top of the tower to be hthe A p flagstaff to be from a distance d. hat of the ter getling a tower by the distance to a, he finds loser thatthe top of the flagstaft just disappears. Prove thathe B sin(B 4sin height of the flagstaff is a) cosa sin(0 B) where (d-a)tan 8=dtan a.
26. A
r
10
Also,
h3
radius
cos
3
h=r=
60
of the pond = height of the pole
'
&0
Here, QR is the height of the tower and the ao. PM =h (say). Clearly, from the figure, MN = QR.
25. A circular plate touches a vertical wall. The plate is fixed horizontally at a certain height above the ground. A lighted candle of a length equal to half of the height of the plate above the ground, stands at the centre of the plate. Prove that the breadth of the
A
Clearly from the figure, the required breadth of the shadow is the length of the chord PQ of the circle whose centre is O and radius = OD. Now, from similar triangles ACB and AOD, we get
OD =3a.
Clearly, the distance of the chord PQ from the centre O is OR = radius of the plate = a.
aA The man is at A initially and at disappearS.
B
when the
QR QR tan 6RB d-a
dl
QR
From (1) and (2), tan 0
-
tan
PQ=2. RQ=2. Hence, the problem.
are inclined to of equal heights a their noon the breadths of
sun, are
BC-b, QR =c. When the 1espectively along the breadths are BD and QS cross-section of the shadows.
P
At that time the second man is atB where 2BOA =B (from the question). Let the height of B BD =x. =
OB = OA
hcosec a
¢
DBC=
from
and 0
if>0,
c
Coordinates and Straight Lines
of the pentagon ABCDE bewhere
A=1,3), B=(-2,5), C=(3,-1), D =02)* E-(2, t).
Being equilateral triangle, the angle between (1) and (2) will be 60° tan 60°=
1+m(-1
or
S.ABCisa variable triangle with the fixed vertex C(1,2) (cos t, sin nd A B having the coordinatesparameter. Find oinf-cos t)respectively where tis a thelocus of the centroid of the
AABC.
LetGbethe centroid in any sition.
2
or
(by angle formula)
N3=
m+1
or
)I
=
V3(1 -m),
-V3(1 m)
(1+N3ym =v3-1 and (1-v3)m = - v3-1
3-1 33-1 1 N3
C-S
. 3-1)2, 3 2
2-v3, the
1)2
equation of BD is y 2x-4.
the
The distance of
As ABCD is a rectangle,
2
PB-PDAC-6-10-
equations of the other two sides are -2)
andy-3 =(2+ v3)x -2) Now, the slope of DB is tan 0 =2. Points on DB at distances v5 from P(3, 2) have h he coordinates
3-2N3 (3-
3
v3)x
v5cos 0, 2t v5sin 0),
the
y=2-x=2-3-1
ie,
3-1
C=[Y2 v3 C=|3-1 v3-1
coordinates of the other vertices are (3 + 1,2+2), (4, 4) and (2, 0).
equations of two sides of a square are 3x +4y-5=0 and 3r+4y-15 0. The line along the third side has a point (6, 5) on it. Find the equations of this and the remaining side of the square.
8. The
each side= AB or AC
Clearly 3x+4y
-5 =0
3X+4y-5=0
-64-23 3-1 6V3-17-
(2)
5T7a
and
4x-3y= 19.
or
whether P lies between = AD = PA + PD may
lead to wrong result lying in the first quadrant, are 9. Two sides of a thombus, length of 0 and 12r- 5y =0. If the given by 3x4 12, find the equations of the the longer diagonal is other two sides of rhombus. the first quadrant and their
As the sides are
vertex O(0, 0) is acute. So the longer diagonal passes through O0, 0.
(5, 1). The middle point
l ot AC
15
whose equation is y = 2x +C.
C(5,1)
y 2x+C
A.
It passes through (6, 5). So 4 x 6-3x5
;:
equation of AD is 4x 3y =9. As BC | AD, the equation of BC is 4x-3y=
the
Let PQ1 BC.
ThenPO= 4.6-3.5-2
distance between the parallel lines
ABandDC
Now, any point on AB 0. is
=7
>»
6130
a
11
theequation of CB, from
(2), is
18 130 5
3r-4y-
condition for the quadrilateral to be concyclic whose consecutive sides are
10. Find the
ax+by +c=0;r=1,2,3, 4. equationss Let L,a,x + b,y + c, =0 be the sides and the L =0, L2 = 0, L = 0, of AB, BC, CD, DA be respectively 3x-4y
= 0
L4
0
=0.
Let 2ABC
=
0; then LADC = T-6.
the slope of OC is
and
La-0
0
side AD is on a line through Pl6, the perpendicular to 3x + 4y 15 0. Any line perpendicular to 3x+ 4y 15 0 has u :
12
equation of AB, from (1), is 12x -5y
the
the angle at the
and tan
The slope of OA is tan
equation 4x- 3y =
a130 65
in
inclinations are tan
00)0
3x+ 4y-15
or5+9)-12
B
v6.
3,2. It lies on the other diagonal BD
A(1,3)
But OB =12;
Letthe length of each side be a.
points (1, 3) and (5, 1) are two opposite vertices of a rectangle. The other two vertices lie on the line Find c and the remaining vertices. y 2r+c.
ie., c--4.
BC is equationi of the remaining side
the 4x-3y-1
5
B-
19
910-i
P(6,5)
7. The
2-2x3+c,
9--+10
2
V3-V24-12v3 124-123
=
are wo
Let
a)-213-3
A = (1, 3) and C
and 3+4y 15 0
these sides parallel lines, their 'm' being equal to be AB and DC in the square ABCD and the point P(6, 5) be on the side AD.
-N
ie,
.
3r-4y14
99a Solving (1) and (2), y=65
19-Al=10
Note As ithas not been verified PQ A and D or not, use of
ie3525
Let
equation of CB is
the
-
2-x=(2-v3)x -1+213
B or
2
2or
erce
1 +2v3 and y = (2 +v3)x or -1 -23. Solving one of these, say, y= (2- 3)x-1 +213 with x+ y =2 we get
or
this point from DC
3x0-4
V3
2+
y-3-2-9x y=2-V3)x
or
Coordinates and Straight Lines
Problems Plus in lIT Malhematics
tan
A(0+acos Similaly, C
L20
0,0+asin 6)=
0+ acos , 0+ asin )=1
0
12a
13
NOw, ABis a line parallel to OC and passing through
tan 0= But tan
2
|1+mm2
theequation of AB is 12r-
125 the
5y =k where
and
ork
when 35
(1)
where
.Mef
equation of AB is 12r -5y=
e equation of CB is 3x-4y
n(t-0)-1mm 1+mm2|1+
mm
0 is acute,
T
is obtuse)
C-10 Problems Plus in IIT Mathematics
-11
Coordinates and Straight Lines
55425 cos(-a)=14
or or
A(0-3)
where cos d or
by4-a,ba
or
a,ba2b
or
Gba)an,
b-abe +
a142 + bba
ab4ab
a,42 + b,ba
+ cos
ag4,+ bsb t
bbi)
+
(agh-abaa,
*
y-2
11. Find the
equation of the line passing through the point P(1, 2) cutting the lines x+y-5=0 and 2x-y=7 at A and B respectively such that the harmonic mean of PA and PB is 10. Let the equation of the line passing through P(1, 2) be y-2
Let PA
m(x-1) PB
=r
.(1)
tan cos1
+cosiA
V146
-1).
and orthocentre of the triangle formed by the lines 3x-2y=6,3x+4y+12=0 and 3 -8y + 12 =0.
B= (1+r2cos 6,2+2sin 0), where tan 6 =m.
3x-8y +12 0
a
(2)
e
3r+4y+12=0a |2x-y-7
1+cos +2+7sin
cos 0+ sin 0)
=
.. )
6-5 0
2(1+r2cos 0)-(2+r,sin
3x-8y+
Tz(2cos -sin
12cos 2
-
6)
7 =0 (3)
r
2cos
-
0
C
C=(4,0) B
=4;
Solving (1), (3) we get y =-3,x = 0; To find
Now, HM of PA, PB is 10
or
12
Solving (1), (2) we gety =3, x
)=7
102PAPA +PB PB
A
- sin
0)+ 35(cos
10(2cos
or
55cos 0+25sin
)
+
8a +
the centroid
'
(1)
a-4
mof AC
we get, BH I AC or
2
4--4
a-4
or
3(B -3);
8a +30B +3)
4a- 3ß 7
t2,
sn21t *h*1s 3
ie, ie,
=
0,i.e.,
8a +3ß =-9
or
6y -9
=
-16x
16x+6y-9=0 Again, E midpoint of AC
.
and 3x + 4y +12 0. 3x-2y-6=0 Any line through A is a line passing through the intersection of the above lines. So the equation of the altitude through A is
(iii)
3x-2y-6+A(3x
+4y+ 12) = 0
Being altitude, it is perpendicular to BC whose slope 8
[from (2), (3)1
sin )=14
So,theslope of the equ side Ac
ACis
To find the circumcentre
ie
Then MA
=
= MB MC.
g
The slope of a)
is-
by condition of perpendicularity,
ME is
on of the perpendicular bisector ME of the
0 = 14
Let M(a, p) be the circumcentre.
..(
Alternatioely The equations of AB, AC are respectively
or
y1), etc:
...()
orthocentre is
the
equation of the perpendicular bisector MD of the the side BCis
iheslopeof ACis
-1
-1
Again AH 1 BC or
-
Solving () and (I) we get a=
is
0
4(a4)
=
0--4)
ofBC
of AH
(ii)
D-midpoint of BC
2
'm of BH=
16
and (i) we get a=
recumcentre
(4,3) =(0,3)
where vertices are (x1
(cos 0+ sin 8)1,
or
8a+12p-16 =0, i.e., 2a + 3B-4 0
8a-6B+7=0
=
We know that centroid G
102
sin
6 +9
or
So,the slope of MD
Solving (2), (3) we get y=0, x= -4;
sin 6
Now,
and
Solving
C-4,0)
B(4,3)
+16 and
6+9=-8a-68+25
Theslope of BCis
.(2)
As B is on the line 2x-y-7=0,
or
6B +9
68 +9=-8a 6B+25 8a
or
Now,
2
-
A(0-3)
Allernatively Take the perpendicular bisectors of the sides BC, CAwhich are respectively MD and ME.
0
cos +sin
AH L BC.
=a+p2-8a -6+25 =a+p+8a +16
respectively.
As Ais on the line x +y-5 =0,
or
a
12. Find the coordinates of the centroid, circumcentre
3x-2y-6=0e
P(12) x*y-5 0/
(a-0)+(+3)2 (a-4)+(-3) or
A=(1 +71cos 6,2+r^sin 6)
Tofind the orthocentre Let Ha, B) be the orthocentre. Then BHL AC
= (a +4)+(B-0)2
Let the sides AB, BC and CA have the equations
Then
=r
5V146
8
circumcentre-12
the (-4,0)
B(4,3)
5146cos 14+cos
iv
0
Solving ii) and (iv) we get x
55-23
fom (1), the equation of the line is
bê)=0.
+7
8x-6y
cos(-a)5146
or
|a44 bab|
=
or + 2), ie, 6y +9 -8(x + 2)
or
3(3+3A) =8(4-2) = 9+9N 32A- 16
(a)
C12
or
Problems Plus in IIT Mathematics
23
the
.
25;
L
equation of the altitude through A is
3x-2y-6+3x4y+12) =0 or
69x-46y-
or
144xr +
138+75x + 100y +300 0
Itis to
or 8x + 3y +9 = 0 (6) Again, any line through the intersection of BC and BA is
k Sin
BC whose
,.
e, + Acos
cos 0(cos
or
cos(-0)+Acos(-0)=0
Puing in (1), the equation of the altitude through becomes or
9+9
the
equation of the altitude through B is 3x-8y+12 +23(3x 2y-6) =0
or
=
8
=
4(21 +8)
=
72r-54y-126
cos(-8
s,-,0
23
+32;
m' ofGO=
Cos
or 4x-3y -7 0 Solving (6) and (c) we get the coordinates of
orthoentre. We get x
=y
and
ortboemtre-(
L,=Xcos 6,+ ysin 6,-P=0;r=1,2,3. Show that the orthocentre of the triangle is given by = Lcos(® -9,)= Lgcos(@,-0 Lcos-0,) =
0, L=0,
L
B
= 0.
L1-0 Any line passing through A, ie, the intersection of the lines L2 = 0, Ly = 0 is
COS
Cos
d2t
COS
altitudes,
and (4). So the orthocentre is given by
L,(0, 0)0
dg
(0,0) and (-1,-4) are on the opposite sides of
a, + sin
L 3x+2y +6=
l23x+2y+6 0
a+
Here, the circumcentre O =(0, 0). So OA= OB
EX-2y-2
NOw
L0,3)-1-6 2
tan y;
slope of the refracted ray =
the
is N5.
53 6 3
21. Find the
y+
8+5a
11
3(5-2N3)x313y +13v3 - 50
:or
23.
0.
ExercisesS
1. Ifa vertex of a
triangle be (1, 1) and the middle points of two sides through it be (-2, 3) and (5, 2) then find the centroid and the incentre of the triangle. 2. If two vertices of a parallelogram are (3, 2), (-1,1) and the diagonals cut at (2, -5), find the other vertices of the parallelogram. triangle has its orthocentre at (1, 1) and
3. If a
circumcentre at
11.
,
Prove that the area of the triangle whose vertices bsin 8,);r=1,2,3 is equal to (acos
2ab
sin--
triangle. does not cut the line segmentjoining the points (-2,-3) and (1,-2). intermal bisector of the the and a triangle 5. ABC is angle A cuts the side BC at D. If A= (3,-5) AAD. B=(-3,3), C= (-1,-2) then find the length of
4. Prove that the line 2x-3y +6
6. Find the point P if Q
The coordinates of the points .A, B and C respectively (6, 3), (-3, 5) and (4, -2) and coordinates of P are (x, y), prove that ar(APBC): ar(AABC) = | x+y-2|:7.
=
0
(-2, 4) is the point on the line
segment OP such that OQ
=
OP,
triple its length, then find the new position of B. are A(1,2 8. The coordinates of the extremities of a rod and B(3, 4). S(0, 0) is a point source of light. The rod AB is parallel to the wall and is midway between the point source and the wall. CD is the shadow of AB on the wall. S, AB and CD are in the same horizontal plane. Find the ends C, D of the shadow. 9. Find the circumcentre of the triangle whose vertices 4) are (2, 1), (5, 2) and (3, 10. Find the centroid and the incentre of the triangle whose vertices are (2, 4), (6, 4) and (2,0).
the
sin
-)sin
are
@-
y
ae-1
15. Find the locus of the point at which the line segment joining the given points (o, B) and (7, 8) subtends a
right angle 16.
LinesL
make an angle 8 with intersect at the point P and a line different from each other. Find the equation of P and makes the same L2 which passes through
A rod of lengthI cm slides on two mutualy perpendicular lines. Find the locus of the middle point
position of a moving point in the r-y plane at time t is given by (ucos a t,1usin oa t-pt) where 1, a, p are constants. Find the equation of the locu of the moving point. 18. The straight line lis perpendicular to the Ine 5x -y =1. The area of the A formed by the line and the coordirnate axes is 5. Find the equation of tne line. 17. The
l
= 5 and 4x-3y 15 intersect at 24. Straight lines 3r+ 4y lines such that A. Points B and Care chosen on these possible equations of the AB AC. Determine the line BCpassing through the point (1, 2). bisectors of the 25. The equations of perpendicular y + 5 =0 and sidesAB and AC of a AABC are (1, -2), find +2y 0 respectively. If the vertex A is the equation of the line BC. BC and CA of a 26. The equations of the sides AB, AABC are respectively x +y =1, x-2y =3 and the foot of the perpendicular from A 2x-y 0.1 to the side BC, and DE is drawn parallel to BA cutting AC at E. Find the point E. 27. The coordinates of the foot of the perpendicular drawn from the point (x1, y1) on the line + my + n =0 are (h, k). Prove that
x-
Hence or otherwise prove that no three distint points of the above type can be collinear 14. Show that the equation of the locus ofa point, whih moves so that the difference of its distances tirom two given points (ne, 0) and (-ae, 0) is equal to 24. is
O being the origin
7. One end of a thin straight elastic string is fixed at A(4, -1) and the other end B is at (1,2) in the unstretched condition. If the string is stretched to
are
triangle have integral coordinates prove that the triangle cannot be equilateral.
13.
10.
angle 8 with L.
12. If the vertices of a
then find the centroid of the
=
between the lines is such that its segment 22 A line bisected at the 5r-y+4= 0 and 3x + 4y-4 =0 is equation. point (1, 5). Obtain its = =ax + by+c=0 and L = lx + my +n 0
the equation of the refracted rayis
7+5V3)(19 +53)208+1303 286 192-(5V3)
lie a unit points on the line r+y=4 that at
2distance from the line 4r +3y
lr
zh-k
m
Iatmyt+ 12+ m
28. Equations of two straight lines are Xcos a+ ysin a = p and xcos ß+ ysin ß = p'. Show that the area of the quadrilateral formed by the two lines and the perpendiculars drawn from the origin to the lines is
2sin(B-2Pp-(p*+p)cos(a- B)]. 29. The sides AB, BC, CD and DA of a quadrilateral = have the equations x+2y 3, x=1, x-3y =4 and respectively. Find the angle between thediagonals AC and BD. 30. Let P, Q and R be three collinear points on BC, CA and AB of AABC. Show that BP CQ AR+ PC QA RB =0, Proper sign for extermal section being taken into account.
5x+y+12 =0
AC: If D is the triangle with AB = perpendicular of the midpoint of BC, E the foot the midpoint of DE, F drawn from D to AC and prove that AF L BE. a and b on the Let the line 1, have intercepts passes through the coordinate axes. Another line 2 on, between the axes. middle point of the intercept the axes respectively then IFh cuts intercepts p, g on
31. Let ABC be a
19.
80+5030
be y then tan
1536-1 13
553+88-8-53
11-
and the line through
3-
A(10/3,-7/3) with the slope
tan y=
8+53 5V3+8-
As
6
Coordinates and Straight Lines
32.
prove that ag + bp = 2p. = 0, 5x -y =6 and Show that the lines 4x+y-9 intercepts on any line of x-2y +3 =0 make equal gradient 2. lines are 7x +y= 16, 34. The equations of three = 0. Any line through 5x-y-8=0 and x -5y +8 the lines at P, Q, R the point A= (c+ 1, 2c) meets harmonic mean respectively. Prove that AP is the
33.
between AQ and AR. 3 off equal that the lines y = m,x;r= 1,2, cut if 1+ m intercepts on the transversal x+y=1 l+mi2 1 + ma are in HP. cuts the lines 36. A straight line through A(-2,-3) B C respectively X+3y 9 and x +y +1 = 0 at and Find the equation of the line if AB. AC=20. meets the lines 37. A line through the point A(-5,-4) = 0 at the x+3y +2 0, 2x +y +4 = 0 and x-y -5 IF points B,C and D respectively.
35. Prove
find the equation of the line.
In the AABC, the equations of AB and AC are 2x+3y 29 and x +2y 16.If the median through A cuts BC at (5, 6), find the equation of the side BC. 39. Find points equidistant from the lines 4x +3y 0 and 3x-4y =3, the points being at a distance 2 from the point of intersection of the given lines.
38.
equations of the lines which pass through (4,5) and make equal angles with the lines 5y 12x+6 and 3r = 4y +7. 41. Find the equations of the bisectors of the angles between the lines x+ y-3 =0 and 7x -y+5 = 0, and indicate which of them bisects the acute angle between the lines. 42. The bisectors of the angle between the lines y= v3x +3 and v3y =r+313 meet the x-axis at P and Q. Find the length PQ. 43. Prove that the bisector of the angle containing the origin is also the bisector of the acute angle between the lines. a^x+ biy + C =0 and ar + b,y +c2=0 provided C, C2 are of the same sign and a142 +bba 0
The equation of a circle whose centre is the origin and radius is a, is
4. Location of a circle
The general equation of a circleis =
be two circles. Let D be the discriminant for the quadratic equation in x (or y) obtained by eliminating y (or x) from the two equations of the circle. Then they are two intersecting circles if D >0 they are nonintersecting (no common points) if D 0
a point circle (i.e., a circle of zero radius) if
no real circle if g+f0 where D is the discriminant of the quadraticif C43
44
Problems Plus in lIT Mathematics
x (or y) obtained by eliminating y (or x)
tonin from L 0, S
0;
=
equivalently, the length of the perpendicular from the centre to the line is less than the radius. The line touches the circle at a point if D=0;equivalently, the length of the perpendicular from the centre to the line is equal to the radius. The line does not cut or touch the circle if D0;
+2a>0
or
-8
or
a4
(a+2)(a-2)>0 (2)
by
-*-
.Utv2a1-2
aE
sign-scheme for quadratic expression, we get
,
by 2x(x-a) + 2y-b)=0, a #0, b#0. Find the condition on a and bif two chords, each bisected by the x-axis, can be drawn to the circle from
where
a
2(01+ 1)
the point
2
centre=
2tr+y)-2ax-
Mis the middle point of PA, CM L PA
or
2a 4
Or
+
Qm
+
1+2a=0
by =0
x*+yax-5y=0
Its centre=
V20t
ine
a2
Here, the equation of the circle is
1137g
and. the Foint
a>2.
9. Let a circle be given
2(m+ 1)
M=(a,-a)
AS
0r
(-,-2) u (2, +«).
2a (+m)r
2
Now, C
must be on the same side of t
*-y+1=0. .(9)
-
point of interscction of (1) and the line
0
or
u,
For two chords like PA, the equation must have two real and uncqual roots. So D>0
ac-2
(5a-9)(a+1) 0.
ly 0
Here the circles
Any line through ,has
the equation
y-m-) y =0
Asatisfes(1) Thecentre
willbe
perpendicular to the line (2).
the
through the
m or
bm+2am -(4am-2b) bm2-2am +2b 0
radius = v1+2-(-20) =5. So, the required circle of radius 5 must touch the given circle externally
( A* )
y=tv-e
centre-0of the circle 4)
will cut the
Y5.5) (a,p)
(1.2)
Let the centre of the required circle be (a, B).
circle (2) at real points.
5
Now, any chord through
required circle is
and
5
a-9,B=8
ym
( +3)3
0
For the given circle, centre = (1,2) and
of intersection will be imaginary and hence the circles will be nonintersecting be within the other or fully will circle Hence, one outside the other. If (1) is within (2) then any chord
- -
theline
is 5 and the equation of the circle whose radius 0 at which touches the circle x+y*-2x-4y-20 the point (5, 5).
13. Find
ifc>0, the points
a-r= distance between the centres
20
circle are obtained by
u0.
c0,
(2)
ux+c=0 of the
x=0
y+c=0
If the required circle has the radius a then
is
and the centre
(1)
Puttingx =0 in (1),
lies on the ircle.
at(m:0
+y x=0;
(1)-(2)
at(a-2m0
As the chord along (2) is bisected
joining
are:+ y*+ Ax +c=0
The points of intersection solving (1) and (2).
.2
(2) cuts the I-ais, ie,
Hence, one circle is within the other if
+y+x
for the circle, centre = (9, 8) and radius
=
5.
.
the equation of the required circle is
or or
or
=
0
.
roots,i.e., 4a-4 b-2b >0 or a-2b>0
equation of the circle having the pair of lines its normals and having the size just sufficient to contain the circle
x+27y+3x+6y = 0 as
two normals of the circle are x+3 =0, x+2y=0.
their
lis
or
A-B
4-
H
+MAx2Hxy +By2+2Gx +2Fy+C)=0 :
the centre of the circle.
The given circle is r*+y*-4x-3y0
(1tm)*+amtHx+
...
(1)
or
m
+2m
a-ma(1
or
xy=0.
a+A =b+AB and 2h+2 2H=0
a-b=(B-A) and"hleRH 1 -AH
4-8 H
2x3-5-1)+1829 radius va2+(29)
+m
=
29
V38.
4c(1+m)>0 This is truefor all real m.
and
The length of the perpendicular from the to the chord
am)-4+m)
coefficient of r= coefficient of y
.
+C=0.
Both roots must be real and unequal. So D >0
As the points of intersection of the two curves concyclic, (1) must be a circle
coefficient of
+2y) =0
intersection|-3,
+ 2gx+ 2fy +c=0 and Ax+2Hxy+By+2Gx +2Fy+c=0 intersect at four
for some
xr-4)+yy-3)=0. Here the pair is x(r + 3)+2y(x +3)=0
the equation of the circle whose centre is (3,-1 and which cuts off a chord of length 6 on the line 2-5y +18 =0.
*uT+C=0
If the curves ar+2hxy+by
ax+2hay+by?+2gx +2fy+c
10. Find the
(r+3)r
Solving (2) and (3),
Any second degree curve passing throug intersections of the given curves is
or a 2b2
V2b
++6x-3y
14. Find 11.
(3)
Two chords are possible if (3) has two real and unequal
or
=5* -9)-8)* 120 =0. 16y+ or+y-18x
225 +6x+9+-3y+44
m
(2u-12-4c)+2-4c>0
Also, the circle (2)
Tadius
2x-5y+18 -0
is real if
-/2-c
Hence, (4) is true for
is real, i.e., 4c20.
allrealm if2A4 1-4c>0 or 23>3+4c; butc>0 44cs0 Au0.
.. (4)
the
required equation of the circle is
(-)1i)= pr
Or
(133)
2-6x+9+y?+2y+1= 38
+y-6x+2y-28 0.
ntre (3, -1)
c53 C-52
Circles
Problems Plus in UT Mathematics
Circle touches the line y =xat a point P such that OP=4V2 where O is the origin. The irdle contains the point (-10, 2) in its interior and the length of its chord on the line x + y=0 is 6v2. Find the equation of the circle.
A
Clearly y =x and x +y=0 are two perpendicular lines passing through the origin O. Let P (k, k) as it is on the line y =x. =4V2
OP=VR+R t2k=4v2 k=t4
a+B+8=0 and a-ß=t
or
2a8t a=1,-9 and 9,1 B)=(1,-9)
If points
or
or or
(-9,1).
the distance
betwee. (-10,2)and (1,-9) is vi1+11> 5v2
So(-10,2)
So (-10,2) will be an intenor
the distance
vi+1
52. So (-10,2) will not be an interior point.
1+(-1)7
V(4v2)+ (3v2)
=
anu
X-
N2
2,Y
=V16+7-2.16-4.7-20
15xV1+2-(-20)
the new coordinates the point (-4v2, in be an interior point e,
-6V)
required circle willbe in the new third
uadrant
the centre=(42s2) and radius= 5v2. tne equation of the circle, in the new coordinates
(X+412)+ (Y+5V2)=(52)
CQ2={{acos
8 -asin COs 0
(asin -a)
-a2+a2-2a(a -asin 6)- 2a sin
+
+a)
0
s |*a+a2-20(a
+ asin 6)
+2asin +a2+sin01
BC
=a22-2(1- sin 0)-2sin 75.
and BQ are fixed parallel tangents to a circle, and tangent at any. point C cuts them at P and Q respectively, Show that CP CQis independent of the position of Con the circle and POQ is a right angle. Let the parallel tangents touch the circle at M andN ESpectively. Then MN is a diameter through the centre O. 17, AP
61 +(asin
AB
15x radius
2--62
and
=225 15. So, ar(ABCD) =2 ar(AACB) =2
5/2.
Also, when x=-10, y=2 we get
the =
=VS(16,7)
Then we have to find a circle touching X-axis cutting off 6v2 from Y-axis.
B-8-a=-1,9. B) =
and
=/sin
asin 0)
*acos -4tasin cos
to the circle
So, take the following equations of transformation:
,
4x7+2016 3
Now, BC length of the tangent from C(16, 7)
radiuoQP or 18(ap-zta-p* 188-^a-p), using ()
asin 66 x t Cos 0
and
C=(acos
2
3)
d,
a-asin 6 y=a and x=- cos
cos
to the chord
:ovy
100;
acos 6 +y asin 0=a2
y=-a
C (16,7)
(a-P)=
I
p-(a-asin 8,.
a-4-1=1
(2)
asin 0).
or rcos 6 + ysin 6 =a Solving (2) and (3),
(2)
3x-4y-20 = 0
,
The tangent to the circle at Cis
are tvo
y=*
y-0
Hence, (a,
20 =0,
0
a +B 8 r The length of perpendicular from Q (a,
From (1)
-
y = ta and the parallel tangernts as (acos is circle the Any point Con
A
Taking P=(4, 4),
or
. (1)
y-7=0
ie, 5y-35=0
as the y-axis
+y=a2
circle whose equation is be the centre of the
Let the centre Q of the circle = (, P).
QP LOP
and MN Taking the centre O as the origin the circle as we can take the equation of
=0.
tangents at the points B(1, 7) andd Suppose that the the point C Find the D4,-2) on the circle meet at ABCD. area of the quadrilateral tangents at B, D are respectively The equation of the x1+y:7-(r+1)-2y +7)-20 0
and
and
10)* = 100
+y+18x-2y+ 32
16. Let A
(r+9)+-1)2= (5v2)2 Second Method Here x+y=0 perpendicular lines.
(+y+8)+ (x-y+
2x+2y+36x-4y+ 64 = 0
r+y-2x-4y-20=0.
pont
the required circle has centre (-9,1), the equation of the required circle is
o(ap)
th
will not be an interior point.
If the centre = (-9,1) then points(-10,2) and (-9, 1) is
(-10,2)
100
10
the centre =(1,-9) then
e.ra
the equation
Taking P=(4,4) as before, we get
(a,
or (4,4).
P=(4,4)
tance bet between If the centre = (l,)tnen the distance th 5V2. points (-10, 2) and (1,9) is Y9+7> not interior an be point. will So-10,2)
0+e
xa2-2(1+sin@)+2sin
&1+sin 02
asin cos
cos
"0)2
a.sin 0 cos
CP2=a*=constant.
*cos
Froviems Plus in lit MathematicS
C55
Circles
Also'm' of PO =m
a-0
-asin
1-sin
cos
'm
of QO= m2*
-0
or As (
=--OS
a+asin
1+sin 9
6 cos sin
m21-sin 01+
hence
0)
65-4
a point on the line 4x-3y=6 tangents are drawn to the circle x*+y-6x-4y +4 = 0 which
or
9x+4x-6)54x
or
25x-150x=0
or
z*-6)=0
between them. Find the
Let the angle between the tangents be 0.
4x
c
radius=CA =v3+2-4=3
+y-6x-4/1+4
tan 2+yi
or or
x(7x-24y-48)=0
25
is
2 +y-6-41
+4-9
+y-6x-4y+
X+yí-6x,-4y +4+9
+y-6K-4y-5 x+yf-6x1-4y1+13
= +-64y +y-6x-4y+13
7x +7y-42x,
(
=
an or
-28y1+91
25x7+25y?-150x,100y
or 125
the
=3.
V3+62 36
=3.
02)
2
=u.
pair of tangents from
4) 36 +36-66-4-6+4
(2)
L-3-6)
a+p*-1+22
5+20-P(-1)=
=0
(3)
LetP(a, B) be the centre of the circle passing through the centres of the three given circles. Then
8a+6B +14 0
+7-0
c32a 9a+ (4a
..(4)
7
-9
+7)+18a
25a+170a+ 280
a+(-22=(a +3)2+(B+6)2
+ 24(4a +7)
= (a
4-4ß 6a +9 +12ß+36 = 6a+16+41 =0
and
6a-8ß-5 =0
+63 = 0
0
5a+34a +56 0 or Sa+20d +14a +56 =0 (5c14)a+4)=0
(1)-(2)
(2)
3p-(4a +7)
--16+7),
-4
-9, D
6a-8
or 6a+-50; 46
the
+36+ 4B +4
(1)
0
-
50
a= 18
centre of the circle passing through centres of given circles is
3,7/5.
TOm (1), the equations of required circles are
+y2 8x-6y +24 0
12a
(2)
24B+46
-12
a-4-14/5
(4) gives
+6)+(B +2), =r)
=
or
or
(3x+4y-26)2
tangents are x-6=0,7x-24y+102 =0.
88+7=0
Puting in (2) from (4),
or
916y2+676+24xy-156x-208 7x-24xy + 60x+144y -612 = 0 (7x-24y +102)(-6)=0 ds
centre (3,-6) and radius =
(4)=(a2+p2-1)+8
a+p+10a-2+21
4a+3p
or
6+:6-3(x +6)-2y+6)+4) 16(+y2-6x-4y +4)
or
2-)
is
+24xy 48x 64y
tangents from (0,-2) arer=0,7x-24y-48
-6x-4y+4
.(1)
the second circle orthogonally if
(3)-(2)
=(-3x-4y +8)
9x+16y+64 7x-24xy 48x =0
a
(1) cuts
or
+4) (0+4-0+8+4)
Similarly, the equation of the
1+tan 2
(0, -2)
16+y-6x-4y+4)
length of tangent = vxf+yi-6x-41+4
tanPA
co
y-2ar-2py+a+p2-1=0 1+2-P)
V6+2-31
=
For the third circle,
cuts the first circle orthogonally if
or
-0+-2)-3(r+0)-2y-2)+43 or
centre= (-6,-2) and radius
be
fusing 28,8, +2ff=ci+ C}
xx+yy-3(x+X) -2(y +yi)4
(+y-67-4y
3y =6
1
(1)
2:-a)
the equation of the pair of tangents from
Now, centre C= (3, 2) and
or
required circle of radius
or
T
N(-2)+5 = 3.
centre (0, 2) and radius =
(-a)+-)-12 y1) is
6x+ 12y +36=0.
For the second circle,
circles touch externally, not internally.
Let the
x+y-6x-4y +4 S-+yi6x 4y+4
(3,2)
x+y+
For the first circle,
d--1+5)+(4-1)
(6, 6).
Ti
x+y-4y-5-0, r+y+12r+4y +310 and
The distance between the centres = V16 +9 =5
where S
P(11
1-tan
circle,
(-5,1) and radius r= V5+1-22 =2.
the
S51
circle of minimum radius which contains the three circles
dnt
)=(0,-2),
:ie, 5(+y)+28xr - 14y +44 0.
-
The equation of the pair of tangents from (r,
Then tan 6=
or
-12(4x1-6) 108=0
-26
Let P(x, y) be a point on the line 4x -3y =6 having the given property.
=
For the second
centre
0,6
coordinates of all such points and the equations of tangents.
and PA
120
and
20. Find the equation of the
(1, 4) and radius r = v12+4-8=3.
centre
18.From
o
Show that the circles +y* +2x-8y +8 = 0 and +y+10x-2y+22=0 touch each other. Do they touch intemaly? Also obtain the equations of circles of radius 1 which cut both the circles orthogonally
circle, For the first
Solving (1) and (2), -1;
P0Q=;
make an angle of tan
)
19.
is on the line 4 -3y = 6, we get
4x-3y16
CoS 6
Clearly, m, m,= Clearly, m1
18x+ 18y-108x1-72y1-216=0 x+yí-61 -4/1-12=0
or
coS 6
radius r= va+(B-2)
C-56 Problems Plus in IIT Mathenatics
C57
Circles
V
+y6r=0
7 36
23725-949
A
radius of the required circle
Here the circle x+y-6x
=
centre (3, 0) and radius
3.
The circle +y+2=
0
5
=3+949 and
has the
0
form
-
x-0
respect to the circle
and sum of radii = 3 +1=4 circles touch externally.
Clearly the inclination of y
(3,0)
and that of y= 3 Also, their point
41- 511+6l+1 0 lx +1ny +1=0 touches a and centre of the circle.
21. If
then show that the line fixed circle. Find the radius
m
51+61+1
It is possible to find a, The condition is
=0 B, a
nd
(1) or
C2mc 2)
or
the line touches the fixed circle, (x-3)+(y- 0) = (V52
+ y-
6x +4 =0
whose centre = (, P)
=
(3,0) and radius =n= Vs.
1
.
6mc+c=9 8mc=8;
3)gives
c
When
.
.
axby
always lies on the line ar = by which is a fixed
line.
equation of the circle described on the and common chord of the circles x*+y-4x-5=0 x+y*+8y +7=0 as a diameter.
25. Find the
Here the circles are 5, = +y2-4x-5=0
S2x*+y
and
+8y +7=0
. **
(1) (2)
(2)
N3,
mc
tV3.
m= 1N3.
from (3) we get
13
and
y
y-a?+Mxx It will pass through
2+y2-4x -5 +r2+y2+8y +7)=0
or
(1+)r* +y)-4x
or
-v3
(
+
yy-a)=0
y1) if
(iyi-ax1+3)=
But+ya?,
0.
for (u yi) is not on the circle.
-1 from (3), the circumcircle of APQR is
4 *y13**1
+8Ay +7
y
(3)
-5 =0
-0. 1+0.
nte-1
3)
(4)
The equation of the common chord of (1) and (2) is
S-S-0
+y-a+f+y?-a)=0 Or
3+2nt3=1; m=1N3 two commontangents äre y
or
Any circle through the intersection of (1) and (2), is
from (3) we get
3-2iV3 1; When c-3,
circumcircle of APQR is a circle passing through the intersection of the circle (1) and the line (2), and the pointP y
he
(4)
1.
c3
Any circle through the intersection of the circles (1) and (2) is given by S, +15a=0 a-1)
P(x1)
(6mc +e)- (c2nic)=9-1-8 or
i.e., a = V5. a-3-4, Also a = 3, B =0,a = v5 satisfies (3)
coefficients,
V1).
equation of QR is xx + yy1-a'=0
(31+ c)=9(1 +m).(2)
(1)
which implies
must be the same line.
(1)
from the point PX,
the
and+=1 b
SM +C
(2)
a =3
--1
+y-a2-0
the length of the perpendicular from (3, y= mx+cis equal to the radius of the first circle
if (1) and (2) are identical.
6 0 -20B2-2-1 01
Tangents PO, PR are drawn to the circle x+y=a2 from the point P(x y1) touching the circle at Q, R respectively. Find the equation of the circumcircle of the APOR.
=c2is
.(1)
ax=by=c;
(y)
of intersection is (-3,0).
QR is the chord of contact of the tangents to the circle
and
+(n2-B2)m2
-2Bim-2ia-2mB-1 =0
B=0,
t
+c)=1+m
a+m)= la +mß+1)
But 4
0)
23.
comparing or
as shown in the figure.
is
we get from the figure that the triangle is equilateral.
equal to the radius of the second circl
m or i-1
+mß+ 1|
(a-a y
Then, the length of the perpendicular from (-1, y=mx+cis
The line Ix +my+1 =0 touches thecircle if
or
Let y= +cbe acommon tangent to the circlesev at the point of contact max
Let the circle be (r-a+(y-B)=a?
or
with the x-axis is
6
Obviously, the point of contact = (0, 0).
-
=-V3
x+y
yc
Now, the line (1)
centres.
n
Xx +
X
In similar situations if the radii of the circles are different then add the largest radius to the radius of the circle passing through, the
a
1 that, for all e eR, the pole of the line on a fixed with respect to the circle x+y=clies line. y1) with Let the pole be (x1. y). Then the polar of (x1,
24. Prove
an equilateral triangle.
the distance between (3, 0) and (-1,0)
the
- yyi =0.
orx+y2-xx,
y=+13
and
3
=0,y
1.
-V3+1+07-4
equation of the required circle is
a)=0
+y-a3-1(x+y
tangent at the point of contact (0, 0) is = 0. 0+y:0+ (x +0) =0;
prove that the lines Now, we have to
has the
centre = (-1,0) and radius =
re-2
the
The
form an equilateral triangle.
-V the
common tangents to the Circles and x+y+2r=0
22. Show that the
ie,
or
+y-4x-5)-(x+
y2+8y +7)=0
4x +8y+12 0
or x+2y +3=0 .(5) The centre will lie o the common chord, ie., (3) will be a circle described on the common chord as a diameter if
C-5S
Problems Plus in
or
2-8
putting circle is
y+
or
++28)x
2xy)-4x+8y +2=0 y-2x
of the circles, passes through a fixed point.
27.
B=(
Va)
A(X11)
i+f
th.
A variable circle passes through the
point
ot
intersection O of two given straight lines and cuts off from them portions OP and OQ such that prove that the circle always m OP+n 00-1 passes through a fixed point other than 0.
Let O be selected as the origin and the lines be y =0 and y=ar where Pis on y=0 and Qis on y = ar.
Q=(0Qcos e, 0Qsin
0)
which is of the form S+ AL =0. circle passes through the intersection of the
and the line 2r +ky =0,both being
fixed.
r
+y+py = 0 cut But the line 2x +ky = 0 and the circle at two fixed points, one of them being the origin O. Hence, the circles pass through a fixed point other
(0,0)
S0
than O. 28.
tan
The equation of the circle with AB as a diameter, is
(-XMr
- x)+(y-y1y-y)=0
0=a)
If two circles cut a third circle orthogonally, prove that their common chord will pass through the centre of the third circde.
Let us take the equations of the
2) =aX
and .(3
Let the equation of the circle be
r+y+2gx+ 2fy+c=0
(x-xr-x)+0-y-y»
The chord of intersection any one of the circles (4) is given by
2gx + 2+c-
S =0
(+X2)r +(y1 +y2)y-xi*2-y1Y2
OP=
.
(.
+y+2g
"
og+00
V1
Thecircles
(1) and (3)
cut orthogonally .(4)
a+c from (4) and (5), 2g(-)
.
=0. But A
2
80
ience, thecentre of the third circle=(0,-f).
-0
he
+a
common chord of (1) and (2) has the equation
S-S=0 y+2x +a)-(r+y+ Or 201-)r=0 x=0
OP+n-00=1
1
0f Hence,
satisfies the equation x = the problem.
S-S
=0,
2gx +2fy +c+4 =0
It has to pass through the centre (0, 0) of the circle (1.
(2)
0
. ()
0+a
1+a
V1+a
+m
=
The circles (2) and (3) cut orthogonaly
0Q=2tfu) m-38)
.(1)
+a
Is-S10 The equation of the common chord of (1) and (4) is
ie,
For a general statement in geometry concerning two circles we can select axes suitably (the line of centres as x-axis and the point midway between the centres as origin) to get the above form of equations. Letthe third circle be x2 + y2 +2gx + 2fy +c=0. 3
2g +39 00
i
-28
and
But m
2gx+2y+-7
****
OP+2g OP =0;
t-1)
-yy-y)a or
C=0
(4)4) 1=0. 1
3 of the fixed circle
+y+
... (
As it passes through 0, P and Q we get
+
two circles as
+y+2x+a=0
The equation of the line AB is
2 circle passing through A and B (the points of any intersection of the circle 2) and the line (3) is given by
.(4)
of a The. circle (4) will cut the circle (1) at extremities diameter if the common chord of (4) and (1) passes (1). of through the centre
Now, P= (OP, 0) and
B(22
(3)
S=x+y2+2gx +2fy+c=0
+y+Py+8(2x +ky) = 0
this
(1
. (2)
Let the required circle be
xty+28x+ (kg+Ply =0
circle y+Py=0
at
Sx+y-4=0 Sax+y-6x -8y +10 = 0 S=x+y* +2x-4y-2=0
kg+p where k and p are constants.
or
0
diameter Here the circles are
from (1) and (2), the circle has the equation fixed pont
x+y-6x-8y+ 10 0
+y=4,
x*+y*+2r -4y 2
l+2 m
+G +92* 2)9-X¥2-yiyat c-o
Hence, their point of intersection is also a Thisproves the result
Let Sx+y+2gr+25+c0 be the fixed circle, and A = (*, 1),
i
11-0
and
+4y +1=0.
26. A fixed circle is cut by circles passing through wo fixed points A(r, y1) and B(r y). Show that the chord of intersection of the fixed circle with any one
circles
This represents a famuly of ines passing through intersection of the two fixed straight lines
8y +7)=0
which cuts each of the and the extremities of a
29. Find the equation of the circle 1
-12-1a +0-
=
+3(1 +) A=1 0; this value in (3), the equation of the required
y4-5 or
Circles
(r+2+28)* +U +2 +2f)y
or
(from (4) and (5))
+3=0
C-59
IT MathematicsS
(5)
(5) C+4=0 The equation of the common chord of (2) and (4) is
S-S 0, (28+ 6)x +(2f+8)y +C- 10 =0. It has to pass through the centre (3,4) of the circle (2) (2g+6)3 +(2f+ 8)4 +c- 10 = 0 i.e,
or 6g+ 8f+ C+ 40 =0 (6) The equation of the conmmon chord of (3) and (4) is
S-S0, (28-2)x +02f+4)y+c+2=0. Ithas to pass through the centre (-1,2) of the circle (3) (2f+ 4)2 +c+2 =0 28-2X-1)+ ie,
or-2g +4f+c+ 12 =0 Now, 5)
6g+8+
22+a) =0 0.
c-4; 36
=
(7)
so, (6) and (7) give
0, i.e., 3g +4f+ 18
=
0
and -28+4f+8 0 (8)-99*
58+10=0 and so f=-3. 8-2 therequired circleis x+y*-4x-6y
4=0.
.
(8)
.
(9)
C-60
C61
Problems Plus in lIT Mathematics
Circles
30. For
what values of Tand m the circle 5(x+y") + ly - m =0 belongs to the coaxal systemn determined by the circles x+y+2r+4y-6=0 and
24+y)-x=
P(a.B)B
Y-B
(*1
. (2) tp)X=a(a +p)+ B(B+g) or (+9)Y+(a origin and the intersection The pair of lines through the degree of (, (2) is found by making8 (), second the help of (2).
0? o(0,0)
If
the radical axis for each pair of the three given circles is the same then the result is established.
S
The centre of the circle is O0, 0). Clearly po perpendicular to AB.
y-x=0 Now,
=0 is
S-S2=0, ie,
+y+2x+4y-
the
5x +8y -12 =0
0B a-0 a
e,
y'--*yz-)5r+21y-2m
= 0
8
Y1)
2p(a+p)
-a-p-4
equation of the locus of P(a, ) is
Find the locus of the middle points of chords of a given circle r+y=a*which subtend a right angle at the fixed point (p, q)
ie, and
or
+p+9-a (a
the
-Bk-B) 2), (a-a)x
(*1-a)h-a)=0 -aa =0.
equation of the locus in X,
aa(k-p)-Bla(h-a)+
+p)* +(p+4))=0
Y coordinates
is
-aa(h a)
21X(X+p)+YY+4)
or
2+y2 =4, m =6.
S+S
in
a2
new (X, Y) coordinates, P=(0, 0)
and equations of transformation are
n-a2
.
of the middle points of the chords of the circle r*+y'=a" which pass through a given point ( y
Let Pa, B)be the middle point of any chord AB thrpugh
the given point (x1, y1).
..
the middle point M of the chord AB be (a, P) Let new coordinates.
(
or
+
(a-a)la(h-a)+PkBI12 ++az(h-a)(a-a)tk-B)Ph-a)
y)=0
2(+y)-(px+9y)
=2a k-6)-plath
or
a+P
AB is perpendicular to MC, the equationofor MC
+(-4a+a-aa)(h -a)+ P(a-a)k- p))2 p 0*-9)-aB(h a)
qyXr*+ y)
a+yi)=0 +(p*+9*-a)=0
+(p+92-a
the
Or
will
2+y-2(px+
2+y)-2(px+ qy)
+y 7+y-pX -qy+g2-a)=0.
-a)+p(k-p))
(a-a)k-B)-B(h a) (aa -p°%k-B)-aßh-a))2
+(p+92-a?xx +y)=0or
(1)
Then 'm' of CM=P1,
be
(a- a)Xk-B)-ph a)
+2(px +4y)lx* + y')-(px + qy))
FP-9)
31. Find the locus
+ B(k- P)}
+
+(p+9-ax+
=0, (
-1). =p+X,y=q+Y. If we can find a e R such that S +152 =0 and the equation of the circle becomes S0 are identical equations then we can say (p +0)+(q+ S =0, is a circle belonging to the coaxal system determined by S, = 0, S = 0. whose centreis C=
B)}
p)*»(Y +4))=0But x7 +y?= 2ax Using equations of transformation (0), we get the alk-B)-Blah-a)+pCk-B)12 equation of the locus of M in old coordinates X, y as (a a(k-B)-B(h-a) 2(T-px+y-9)y+24px +qyi{(r -p)x+y-9)¥}
+p+9-MX
Note The system of circles coaxal with the circldes S=0 and S2 =0 has the equation of the form
B(k-
+(a -a)lath-a)
+2p(X+P)+40Y+ 9MX(X +P)+Y(Y+9))
l
(3)
+Bys
From (3), (h-a)x +(k-By1-a(h-a)-p(-B)=0
+24p(a +p) + q(ß + q)}la(a +p) + B(B + )}
(1) and (2) must be iderntical. So, comparing them,
(using perpendicularity) =-1using
1-ah-a +
From
2(a(a+p) + B(B+ 4))
P.q)
. (2)
(h, k).
-a-p-qla(a +p)+(B + +)| or
B) be the foot of the perpendicular to (1) fromn Then (a, B) satisfies (1), (2) (-o)a + y1B-ax =0
Let P(a,
(a+p) (a(a+p)+p(B+9J
2g( +9) a(a +p)+B(B+4)
Change the origin to the point P(p, 4).
-0
or or
32.
Thk)
orala+P)+P(B+9)
The equation of the radical axis of circles S, = 0, S3 =0 is
S-S 0
. (1)
+ y1y-ax =0
between the lines if
-)+v-)=0.
(1)
+ yy1= a(r + X1)
(1-a)x
a-p?a(a+p)+(B +4
aa-z)+B(By)=0
or
XX1
the
The pair has a right angle coefficient of X* + coefficient of Y? = 0
-.-1
6- r
orx+4y-6-0 or
of OP
equation of y1) be any point on the circle. The tangent to the circle at (x, yi) is ie.,
(a+p)x
a-X
m S2
:
x+Y+2pX+q).t9Y+ a(a + p) + P(B +9)
is
m'of ABPi
SyygyThe equation of the radical axis of circles S, =0,
+y=2ax. Let(x
homogeneous with the equation of the pair PA, PB is
Let the circles be
S=x+y+2x +4y-6=0
perpendiculars from of the feet of the to the circle the point (h, k) to the tangents
:33. Find the locus
-a)
2al(aa
x
0)
the
k(a-a)-
B(h-a)}
equation of the locus of the feet is
(lax-y-y)-ryh-x)}
+lx(r-2a
h-x)+y«-«)k-y)N2
C62
Problems Plus in IIT Mathematics
-y xk- y)-xy(h -x)Wk(x-)-yh-a)).
2al(ax
34. Show that the locus of the point, the tangents from which to the circle «+y*a^ include a constant angle a is
y-2a*
Then
Tet AB
y=a
(1)
be inclined at an angle a.
Let
+y=9.
to
+y-9%*+yí-9)= -9-9+yh
-9
(*x, +yy
S
S,
=
-h
(2)
-3m
k-mh
Dr
T
-9
2x1y1xy-18xx, 18p
a(1+m)=(k-mh)
or
a2-h ym2+2hkm+a2-k2=0
.
-m
mm)
or
(1+mm)* tan'u = (m, +m)*-4mm
(1+mm)
+yi-18
m
x2+y4=18
V+m ..(2)
+y+2Ax +41y 18(+1)=0
(3) is the family of circles
(2a-h?-k)tan
'a=4hk*-4(a-hKa?-k3) 4a h+k2-a)
equation of the locus is tan
a=4ar+y2-a
Second Method Any tahgent
to +y=a*is
y = mx +avi + m2
It passes through P(h, ) if k=mh +avi+
or (k-mh)*=a (1
+ m)
which is the same as (3).
the
C PN
touching (1) at (12, 4).
+y-2x-4y-200
the
angentially at leaves it point (-2,-2). After getting reflected the line it passes through the centre of the fromnd ciredicula equation of the straight line if its perpe distancefrom Pis the
ball
8+63
-
8v3 +6 + c(4 -
or
0
2cos
CoS
0
cos 20
20= sin
Now
'
14-213+c(4 +3v3) = t.10-+25 4+33
=ttan 20
As the ball is projected from a point in the third quadrant while moving in the anticlockwise sense, clearly the slope of the AB will be positive and it should cut negative intercept on the y-axis.
2sin6
cos 0
c23-39 4+33
8
the
cos(90°-0)
20
90°-; ofPC-2
+33)
16+ 27+2413+48+9-2413
c2V3-14 +25
1+2+2+2 92cos Sins 20 20
in anticlockwise direction
is P = (-2,-2)
or
tan 20
sec
the circle
Here, centre C=(1,2) and the point wne leaves the circle
5 . sec
Fromthe right-angled ACPN,
required equation of coaxal circles is (5)
36. A ball moving around
4N3-3
From the right-angled APLN,
2x +4y-18=0 from(3))
C+*5
-24+33
Now PN1 PC and PLLAB. So ZNPL=ZPNM = 0.
PN PLsec .. (3
4+313
4+3V3
2-2m1 +
.. ()
As (2) touches (1),
Dr
=-43+4V3 43 3
Let the equation of AB be y = mx +C where
P-2-21R
Also any two circles of (3) have the same radical axs
+y-24
4-313 4+313
=m3v3 -4
thetangent at (l2, 4) to the circlex*+y^= 181s
or
=
4v3m
3 43
AB "m' 'm' of of AB
+y-18 +(2x+4y18) =0
tan *a
=
3+413 3-413
The equation of circles passing through the intersection of (1) and (2) is
a1+ma
or
the
4733 = 3m t 47313 m=
Clearly/ (2, 4) is a point on x?+y2= 18.
This is quadratic equation in m whose roots m, m, are the slopes of the two tangents. As the angle between the two tangents is a,
or
4-3m
or
C( 1,2)
y-9+-9=0
y-18
or
t V3(3 + 4m)
or
As the pair is at right angle, a + b=0,
or
t 13
3+4m
equation of the locus of point of intersection of the perpendicular tangents is
V1+m
tan
4-3
ie., sum of coefficients ofx and y is zero
radius of (1) = length of the perpendicular from (0,0) si to the line (2)
=
60°= tan 60 3+4m
c1.2)
-
Any line through P(. ) is y-k=mx (2) will touch (1) if
tan(180°-90°+30)
2)
the equation of the pair of tangents is
= m
So ZPNM= 2CNM=0 (say).
(7 ) be the point of intersection of erpendeular
or
Let 'm' of MN
reflection.
tangents.
.
line and N be the point of
Let NM be the normal to the line AB at N.
At first we find the locus of point of intersecti ton f mutually perpendicular tangents.
ie,
(hay
of
h
intersection of mutually perpendicular tangents the circle
be the required
dence. PN is the line of incidence and NC is the line
35. Find the equation of the system of coaxal cirel are tangent at (2, 4) to the locus of
tan'a=4a'ta+ y-a').
Let P(h, k) be a point from which the tangents drawn to the circle
C63
Circles
proceed as above.
(-2)
30°
equation of the line
isy4+3134+313
Similarly, we can obtain the equation of the line when 'm of AB is 413 which is also
313-
positive.
C4 Problems Plus in iIT Mathematics
C65
Circles
Exercises Find the equation of the family of circles which touch the pair of linesx-y+2y-1=0. = 0 and 5x -12y- 40= 0 touch 29. Lines 5x + 12y-10 C lies in the circle C, of diameter 6. If the centre of C2 the first quadrant, find the equation of the circle which is concentric with C and cuts intercepts of length 8 on these lines. 28.
the equation of a circle whose diameter has the ind length 20 and the are 2x+ y=6 and
2.
14. Find the range of values of r
equations of two of its diameterss
the line
3x + 2y=4.
Find the equations of circles touching the y-axis at (0,3) and making an intercept of 8 units on thex-axis.
circles
5.
6.
1
9. Of the two
concentric circles the smaller one has the equation x+y*=4. If each of the two intercepts on the line x+y=2 made between the two circles is 1, find the equation of the larger circle. 2,3, 4 are
1,
3x + 4y +15
=
+y+2gr +2fy+c=0. 25. Find the
26.
0.
the four lines 47 -3y=5, 7x+y= 40 and x+3y+10=0 torm
13. Can
x-2y=10, the sides of a
cyclic quadrilateral? Justify your answer
b
contact. 21. Two circles, each of radius 5, touch each other t (1, 2). If the equation of the common tangent 4x +3y 10, find the equation of the circles. 22. Prove that x+y2=a2 and (x-20)2+y=a' ar wo equal circles touching each other. Find the equation of the circle(s) of equal radius touching both the circles. 23. Find the equation of the circle which touches5 circle r+y-6x+6y +17 0 externally ana which the lines x-3xy -3x+9y=0 are norm Find the area of an equilateral triangle insct ibed it 24. he circle
is
concyclic then prove that the product of their ordinates is equal to 1. 11. Find the angle that the chord of the circle x*+y-4y = 0 along the line r+y=1 subtends at the circumference of the larger segment. 12. Prove that the circlesx +y-9=4r*;r=1,2,3 cut off equal intercepts between the circles on the line
are
0.
line 31. A circle touches both the x-axis and the 4x 3y+4=0. Its centre is in the third quadrant and lies onthe line x - y -1 =0. Find the equation ot the circle w 32. Obtain the equations of the straight lines passing through the point A(2, 0) and making an angle 45° with the tangent at A: to the circle Find the equations of the 3)=25. x+2)+ circles, each of radius 3 whose centres are on these straight lines at a distance 52 fromA. 33.
The extremities of a diagonal of a, rectangle are 4 and (6, -1). A circle circumscribes the rectangle and cuts an intercept AB on the y-axis. Find the area of the triangle formed by AB and the tangents to the circle at A and B.
area of a square circumscribing the c rcle
30x+y6x+8y =0.
by Show that the area of the angle formed cinche tangents from the point (4,3) to the xty9and the line segmentjoiningtne points of contact is 7square
units.
:
27. Find the
3)
equations of the circles passing through 2 and touching the lines x + y=2 and x
x+y*=8
:44.
4
o'
1
(. 10. If the four distinct points m
th.
whether the circles x +y-21 - 4u( andr+y-8y-40 touch each other externall or intermally. Also find the point of tangency. 19. A circle of radius 2 lies in the first quadrant and touches both the axes of coordinates. Find the equation of the cirde with centre at (6, 5) and touching the above circle. 20. Show that the circles x+y-10x+ 4y-20 0 and x*+y+14x-6y +22 =0 touch each other. Find the equation of the conmmon tangent at the point
0.
2
the 32 but not
18. Examine
the points of intersection of C with the r-axis.
3
=
those values of pe Rfor which t chords can be drawm from the point (p, p) to the circle(r-p)+y=pboth ofwhich bisected
thelinex-2y+2
the equation of the circle passing through the points A(4,3) and B(2, 5), and touching the y-axis. Also find the point P on the y-axis such that LAPB has the greatest magnitude.
30. Find
values of m such that
17. Determine all
be two fixed points
1,2,3
un
h
largest circle
circle+y=4.
7. Find the equation of the circumcircle of the triangle formed by the lines y =x, y= 2x and y = 3r +2. 8. Prove that the equation of the circle passing through
Y;r=
ween
with thecentr e and lying between the Im
y=mx+4 cuts the circle x* +y*-4x
in a plane. Let C denote a cirele with centre B and passing through A. Prove that the real roots of the equation x+px +q=0 are given by the abscissae of
thenoncollinear points (z
2x+y3
16. Find the possible
The abscissa of two points A and B are the roots of the equation r2+2ax-b* =0 and their ordinates are the roots of the equation r+2px-q2=0. Find the equation and radius of the circle with AB as diameter. 2
points
lie
and x*+y?=2.
-y=1andr-y+3=0.
0.
Let A = (0, 1) and B =5
+=1
on the line
points (1,-2) and (4,-3), and whose centre lies on the line 3x +4y =7. 4. Find the equation of the circle passing through the intersection of the lines 3x + y=4 and x-3y +2=0 and concentric with the circle
20x+y)-3x +8y
3
15. Find the equation of the
3. Find the equation of the circle passing through the
for which then.
-2)=20-1)=r
the circle tangent drawn from the point (4, 0) to the first touches it at a point A in B on quadrant. Find the coordinates of another point the circle such that AB =4. (x -a)+y=b? 41. A tangent is drawn to the circle circle and a perpendicular tangent tothe their point of (x+a) +y= c2; find the locus of.bisectors of the intersection, and prove that the of angles between them always touch one or other two other fixed circles. circles 42. Find the coordinates of the point at which the 8y + 43 = 0 r+y-4x- 2y=4 and x+y-12x of common touch each other. Also find equations tangents touching the circles in distinct points. 43. A circle Cj of radius 5 has its centre at the origin A. Circles C2 and Cg with centres at B and C, and of radii 3 and 4 metres respectively touch the circle C and also touch the x-axis to the right of A, C2 being in the first quadrant and C3 in the fourth quadrant. Find the equations of the common tangents to the circles C2 and C3 except the x-axis. 40. A
at the 34. Tangents are drawn to the circle x+y=12 points where it is met by the circle whose equation x+y-5x+3y-2= 0; find the point of inter section of these tangents. 35,
circle 45. Find the equations of common tangents to circles +y=1 and (r- 1)2 (-3)2-4.
the
is
46. From any point on the circle + 2fy + c=0 r+y+28r
Find the equation of the tangent to the circle 60y +2100 = 0 at the point nearest to the origin.
tangents are drawn to the circle x*+y+2gx +2fy + csin'a+ (8+/cos 'a = 0. Prove that the angle between the tangents is 2a. 47. The chord of contact of tangents from a point on the circle 2+y=a2 to the circle x2+y^=b2 touches the circlex* +y*=ciShow that a, b, c are in GP 48. Show that the polar of the origin with respect to the circle +y?+2gx+2fy+c=0 touches the
r+y80x36.
A line segment AB is divided at C so that AC= 3CB. Circles are described on AC and CB as diameters and a common tangent meets AB produced at D. Show that BD is equal to the radius of the smaller
Prove that the line x+y =22 touches the circle +y=4. Also, if the point of contact be P then find the equations of the tangents to this circle which are at a distance 2 - 2 from the point P.
37. Prove that the tangent to the circle r+y*=5 point (1,-2) also touches the circle
-8x
+6y+ 20
at the
0
and find its point of contact. 38, AB is a diameter of a circle. CD is a chord parallel to AB and 2CD =AB. The tangent at B meets AC Produced at E. Prove that AE = 2AB. the point P(1, 0), a tangent PA is drawn to the circle + y2-8x + 12= 0, A being the point of contact. Find the equations of the tangents to the Circle from the middle point of PA if A is in the first
rom
quadrant
circle
r+y=a2
ifc2 =a?+s)
49. If the pole of a straight line with
+y=c
respect to the circle
lies onthecirclex+ y=9c* then prove that the line is a tangent to the circle 9x+9y2=c2
50.
Tangents are drawn to the circle x*+ y2=a2 from the point (h, k). Prove that the area of the triangle formed by them and their chord of contact is
ah2+k2-a ih2+k2
-66 Problems Plus in
51. Let C be the centre of a circle. Lines L and L are the polars of the points A and B with respect to the circle respectively. Perpendiculars AM and BN are dropped from A to the line L and from B to the line L, respectively. Prove that CA : CB =AM: BN. 52. Find the pole of the chord of the circle
x2+y2=81,
53.
the chord being bisected at the point (-2, 3). Find the equation of the circle which has for its diameter the chord cut off on the line pr + 4y-1=0 by the circle x*+y2-a2 =0.
IT Mathematics
circles. 64.
y+ax+
1-a bbc-c'
ABC
66. Prove
through two fixed points. Find the fixed points. x*+ y*-6x-4y +9 =0bisects the circumference of the circle x+y-8x -6y +23 = 0. 57. The polar lines ofapoint P with respect to two given intersecting circles meet in Q. Show that the common chord of the circles bisects PQ. 58. Find the equation to the circle whose diameter is the common chord of the circles (x-a)*+y2=a^ and x+(y- b)=bi Also find the. length of the common chord. 59. Prove that the length of the comnon chord of the
56. Prove that the circle
circles
+y+2hx
+a=0
and+y-2ky-a=0is
that x+y*+A(y-x-212) = 4 represen circles touching each other at a common point tor all real Also find the common point.
.
that the equation x*+y-2x-2y represents a circle passing through two fixed -S=0 points A, B for all real If the tangents at A, B to the circle meet on the line x + 2y +5= 0 then find the equation of that particular circle. 67. Find the equations of circles passing through the intersection of the circles x*+ y = 4 and
.
i
x+y-2x-2y+1
0
whose centre is at a distance 2v2 from the origin. 68. Prove that the common chord of the circle +y+6x+8y+7=0 and the circle which alwavs touches the line y = x and passes through (0, 0), will always pass through a fixed point. 69. Consider a family of circles passing through wo fixed points A(3,7) and B(6, 5). Show that the chords in which the circle x +y-4x-6y-3 = 0 cuts the members of the family are concurrent. Find the coordinates of the point of concurrency. 70. If the circle x+y2=16
60. Find the
equation of the circle passing through the points of intersection of the circles x 2+y2=4a and r*+y-2x-4y +4 0 and touching the line
61.
0.
intersects another circle C 0
radius 5 in such a manner that the common chord
maximum length'and has a slope show that the coordinates of the centre of the circle are
x+2y=0. Find the equation of the circle which passes through the points of intersection of the circles = +2y +4 -6 0 and +y*+2x-4y
x+y-6x
and whose centre lies on the liney=x. 62 Find the equation of the circle whose diameter is the common chord of the circles x+y+2x + 3y + 1 =0 and x+y*+4r+3y +2=0. 63. Do the circles
x+y2-8x
+y-6x-4y+9=0
and
-6y +23 =0 intersect each other? If they intersect, find the length of the common chord. touch, find the common tangent at the point If they
87.
+y 1
2 to two circles is 77. Prove that a common tangent bisected by the radical axis of the circles. the equation of the circle coaxal with the circles 78. Find
2x+2y-2x
and+y+4x
+
6y-3=0
+2y+1=0,
and whose centre is on the radical axis of the circles. 79. Three circles are given:
r+y=1,x*+y*-8x+ 15 and +y+10y +24 = 0.
0
Determine the radical centre (i.e., the point such that the tangents drawn from it to three circles are equal in length). From a point P, tangents drawn to the circles
= x+ytx-3 0, 3x+3y-5x+3y =0 and 4x2+4y2+8x+7y +9 =0 are equal. Find the
equation of the circle through P which touches the liner+y5 at the point (6, -1). 81.
82.
Prove that a common tangent to two circles of a coaxal system subtends a right angle at either imiting point of the system.
LetA=(-2,0) and B =(1,0) and Pis a variable point Such that LAPB = 60°. Prove analytically that the locusof Pis a circle. Find its radius and centre.
the
71 Find the equation
0
orthogonally
circles cutting the 76. Prove that the general equation of two circles x+y*+28X +2fy + C, = 0; r=1,2 orthogonally is
5
ofthe
86. The
the family of circles whose centres lie on 75. Prove that the line 2r-2y +9 =0 and each of which cuts the circlex+y=4 orthogonally, passes through two fixed points. Find the fixed points.
80.
of the circle passing througn u origin and cutting the circles + +y-4x+ 6y 10 0 and 2+y2+ 12y +6 at right angles. kRU 2OY 72. Find the equation of the circle which passes rougn theorigin, has its centre on the line x + y +4 cuts the circlex+y24r + 2y+4 0 orthogona 73 Prove that the two circles which pass throu the
onts (0,
a)
ca
(2+m ).
touch the n each other ortho8o lly
and (0,-a) and
mx t+c,will cut
83. A circle passes through the origin O and cuts two
lines r+y=0 and x
-
y
=
0in PandQrespectively.If
thestraightline PQ always passesthrough a fixed Point, find the locus of the centre of the circle.
8
Apoint moves such that the sum of the squares of its Cistances from the sides of a square of side unity is Equal to 9. Show that the locus of the point is a circle. its centre and radius. wo rods of lengths a and b slide along the coordinate axes in a manner that their ends are always concyclic. Find the locus of the centre of the Circle passing through these ends.
ind
5
any point of the circle point A(1, 5) is joined to middle point of AP Find the locus of the as P moves on the circle. the point P(1,2) and A variable circle passes through locus of the other touches the r-axis. Show that the end of the diameter through P is (r-1)8y. From the point A(0, 3) on the circle
S=0,5=0 have radii r and r'respectively. S S' circles =0 ill cut each other Prove that the Circles
74.
c=0 I
Prove that the equation 2(3 +p)x + 2(3-p)y +4=0 represents a circle for all values of p, passing
by +
Rand S are concyclic then show that
65. Prove
x+y+
e
in Pand Q.Another lineA *+B y+C'=0cu ircle 4y2+a'x +b'y+c=0 in R and S.. Ifts he P.O
equation of a chord of the circle +y-2ax = 0, prove that the circle of which this chord is a diameter has the equation
(1+m(x+y)-2a(r+ my) =0.
55.
the
The line Ax + By +C=0 cuts the circle
54. If y= mx be the
x
C-67 Circles
of contact. F they do not intersect or touch radius of the smallest circle touchined
+y=4.
88.
*+4x +(y 3)
= 0,
M such
to point a chord AB is drawn and extended the locus of M. that AM = 2AB. Find the equation of
a
circle origin, chords are drawn to the of the middle points locus the 1. Find = (r-1)+y of the chords. a fixed point 90. The base of a triangle passes through a, b) and its sides are respectively bisected at right y -8xy-9x=0. Prove that the angles by the lines locus of the vertex is a circle. Firnd its equation. O and 91. A circle of radiusr passes through the origin Cuts the axes at A and B. Find the locus of the
89. From the
centroid of AOAB. circle of radius r passes through the origin O and cuts the axes at A and B. Let P be the foot of the perpendicular from the origin to the line AB. Find the locus of P. 93. Find the locus of the middle points of the chords of thecirclex+y=a*subtendinga right angle at the point (a, 0). 94. Through a fixed point P a line is drawn perpendicular to any diameter I of a given circle to meet the diameter m at the point Q where m makes 45° with l in anticlockwise sense. Prove, that the locus of Q is a circle passing through the centre of the given circle. 95. The point A is one of the points of intersection of two given intersecting circles. Any line is drawn through A to cut the circles again at P and Q. Prove that the locus of the middle point of PQ is a circle. 96. Prove that the locus of a point which moves such that the sum of the squares of its distances from three vertices of a triangle is constant, is a circle whose centre is at the centroid of the triangle. 97. The tangent at any point P on the circle x+y?=2 cuts the axes in L and M. Find the locus of the middle point of LM. 98. A triangle has two of its sides along the axes; its third side touches the circle x*+ y*-2ax 2ay +a2=0. Find the equation of the locus of the circumcentre of the triangle.
92. A
na tne locus ot the foot of the perpendicular drawn from a fixed point on the r-axis to any tangent to the circle x +y=a2.A
C-68 Problems Plus in IIT Mathematics
C-69 Circles
00.
Let
S
=x*+y4+2gx + 2fy+c=0 be
a
The centre of a circle 1s (,1 and its rad units. If the centre is shifted along the lin through a distance V2 units, find the equati circle in the new position.
114.
given
circle Find the locus of the foot of the perpendicular drawn from the origin upon any chord which subtends a right angle at the origin. 101. Show that the locus of points from which the tangents drawn to a circle are orthogonal, is a concentric circle. 102. Find the locus of the point of intersection of tangents to the circlex=acos 0, y= asin 0 at two
Objective Questions
129.
cincles 130. The
115.
The circle x+ y- 4r 4y+ 4 =0 is inscribed in a triangle which has two of its sides along the axes of coordinates. If the locus of the circumcentre of the =0,
117.
118. The equation of
the circle which is concentric wth the circle 2+y2 5x-4y-1 0 and touches the 0 is . straight line 4r -3y-6
119. A circle
of:
(x- a)+y-B)=b2.
touches the z-axis at (2, 0) and has an intercept of 4 units on y-axis. Then the equation of 2 the circle is 120. The line L passes through the points of intersection of the circles +y=25 and +y2- 8t+7=0 The length of perpendicular from the centre ot the second circle on the line L is-
121.
109. Two equal circles touch one another. Find the locus of a point which moves so that the sum of the
tangents from it to the two circles is the constant k. 110. Find the locus of the centres of circles which cut the
112.
113,
- 6y +9 =0 +y+4x x+y*-4x +6y +4 =0 orthogonally
Find the locus of the point of contact of parallel tangents which are drawn to each of a series of coaxal circles.
Prove that the cirde (r - a)'+(y-a) =a touches the x-axis. If the circle is rolled on the x-axis in the positive direction through a complete revolution, find the equation of the circle in the new position.
The cirele z+y-4r-6y+ 16-0 rolls up the tangent to it at (2 +V3,3) by 2 units in the direction of increasingx. Find the equation of the circle in the new position.
The equation of the locus of midpoints of the chords of the circle 4x?+ 4y?-12x+ 4y +1 =0 that subtend an angle of
122.
at
its centre is
The sides of a' square ae x = 4, 1 = 7, y=l a The equation of the circumcircle of the squan
4
twocircless and
The point of intersection of the lines ax + by a =0 and bx-ay +b=0 is P. A circle with the centre et (1,0) passes throughP The tangent to the circle at P meets the x-axis at (k, 0). Then k = -
externally.
the origin O cuts the circle **+y*=2(x + y) at A. Prove that the locus of the centre of the circle drawn on OA as diameter is also a circle which touches the given circle internally 107. Find the locus of the point from which the chord contact of tangents drawn to the circle x2+ y2=16 subtends a right angle at the centre of the circle. 108. Find the locus of a point whose polar with respect to the circle r+y*=a* touches the circle
The equation of the circle which touches the cirel 2+y2+2x+2y-7 at (2,-1) and pasts through (1,0), is
0
find A.
system of circles is drawn through two fixed points. Tangents are drawn to these circles parallel to a given line. Find the equation of the locus of the points of contact. Find the locus of centres of the circles which touch the two circdes +y*=a2 and x2+y2=4ax
106. A line through
111.
The point on a circle nearest to the exterior P(1, 2), is 4 units away from P while the point Point farthest is (-3,-2). Then the equation of the circle
-0)+-b)*=
is
123. If (4, 1)
be an extremity of a diameter of the circie +6y 15 0 then the other extreu of the diameter is 124. The length of the chord 4x-3y = 5 of the circe The line 3x + 4y=k touches the circle x+y Then k = and the point of contact=126. The line 7y-r=5 is a tangent to the circle x+y-5x +5y =0. The other parallel tangen 125.
127. The length of
the tangent to the circle
x+y+6x-4y-3 0
+
c
(x-b) +(y-a)2=c2
and
a circle passes through the poinis of intersechon of thecoordinates axes with the lines Ar y + =0 and
131. 1f
r-2y+3=0 then the value of
is
through the point subtends an The chord whose equation is y-x=3, circle angle 30° in the major segment of the
+y=1passes 145.
x+y=k3 ifk=-
will touch each other if
Choose the corect option(s).
1
_
132. The area of the triangle fornmed by the positive r-axis, the normal and the tangent to the circle r2+y4 at (1, V3) is (2, 0) 133. The equation of the circle passing through (0, 4), and having the minimum radius is
146.
The radius of the circle
4x+ 4y-10x
+5y +5 0
is
(d) 5
(c) 3V5
-
and
of the circle which bisects the chord cut -2)+0+1)=16 line x - 2y-3 = 0, is offby the circle on the 135. The equation of the line passing through the points of intersection of the circles 3x+3y2x+ 12y-9 =0 + 2y -15 = 0 is. and x+y+6x 134.
136.
The equation of the diameter
The area bounded by the circles 2+y 4 and the rays
2x-3xy-2y=0,
y >0
is
x2+y=1 and given
.
by
137.The set of values of for which the circles -4y +a =0 have and +y2-4x +y2 exactly three real common tangents is 138. If the distances between the origin and centres ot r=1,2,3 the three circles x+y2-2,x-a=0; are in GP then the lengths of tangernts to the circles frpm any point on the circle y* =a are in -
z+
lines 4x-3y+2=0 and 8x = 6y +5 are two angents of the same circle thern the radius of the
139. 1f
x+y-2x
x+y+3x=y+ 10 is
the circle 16(x y)+48x-8y-43=0 nearest to the line 8x-4y +73 =0 is. tangents drawn 144. The chord of contact of the pair of + to the circle from each point on the line 2x y=4 143. The point on
+9 = 0 is
+ 6y
-
116.
104. A
105.
the circle r+y+4x
-
triangleisx+y-xy +Yr*+y*
(5, 1) is_ drawn from the point to the circle x*+y?=4 The equations of tangents which are inclined at 60° with the positive direction and of the x-axis are The pole of the line 3x + 5y + 17 = 0 with respect to
Fill in the blanks.
points whose parametric angles 8 differ by 103,
128,
147. The line 3x -4y =k will cut the circle
x+y2-4x - 8y -5 (a)k-35 ork> 15
= 0
at distinct points if (b)-35 15 oa-ycos =a touches the circle 148. The line xsin x+y=a, Then
a
(b) ae
[0,
T] (d) a is any angle (c)ae 149. The equation of a tangent to the circle = 0o +4y
T
-3
x*+y-6x
which is perpendicular to the liney
(a) x +2y +
=
27-1,is
t 4N5 = 0 (c)x +2yt (d) none of these 0 150. The equation of a tangent drawn to the circle x+y-2x +4y = 0 from the point (0, 1) is =0 4Y5 = 1
(a)2x-y+1 =0
(b) x+2y +
1
(b) 2x-y-2=0
c)x+2y-2=0 (d) x + 2y +1 =0 151. The centre of the circle passing through the origin and cutting off intercepts a and b on the r and y axes,is
circle will be 140. The
limiting
points
x+y+ 4x +9 = 0 are
of
the
coaxal
a(a) The circle r2+y2-2ax -2ay + a
circles
=0 touches
the axes of coordinates at. and. (6) The line (x-1)cos 0+(y- 1)sin 6 = 1 touches the circle whose equation is for all values of 0. 2
The equation of the circle passing through the Points of intersection of the circles xi+y'=6 and x+y6x+8= 0, and the point (1,1) is
(d) (a, b) 152. A square is
inscribed in the circle
x+y-2x+4y +3 = 0 and its sides are parallel to coordinate axes. Then one vertex of the square is
(a)(1+v2,-2) (b) (1- 2, -2) (c)(1,-2+v2) (d) none of these 153. Two circles (x-1)*+(y -3)= r and
2+y-8x
+ 2y +8=0
C-70 Problems Plus in IIT Mathematies
(a) 2 0.
i.e.,
,2a 20is
Selected Solved Examples
m
y mx+ 6. Condition for general equation of the second degree to represent a parabola
1.
.The line y= mx +c touches the parabola
The equation +2hxy + by+2gx+2fy +c =0 represents a parabola if h2= ab, ie., the second degree terms form a perfect square provided
The equation of the normal at (ri, y) is -2a
equation of a parabola into standard
form
equation (y B) = 4a(x a) can be reduced to the standard form by the transformations X, y-B=Y. The equation becomes Y= 4aX. which is the standard form in X, Y coordinates. (y-B)4a(ra) is the form of equation of a parabola whose axis of symmetry is parallel to the -
-
-a
:
V+T*
Note The lines ax +by+c= 0 and bx-ay perpendicular to each other]
X
The equation becomes Y'=J
+c'=0 are
axx +hry1+7y)+ byyi+8(r +X) +Sy+yi)+c=0 equation of the normal at (, yi) is The .Ash 1 (-x1)
y-(dy d
y1
daT
10. Chord of contact, polar line, pole
Let the equation of a parabola be y*-4ax = The chord of contact of tangents from the exterir point P(r, y) to the parabola is
TEyy-2a(a+z)=0
which is the
8. Parametric equations ofa parabola
T=yyh2a(x +X)=0 .The pole of a line L=0 with respect to the parabola
y=2at
are the parametric equations of a
the point (r y1) whose
parabola.
Any point on the parabola coordinates
y'=4ax
polar is the line L = 0.
has the 11. Chord with given
middle point
The equation of a chord of a second degree curve whose middle point is (z1, y1) is S, =T. So, for the parabola y- 4ax =0 it is 9.
Tangents and normals
i-4ax,= Yy1-2a( +1)
Let the equation of a parabola be y= 4ax.
.The equation of the tangent at (u y) to the parabola is
Y912a(x + x,) .The equation of the tangent at («t*, 2at) is
ty=x+at
**
-1)=-1, ie, a =B
Hence the equation of the parabola will be of the form
when a, rectum. (1)
S(1.1)
Solving (2), (3) we get Let
M-yi).
=
p=
So
12.
Diameter of a parabola
The locus of the middle points of parallel chord parabola is a line which is called a diameter
parabola.
=4a(4-B),
a= 4a(4 P)
.2
and
(1-a)" = 4a(9-B),
and
ie., 1-2a +a= 4a(9-B) (-2-a)*=4a(6- P),
V=f
i.e., 4+4a +a= 4a(6-B)
0, As Mis on
=0. So M=(0, 0). the directrix, (0, 0) satisfies (1).
Hence,=0. the equation of the directrixis x +y=0. Using focus-directrix properly, the equation of the parabola is
(3)-(2)
1-2a = 20a
(4)-(3)
3+6a = -12a
3+6a-12a
or
-3+6a
or
24a=-18;
(2)
equation of the parabola whose axis is to the y-axis and which passes through the points (0, 4),a,9) and (-2,6). Also find its latus rectum. ASthe axis is parallel to the y-axis, it will be x - a =0 for and the tangent to the vertex (which is
a
5(3+ 6a)
15
+30a
a-
a-
16 4-P or
-4xVS4V the ind parallel
=
1+-20a;
(5)
+4 0
The latus rectum
=
or-3(1-2a)
or 2( 1)+0-1))=(*+y) or 20-y-2-2y+2)-x+y?+ 2xy
or y-2xy-4x-4y or +y-1) y=4(x
4
.
1-2a204
(1-1-7
9-4-
ie,
a
51 40
8
4-p-9 L
from (1), the equation of the parabola is
Or
or
(1)
unknown constants, 4a being latus
a are
passes through (0, 4), (1, 9) and(-2,6). So
ie,
a
B,
(0-a)
As MV= VS, V is the middle point of
.
(-a)-4a-P)
(3)
MS.
The polar line of the point Pr yi) with respect to t parabola is
at
y-p=0
directrix
V=(a, p). Then a + B=1
78
standard form in X, Y coordinates.
r
Now, V is the foot of the perpendicular from S(1, 1) to
Let
some p.
=0 for
Y-0
theline x+y=1 and
B
.( (1)
directrix
Ser
a+b
the directrix will be of the form x+ y=a
-
x-a
directrix is parallel to the tangent at the vertex V. The edi
The equation of the tangent at (r, y) is
-axis.
The equation (ax+ by +c)'= bx-ay+c can be reduced to the standard form by the transformations ax +by+cy bx-ay +Cy
,
The equation of the normal at (at 2at) is y+ x 2at + at3 Let S ax2+2hxy + by+28x +2y+ c=0be a parab
y perpendicular to the axis) will be
Jatus rectum.
u
yma
abc+ 2/fgh-af-bg-ch0.
The
if
so any tangent to the parabola can be taken as
ax
7. Reduction of
Find the equation ot the parabola whose focus is (1, 1) x and the tangent at the vertex is +y=1. Also find its
-2=0
5)
C78
Mathematics Problems Plus in IIT
2x+3x- y+4 =0 and its latus rectum = 4a = 4
;
The point (E* rcos
6, Sin 8) is
on y= 4ax if
the vertex, focus, axis, directrix and latus rectum of the parabola y*+4x+4y-3 =0.
Now, if
y+4y +4 =-4x +7 or+2)4
Y, -1
ie,
o
there
are perpendicular lines
X, Y
vertex= (0, 0)x, v ie, in X, Y coordinates the vertex is (0, 0
when X-0-*=Xgivesxwhen Y
or and
=-43-
x
16x;
1,-x=X gives x
Xta=0,
when Y=0, y+2= Y gives y= -2
or
The equation of the axis is Y = 0, i.e.,y +2=0.
100
ie.,
thedirectrix is
= 4
x1 =4.
that
represents a parabola. Also find its focus and directrix.
.the
=
(-12)-9 16 144-144 0. Also,A#0
equation represents a parabola.
Now, the equation is (3x -4y)* = 5(4x + 3y + 12). Clearly, the lines 3x-4y =0 and 4x +3y + 12=0 are perpendicular to each other. So, let
v43=X
The equation of the parabola becomes
y-X4x
2ak
sin 6 fk=2a 4a2
IP
(
(2)
atj.2at,)
4x +3y
is the same for all positions of the chord
a(atž.2at2)
Solving (1) and (2) we get the coordinates of R. 6.
For what real values of a, the point (-2a, a + 1) will be an interior point of the smaller region bounded by the circle+y=4 and the parabola y? = 4x?
The point P(-2a, a + 1) will be
thecircle+y-4
53
+0.
0
and
an interior point of both
the parabola y-
4x
=
:
(1)-(2) or
chord PQ passes through K(k, 0) where :Suitably taken. Any point on PQ is (k + rcos where tan 0 is the slope of PQ. The equa parabola is y2=4ax
ie.,5a2+2
k
), Sin
*
a
(2),
0-xt-
h) +atzl2t1 -t)
x=at
f:th+t}
R= {at,t,
at
+ #)}
the area of APQR
(-20)+(a+1-4
positions of the chord. he
x (1)-
C:
0-x-4)+ alih-tšt)
y24
a is the same ur
a(t-)
y-t)=
y a(t, +)
From
0.
Prove that for a suitable point K on the axis of K, am parabola the chord PQ, passing through
and
b) cut
23.
Ellipse and Hyperbola
65.
is
The line 3x +5y = k touches the ellipse 16x+25y'= 400 if k is (a) + Vs
(b)t V15 (d) none of these 66. The equation of a tangent to the hyperbola 2-2y= 18 which is perpendicular to the line *-y=0 is (c) +25
(a) x+y=3
(6) x +y+
(c)X+y=312
(d) x +y+3v2 0
3= 0
C-110 Problems Plus in IIT MathematicCs
67. A
point on the ellipsex2+3y2 =37 where the normal is parallel to the line 6x 5y (a) (5,-2) 2,is -
(b) (5,2) (d) (-5,-2)
()-5,2) 68. If the
ordinate of the point of contact be 2 then the equation of the tangent x2+ 4y2 =25 is to (a) 3x +8y =25 (6) 8x + 3y = 25 (c) 8y 3x =25 (d) 3x-8y= 25 69. A tangent to the ellipse 16x2 +9y2 =144 making equal intercepts on both the axes is
(6) y=x-5
(a)y=x+5
(d) y=-x-5 ()y=-x+5 ellipsex*+ 4?1 6 to 70. If the tangent the to the circle + y' hep P' is a normal 8x to -
then o is equal (a)
n/2
b)
/4
(c) 0
d)-/ .
Das (3, 1) and (1, 1)passes An ellipse having foci thtog the point (1, 3) has the eccentricity
71.
(a)
v2-1(b) V3-1 ()
Answe1S 1.
5x+9y2+30x
2.
9x2+ 4y2 +36x -24y +36 0
3.
12x-4y2-24x+32y
5.
-
18y
41.v2 42. 5x2xy+ 5y76x
126 =0
-
127 =
0
vs+1,2)
6. (1,0), 23
6,23
33)(3v3)
2
10. a-2,
b=4
47.
hyperbola, (1, 2)
2x 3y -3v3 0 51. 2bva2-b2 unit2 49.
16. y-2
xt
tan
=0, 8x +5y -26 0;
aa 2a2-b7
b 26-a)
V2 + V13
y +2+ V13 =0
x+y2 =a2+ b2 29. c(x-a)-2xy = 0 30. 32. (a) 33.
-10a'y)+o*+a'y)-o 16x+ 10xy +y'=2
=4
37. (b) (x+y)*=16x2
8v2
5
espective
1
48.2 2x-v3y 50.
= 2v3
tanvb/a
52.4
53. 1
26.
(x2
82
= 0
(V61 45
5
24.
4
1
43.
88y+506
55.
a12-bm=n2
57.
rectangular hyperbola, v2.
58. 60.
59.0
25, V5
-92 40. hyperbola, (-2, 2)
61.(C)
(i), (a)
62. (a), (b),
34. 4x-y'=
56.
(), (d)
63. (c)
64. (a)
65. (c)
66. (a), (b)
67. (b), (d)
68. (a), (c)
69. (a), (b), (c),
70. (a), (c)
71.(a)
()
CalculuS
1.
Function
Recap of Facts and Formulae 1.
Function and its value
But F(x)-log x", x e Rand t(x)
BR is a real function then IEf:A B to which, corresponding to each fis a rule according a realf(x) € B. is unique xcA there fla) is the value of the function at x = a, i.e., x =a eA corresponds to f(a) c B.
4.
function may be defined in any one of the following ways: (1) uniform definition G1) piecewise definition (ii) general definition given by a property of the function.
B then x is the y=f(x), i.c., x e A independent variable and y is the dependent variable.
Domain and range of a real-valued function
Ify =f(r) be a real function then domain off= the set of real x for which fx) is real range of f the set of real values of f(x) for
For example Let (a)f(x) = x2+1
(b)g(r)= 2x -1,x (c) hx+
In (b), the definition is piecewise. For negative values of x, gtr) = 2x - 1 is to be used while for non-negative values of x, g(x) =X +3 is to be used.
¢(x)+ v(r) is D,nD
In (c), the definition is general. The function is described by no rules but by a property of h(x).
DnD-E the domain offt) =is y(x)
Clearly h(z) = e obeys the property, but there may be other functions satisfying the same property.
where E= the set of zeros of y(x). 5.
Equality of functions
flr)
function flx)
i.e.,
d(r) for all x e the common domain For example = x> 0 f(x) = log x2,x>0 and o(x) 2log x, same are equal functions because they have the +o«) we have domain (0, +oc) and for each xe (0,
f(r)
Some special piecewise functions
Modulus
Two functions f(x) and ox) are equal if () domain off= domain of and )
=
flx) =
=
=
hty) for all x, y eR.
xe
the domain off(r) = ¢(x) x y(x) is D, nDa
3.
y) = h(x)
In (a), the definition is uniform. For every R.f(x) = x*+1.
continuous functions. If the domain of ¢lx) be D, and the domain of y(x) be D then =
0 are not equal because they do not have the same domain.
where A c R,
=
Ixl,
x>0 0,x =0 x,
-X,
Signfunction
fx)
X
= 1,x
0
0,x =0
-1,x
Bthere is no a eA such thatf(a) =
function. Forexample
positive constant. If k is the least possible positive constant then k is the period of the function. For example cos x, Ix1,x2 1are even functions
Function Period
e
R to R
functionf(r) is evenif f-x)=f) for all x e domain functionf(x) is odd if for all x e domain f-)=f)
functionfx) is periodic if
A
B
If there is no such p e B thenfis an onto (surjectiyel
Even, odd and periodic functions A
Funcion
functions 8. Into and onto (surjective)
Ilx]-31llx]+2]>0.
Using sign-scheme,
Composite function
If fx) be a function whose domain is A and range
Is
and gy) be a function whose domain is B and the range is C then (g f) is a composite function whose domain A and the range is C such that (g z e Cwhere
is
f)=y
and g)-
toe
of))=
Forexample Lety=fa)-2+3 and gt)=sin then(g fM*) =glf(r)N=89) g(2x+3)= sin(2x +3).
2 ebefunction satisfying o
and f)-
xl-2 65 then find f6%
But
xl2
or
l>3.
x]= -3,-4,-5,..
x0
8a0,
= 3.
satisfies the relationf(r + y)=f)+f()
Iff)
from (1
FO=2-+N1+
2-212-1)50
0
and 9
6. Find the domain and range of the function
O
Now ain
Hence its
ie,a246)2-4(9+8a)s0
of)RE
20 32-(22-220 (25-3.2250 or
8.
This has to be true for all real y.
and1
range
Now 3-2-2120
or 91-2y+y9+8a +(a+64)y 8ay220 or 9+8ca)y+(a+46)y + (9 +8a)20.
(See Limit)
+3
or 5x-6
the function is not one-to-one for a
36(1-)+ 4(a +8y%8 + ay) 2 0+
forreal x,f() can aftain all real values except
and
D-7
ie., x=X
Asris real, D20
all real y(# 1), z is real, and
fO=lim
Let dom=D, domy=D Then domfsDoDa
So,
t8y)}+6(1y)x
3+1 for
- (8 + ay) = 0.
+3b (f()-ift)P13
for all x e R then prove thatf(r) is a periodic function.
Here,
Ifa + x) b3 -
=b+1-3bF)+3b1f(*}-{f(*)°
=1-fy°-3b
1fr)+362 f)-b}
=1-f)-b3
fla+x)-b+ f)-b=1 This is true for all x.
Putting a +x for x in (1), we get,
.()
Function Problems Plus in !IT
D-6
Mathematics
vsin.
fr)-v3-2*-21 and vr)=Vsin
for
D,, dom y = D2. Then domf=D,
Let dom
6.
or
3-2-(2)-220 (2)-3.23+2s0
or
(2-202-1) By
Find the domain and range of the function
s0
2-x20
rs2
1+z20
r2-1
or or
2e D
Now, sin
ie., 2°s2
s2;
0Srs1
r20
0ssin
zs
the principal value of sin"r is positive in the first quadrant}
When
domf
D,
5. Find the domain and range of
fa)
x-3x+2
+-6=0
or(r+3)Nx-2)=0,i.e., x =2,-3 off= R-(2,-31.
Now yf)
(*+3)(x-2)
f-3)=*
and
2
(-2-1-
6-)8(r-
=2
(6-8x
Or
-N(6
fis
Asf:R->R, the function will be onto if
OC
+6x8x2
-8)
-(-24x -30x? + 45x, +18)
a+6x-82
+6x-8
-8
N-24x-30x7+ 100x-48
a6x-8 Let
6(-) x)
(3x+6)03+6x1 -8x)-8x,(3x+6x-8) = -3x,03+ 6x -8x,)) 0
onto. Find the interval of values of a for which thefunction one-to-one for a=37Justify your answeh
for real values of x.
-(-66x, +55)] = 0
I55x66-z2(-66x,
or
55x,
2 5n
=
Now, putting x
81,N3x? + 6x, -8) +6(3 + 6x,-8x)) =0
-8x,)(3xj+6x
f(n)
1)
=5r
=
5(1 +2 +3+...
+ m)
+ 1)
{m
(3r+3
art6rR>R is defined byf)=a+6x-8r
can assume any real value
2
1363-)+6(- )(3+6x-8x)
7. A functionf:
X+3
=f(1)+5(r-1)=5 +5(r-
5 n=l
601)-8(«- xM3N? + 6r, Or
range offr) =[v3, v6].
x*+*-6
domain
x
the least value of flx) =3
Hene,f)--3x
flx) is not defined for those values of r for which
3t-+
I2
the greatest value off(r) = v6 and
*+X-66
-fr-3)+3.5
=f(r-r-1) + (7-1)5 because by (2), a
(1)
=...
(using ratio and proportion)
when-=5ie, =
+64)
3+6=9 R a-3,f)=
the least value of f(t)} =3 +0 3,
11.
. using (1)
=fr-2)+2.5=fr-3) +5}+2.5
37+6x-8 3+6x-8 3+6x8xi 3+6x-8x when
n D, = [0, 11n [O, 1] = [0,
fr-2)+5)+5,
=
fusing the given property)
1)+5
fr)=ftr-
Letfn)-fr). Then
thegreatest value of f(a)=3+2.5
D= 10, 1
sin).
"
0Sxss1
.
=fr-1)+f(1)
Bysign-scheme, 2 Sa s14.
-3+2
f:
8
a-16a +28) 50 (d+8)a 14) a-2)s0 (a16a
=f)+f(Y)
Heref(r)=fir-1+1)
... (2)
a-8
3+212+-
10, 1.
.
(a-14Xa -2) S0
domain off(x) = [-1,2]. Again, fz)}=(V2-x+ V +x)2 3+2N(2-rX1 +*)
21,
a>
1e,
+ y)
Also prove thatf(r) is odd.
from (1),
Domain of flr) = {x 12-x20 and 1 +r20).
sign-scheme,
1,
9+8a >0,
relation f(r
for all x, y e Randf(1) = 5 then find
s (a446+18+16a) a+46-18-16a) 0
fr)=2-r+vi+z.
Now,
8. 15 fxr) satisfies the
n-1
(a+46)-4(9 +8a)'so
ie,
Now,3-2-21-*20
3-2-20
the function is not one-to-one for a =3.
true for all real y.
Hence its D s0 and 9 + 8a>0
and
and
or
=
ay) 0.
+y)+
This has to be
rangeofi)-R-
={r eR1 sin z201.
(8 +
or(9+8a)y+(a*+46)y + (9 + 8a) 2 0.
(See Limt
real x.flt) can attain all real values except
for
-
36(1-y)+ 4(a +8y)(8 +ay) 20 +8ay 20 8a + (a +64)y 901-2y
or
1
= lim
nD
D, =(r eR I3-2%-2-*20} and D
all real y(* 1), x is real, and
fo)
v)
fr)-)+
So
So,
.
Ao y-3
)-V3-2-21-7
Let
at8y)}x* +6(1-y)r Asrisreal,D20
SeeLimit
4. Find the domain of the function
+55)] =0
-66 =x(55 -66x)
=
0
andy =0 in the given relation,
f0)=0. f0+0)=f0)+f0)% Also, putting-r for y in the given relation
for-)-f() +f-) +f(-r)
f0)=f) 0=fx)
+f(-x), ie.,f-x)=
-fx)
So, the function is odd.
9.
If a, b e R be fixed positive numbers such that
fla+
=b+ b°+1-3b fa)+3b (f*))*-1ftr)}°13 for all r e R then prove that f(r) is a periodic functio
fa +x)- b3 -b+1-3bf)+3biftr))2-1fN =1-f))-3b 1f)* +362.f) -b)
Here,
=1-ft)-b)
fa +)-b)3+ift«)-b)3=1
This is true for all x.
Putting a +x for x in (1), we get,
. .(1)
Problems Plus in
D-8
.(2)
f(2a +x)-b}+1fla+x)-b}=1
and glx)=Vx+1 forallr
f(2a +x)-b-iftr)-8}3=0 = or f(20+x)-b} lf«)-b13
f2+)-b=f)-b f2a+)=f)
or or
f)
is a periodic function.
MfetrM= hr)=x
because x20 lfrom definition of hril
Now,flst))=x for allx gK)=x for allx.
Asx20,(f8*)
the function h is the identity function.
Herefa)= -1
z
So
1
hat'm
ysin(-8)
18.f)=log-s3x-10)
F
is not
21.
Find the natural number a for which
fla
=xis an identity
the function f satisfies the relation +y)=f() fY) for all natural numbers x, y and alsof(1)= 2.
fx
Find the domain of the followingfunctions:
st)= log(1+v1+r)and ht)=f)-8)
then prove that h
7.
4. In the figure, BC = b and AH =h where AH L BC. If EF =x where EF BC then express 1 the area and
perimeter of the rectangle DEFG as functions..
9.f(x)=
VIxl-
+
INr-3
2.f)-snoa yscos4sin
14.f)-151xl--6
if f() is any function, f() +f-) is an even function while f(x)-f(-r) is an odd function. Hence, prove that every function can be expressed as the sum of an even function and an odd function.
fcos x).
27.f)=lo8103x4x +5) 28 f)
31.
x-2
is a bijective function.
product of two odd functions is an even function while their sum is an odd function.
41. Prove that the
42. (a)
f
1+
(b) Prove
Ix-11
is an even
thatf(r) = v1-x)
+
(1+)
43. If the functionfsatisfies
the relation
fr+y)+fr-y)= 2f)
,
f)
for all ye R and f0) + 0 then prove that f(r) is an even function.
2x
2 fX-3x+4 +4T 1)
and domain
=
R-11,4)
trenfind the range off(7). .
5If00,n eN then show thatfiftr) = x. 3. If fr)=log{a + v+z),
2
Check whether f(x)1+
onto or both or none.
function of r and indicate the domain of definition.
k=1
2
8x
+ Vsin(cos )
sin+2-x-
24.f)=cos
cx +d
Ifd=-4, show that ff}
+ 18 an one-one function. 37. Verify ifflr) iff{7) = + 37. =2 4x +30is
39.
23.
25, 1f log,y 6.
+2
all
y
(ii)f(2) *a.
(ü)f(y)=b
Find the functionf.
38.
1
22 fEcos
Exercises
1. Let fx)=
{x, y, z} > {a, b, c} be an one-one function. It is known that only one of the following statements is
true:
f)=3-+cos
But, by definition h(r) # x for x 1
)fz)
=
*z, 0 Srs1 -1.x>1
periodperiod
5
b) 15fx)- 1, x>0 0, x=0
.3/w2] espertively 72 R-2 spectively 74-1,1 73. (4, )
83. (c)
81 (c) 85. (b) 69.0(b)
92. false
3. true
84. (b), (c)
82.
(b), (c)
86. (c) 90. true
. true
then which of the tollowing are true?
-1, I0
[1.+)
80. (a), (c)
49.period
range
-.
there ate multupe optans Choe the correvt optiun(s) ne Nba perwdic function w hoe perioad is D none of these A. indeterninate B1 C.
9
47.period
off.
domain
of folkowinyg,
7
75.
5ten
inteu
bexotes true
(o) fu)1-bl*n,
ifr=-1,-350. Also find the set of
a is and a is a constant. cons and
8. Letybe a funetion of r givenby 101- B1y -2-4r- 2y Two-one and among the two eal values of
x
5. Prove that the function is that correspond to a real value of y, oneis
s-and the other is 2 9.1fr-41+3
0 where r
=
Vx*
+vi,
*e
R.
ye
R
then find the greatest and the least
values of r 10.Let ftr) =21 x,xS1. Prove that the function f"(r) exists and find the solution-set oftheequationf "() =f). -
Problems Plus in IUT Mathematics
11.
=x, *0. Find the domain and
a Letf be function defined by 2f(x-1)
range of
the functionf(r).
2. Differentiation
Recap of Facts and Formulae
1.Definition of
differential coefficient
d(tan
14
dx
1fy=f) be afunction then
differentfal coefficient ofy wrt. r is
o
f-
lim
+h-f)
or
fa)
= lim
a
_xl VE=1
d(sec
0
d
o diferential coeficient of y wrt. x atr =a is
+h-fla)
dcosecx ax
)dcos
Note
2. Differential coefficients of standard functions
(1/1) dx
using chain rule
de")
xV1-(1/**)
alOg
wherex>0 3. Rules
If a where a > 0
log
for differentiation
y=u() + v{r) t w()t... then dy du(a)+ dv(x) , dtu) (term-by-term differentiation)
cos Differentiation of a term:
dcos )
2sinx
1.
(a)2
2. (a) B
(b)[,41 ()R-rlx= nm+(-1)"2+sin (b)A,
B,
C (¢) B, C
6.
= domain ={b, al, range [Na-b, V2(a -
9.
greatest value = 3, least value =1
11.
R--11.
R-
b)1
respectively
-
10. 10, 11
ax
2) an2sec
Answers e
6ec d
dcot23
d cosec )
O
sec
du)v(r)}
tan x
COsec
osecX
=
d
du(?).
dx
d
D))
cot x
v(r) + ucr).
T-
otwr D-15
do(x) (product rule)
v()-
dv(x)
u()*dx
quotient rule)
tvcx))
dulx) lutvon dulvlr), dv(r)dx
dGin
ax
x2vx-1
(chain rule)
dulv{wtr)}1, dolw(r) dw(x) dofw(r)} dux) dx
-16
Problems Plus in lIT Mathematics
Diferentiation
of f)} *
dx
7.
wrt. x taking b) asa
constant+ d.c.of (ft)* * wtt. takingf(x) as aconstant
Higher derivatives of a function If y=fx) then the derivative of
x
i
D-17
wrt.
x is called tha
1
second derivative of y w.r.t. x and it is denoted by
0
lim
4. Differentiation of function represented parametrically
using
If yis a function of x given parametrically by ai y= 40), x=V) then
is a function of x such that x= 40), y = v() where t is the parameter then dy
1log
2.
im08 cos
5.
If
y,
.
andz = v() then
dzde
ie,
:fy=P()
1v-
a)
a function
given in the form
of a
d
=a"nx"-1-ny"-1 4 dx
P) A()
2h
w'x)
4) H(*)
rx) Vr)
u(x)tv(x) P ) 9'() A(r)
r 21-1
ieh wx)
r't v{T)
H(X)
u(r)vr)
w)
(r)
v '(1)
P(x)9(x) (r)
or
im
h0
r()
0g
cOS
los cos 4h-
Selected Solved Examples
los
at
25in
Cos f() in
h0
cos
Now, we have
2+2
or
4h213cos 4h 4h
sin 4h
or
-sin 2h+V3 cos 2h
lim
Now
Vi-x2n
V1-27
=y-1V-y V1-y2
sin4h
logcos 2hsin 2
m
*from the first principle.
x=is
-1,ta".
differentiation can also be done columnwise.
The
where h is a very small positive quantity, remains the same. So, not a turning point of definition.
2ny
fim V3 sin h
t sdtsis
u'()v'()
')
Clearly, the definition of the function
-y
-
hsin
0
r)then
A()H()v)|
fr)=logec z lcos 4xl +lsinxl then find
V1-y" =a"a"-y"» prove that dy=x"-1 dr
Differentiating both sides w.rt. X,
lim
u(X)(r)wr)
6. Relation betweenand
1. If
If V1-H +
2-1
Differentiation of one function wrt. another function
y=)
cos
Now, Hm 0 Differentiation of determinant
-log 2
d
ie, 8.
log,
definition of mod)
cos 3+4
Ify
dz ddxv
6v3
imsin
dx
+for)*).log ft).49) (rule of logarithmic diferentiation).
4v3+213
m 0
= lim h0
-E-1-in
cos 2h + sin 2h
22
2
in
sin 2h+ v3 cos 2h v3 +*-cos 1 2h sin 2h 2
(using L' Hospital's rule)
V1-x2"+V1-y" =a"x"-y") (1-)-(1-y) V1-y -a"-y"riy2-3a "(a"-y"V1-x-V-y
V--v-y or
+x")
a1-x+"=a" v1-y
-y"
Using (2) in (1),
V
Writing in the differential form,
V-T
dy=x"-
Vi-yd.
2)
Problems Plus in IIT Mathenatics
-18 3. If
Diferentiation
Sinyo
sin x and g'() = cos *
u=fr*), v=gt*), f')=
cos
du
(osin yo)
sin(r) 2x
d=f')-2r
:f'l)=sin
f'r)=
x
sin(r*)
or
x'h" 7.
cos(r *)}
h)-2F) Now,
3x cos x3
3r2
(1
diff.coeff. of u" wrtxwhen v is
If fgh are differentiable functions of x and
A)= |(
a constant +diff.coeff. of u"wtt xwhen u is a constant.
8
(g
8. A functionf: R ->
in +(sin y)2.log
hyprove that
express y as an explicit function
6.If 2=y+ythen x
of
and prove that
dy
-1)
2
dx
A)
sin ycos
f+f) 2f+ 4sf'+*f"
g+8
Differentiatingftr + y) =f)
h
2h +4xh'+x*h"|
(=2f+xf ("=2(+ f)+21f'+x
+2))
2 sec log +2)
0
xh' 2xyf
2g+g"
sin yo
20
70
F)
xhRR-2R2
or ... (1)
wrt 2,
Nx-1)
taking the common factor x from R;
5y
fusing
-1
f(x)
But f0) 0
(1)
f0) =
0.
(given); sof0) =1.
Puttingx= 0 in(1),log 1 =0+c;
-25
f)
integrating, logf(«) =2x+c Again, putting x =0, y=0 in the given relation, fof0)-1)
()
=f'0)
f')-2
-
dfferentiaingagain
So=0. f)
t:f'0)=2
f')=2F)
f0+0) f0) R,
20.ysin yo)
fY) wxt. x,
Putting x = 0 and x for y, f')
2xth'+xh"
RR-2R2,R2 (sin yo) log
ye21 -yj-i-y) Differentiatng
=f)
Thus we get ft+y)
2x2-1
h
Puting r=-1, we get (assuming y =Yo when r=-1)
But y is independent of x.
4x-4fyi-2
2 2g+ 4xg +x'g"
4
+2Mog 2 tan log,
=0
Rsatisfies the equation given by
Fa+y)-1+-Fs)+f
+ xh
12x1
f")= -f)
fr+y)=f) fy), for all x, y in Randf() #0 for any x in R Let flr) be differentiable and f'(0) =2. Show that f')= f(*) for all x e R. Hence find f().
differentiating the given rela tion w.rt.z,
2cosy
t:
h 5.
y)
'()
h
hx)= 11 h(10) 11
We know,
sin(sin
.. (1)
'
t)
)=2f0) 80)+2g6) -f) hx)=c;buth(5) = 11. So 11 =c.
h'
+sec(2)+2"tanlog(a + 2y=0.
duax
2gt) g
-f)-8
Hence,d
Findatx=-1 when
(sin
f)+
f')=8)f")=8
h
dx
y
find h(10).
sinyo
cos(r*)
Letfbe twice differentiablesuch thatf"(a)--f) and f')=gta). If ho*) ={f*)}+{st>)* where h(5) 11,
Differentiating h(r) = 1f(x)}*+ lg(«)1* w.rt. x,
Jo
sinx
sin(),2x2
do
"
(sin yo)
Sin
8'(x)=
cos
8')= du
0forthenyis
e-2+tan log,1 =0 h'
83r=cos(r)-32
4.
-125
Now, putting r=-1 in the original relation,
Differentiating u =f(**) and v =g(*) wrt. x we get
do
D-19
or
h
then find
dudx
h'
c0.
logf)-2x
w.rt
fo)=e dx
Note If the differentiability of fx) is not given, thee correct method will be as given ahead.
Problems Plus in
D-20
ft)=
lim
f'(10)=1.
h
flrNf(h)- 1 10. Let
h0
=f(r)*lim
f0)
f(x) lim
: fr)
f29*
F0)= 1as obtained above)
f0)=2f(r)
fa
(takingx
lim lim
if x>0,f(r) 20
7Sigith
Differentiating w.rt.x,
2
Integrating
f-0
flr+ 1) =flr)
Thus (3)
-
=
f0)
bgiven) (1)
f) +f inf2 2e
Again, putting y = 0
f)+1)+3
-f)+4 1+4-5
2k+
-
(2k- 1)sin(2k-
sin(2k
-
1)sin(2k-
1)x
1)x
1}x -
1)xl]
(2k- 1)sin(2k + 1)r|.
-
cos 16.
a
Show that
v=+Bis asolution of the differential
.=
equation
Fndthederivatives w.r.t. x from the first principle atthetndtcated points: 7. xanr at a=1
18.
atx=2
lo
**erengusa
a)Cr
'
y= Va-r+log
20. sin
21.
b cos
-
Findif yisequal to:
l*,x*0 findf'. show that
ntati d
Cx
(x
ax showthatdi
atx=e log
-b)
-t
rthat
19. If
+b
f')af"(t)=0,f""()=0, F_0.
Ify =1 +
where A, B are constants
a VaT
log
ty
8 log xl atx-2
5.1 f)
0,
equatodr3rdr
e
12. sin(log )
2cas a, where
A, B are constants.
x
from (1),
r is a solution
dy of the differential equation y +
11.1)+ atr
fr)ax
+ 1)x
+
15. Show that y = Asin x + Bcos x+ Xsin
rcos
10. cos ata
we get
Hencef(1) = 1.
f5)=f(4) + 1 = {f3) +1)+1 f2)+1) +2
+.
dx
9x-1+1-3]
ie., fx) is identically zero.
fa)0.
sin 2kx sin *
prove that 2y*
wrt
f(l)=f(2)=f(3)=... =0, Sofrom (3),fr+1)=f)+1
1)x =
t cos(2-
7.
b=c
r
-Isin(2k+ 1)x + sin(2k
X
(given).
When 0,f(0) =0 +c, ie.,
Putting y 1 in (1), for all real x,
fa+ 1)=f)+ f1n"*1
5.
0)=a
sin
2kx
2sin
Vsin
r
ts. (2)
or 1, {from (2).
cOS KX
Diferentiate the following functions w.r.t. x from the first principle:
3. cos (eYan
=0
fu)-f)*"]
f)=0
2
Exercises
Taking = 0 and x in place of y we get
Puttingr=0,y= 1 in (1), 2n*1 Ta) =f0)+if(iy
12kcoS 2k Sin x- sin 2x cos
=x, x^ =y, 1 =2}
t
0
+(2k- 1)sin(2k 1xl1
x,
This holds for any real x, y. So y is independent of x, ie,
f0) =f(0) +f(0)
)-F0),,
1sinr +3sin 3x + 5sin 5x
et 2sinksin(2k X
Sin
1.
F)20
-
1)x. =
sin
cos r+cos 3x + cos 5x +.
f)+f2
Differentiating w.r.t.x,
+(2k-1)sin(2k- 1)x.
sin x +3sin 3x + 5sin 5x
sinRx
= b,
fo)=0.
or
S
From the question,
Letf) be a real function not identically zero such that +y)-f)+FYN**neN and xy are any real numbers andf (0) 20. Find the values off(5) and f'10). .. (1) Here,fr +y"* ')=flx) +1fty)) 7*
t..
D-21
sin
+ft,
where x are any real numbers and n eN. 11f)is diferentiableand f'(0)=aandf) find
9.
Putting x=0,y=0we get
+
Let S cos + cos 3x + cos 5x +..+Cos(2kHere the angles are in APwhose first term common diff. = 2.
f)+f(+f)+.
fh)-1
h+0
11. Find the sum of sin 3sin 3r + 5sin 5r
f=1
f)=*
f)-fh)-f
n0
Differentiation
Also (4) = ft)is a function whose value increases by 1 when variable r is increased by 1.
+h-fr) h
lim
IT Malhematics
dx2
(a
+b
cos )
VI -*
-va.vT-
cos+12N1-x
22. (xlog
x)og
log x
where
00 for all x then exists a functionftr) such that (a)ft) «0 for all x
b)-1f") (a0
)stt)=0
log a/1+
s+log
(d)
for any realx
for the cuve 2y=3-ris
37.
2yx V1-r
)-7
(a-
then (1)=-
90.
(a)1
+y-5=0
-2ry+5r
Thenaty=1is
Choese the correct option(s).
The value of
Dittrentiation
and at x=1,y=1. t
85. (
(a) 93. true
83.0
86.(
87. (a)
90.(
91.(a
94. true
95.
true
D-27
Diferentiation
Chapter Test Time: 60 minutes
resulting sentence becomes true. 1, In each of the following, fill in the blanks so that the (a) If (b) If
a
_.
andf)-sinz'then
,
f(x) = (ax + b)cos x + (cx + d)sir
()Iff(x) = og
b=_
and f"(x) = xcos x is an identity in x then
x
,d=
,C.
=and g(x) =fVx)) then g '(4)
|x-1|
r
(d) 2.
d(log x)
Findin terms of xift= 1+x y=x2+t2
function offandf'(x) 3. Ifg is the inverse 1
4.
Ify (1+x)(1 +x1+x)..
5. Ify =fx), p=and
q
6. Letf(r +y) =f(x).fy) xE
(1
=
+r)find
=then
inthesimplestform.
prove thatf "(x) can be obtained at all for allx, y and f(0) # 0 then
equation 7. Transform the differential
and z
prove thatg "() =1+{g(x)]°.
dx a function ofp and q. expressas dy2
Riff°(0) exists.
& Lety
-
dy
(1-x9+y=*by
=Provethat d
substitutingx = cost.
D-28
P'nWems
Pus in UT Mathematics
AnstveTS
-r) 2.sin
1. (a) (a) (c)
2.
4.
1
(d) 2log
rrg
b)0, 1,1,0 respectively *
2x1+x 2x
1-
A
3. Limnit, Indeterminate
Recap ofFacts and 1.
value. Indeterminante forms of values of a function at á indeterminate forms of
The
=ea
x+0
0, o -co; ,0,o 1
X
2.
Formulae
lim (1 +ax)
point are
Form
limits 3. Properties of 1f(x) + ¢(x)) =lim f(x)
lim
Standard limits
x>a
1 lim=0
.lim
f(x) x dlx)}=lim f(x)
x0
where n> 0 lim xl "=0,
lim(x)
x lim o (x)
f(x)
lim
lim
tlim d(r)
lim o(X)
ir
lim logf)
0
lim x"=0if lxl1 lim Ix1"=o if
lim fx))°=
1-
lim
lim o);
fx)*
X
a"
lim
na
1,
.lim
lim flo (r)) =f( li
r
where is in radian
lim
4.
If
x= 1
L'Hospital'srule
f(a)=0 and ¢(a)=0 then
x0
lim 0
lim
1;
where
is in radian
lim
measure
im f
"
X
then If limf(x)=o and lim (x)=0
-1.1
x0
a lim
is continuous
measure
0
cos
())iff()
f)
(x)
log
x0
D-29
lim f)
Problems Plus in lIT Mathematics
D-30
Lirmit, Indeterminate Form
Selected Solved Examples 1.
Evaluate lim +-Vs+r7-Ng +x+
limit
is a rational number. Xwhich n!
.from
(2), limit = 2 ifr is rational. Again, I cos (n! nx)l 0
2.
@X-a Sin
Evaluate:
.
+8-V10-x2
then find the
25.1Efr) is differentiable at x = a, show that
27.Ift,
lim
(vr-x+1+ax-b) =0
Objective Questions
cotx) an
24 l
Exercises Evaluate the following:
-
Fill in the blanks.
y+2y+3yy243
3a-2x4a +3x3a =12;
.(3)
along the
we get x =y+1. As (x, y) tends to (1,0)
When y= x-1,
along
2+a+C-2c=0,
()-0)
y=X-1.
b=a+C
(2)
_1as
Find lim y0
6
(1)
21. lim (2-)
lim
constants a and b.
(given)
12
34-2b+3c
33. If
+ce
x-2x sin x -x°cos x
2a+-20 *20T=2
D
Thus, a =3, b=12, c9.
(1+)
Then, limit = lim
Limit, Indeterminate Form
c3a=3x3 =9.
a(2+x)e+ae 4
E.(L + 2h)-2log(1 + h)
0
n+1abe
finite, find a e R and the
n'cos(,
o0then
lim
State whether the
(b) 0
(a)
1 2
(d) none of these I0
(a)
tncOS )-n sin x
54. If lim
,Tn-
Y)
Iff(2) = 2,f(2) = 1 then lim
a+ bxsin X+CCOS-2 then
56. If lim
(b) 3p
58. If 59.
lim
67.
(40
16.
(d) none of these
20
If lim
9r
=1 +p
(a)
a8 ()8
0
=m
and g ()
Ifftr)= cot then lim
and lim
-
=
x2
cos+
@,0a
3 2(1+x
94+p
(b)
201+a
1
(e) none of these
(a) none of these
59.
lim h0
61.
16fcx)=Vsthen lim f) cos Xt
(a) 0
(c)1
62 lim
x-log(l* 2
is
X (b) 3 (d) none of these
() 0.
lim 0
is equal to
22. e
23. 1
a
27.-
2in-
3h(N3cosh sin h)
(213
is equal to
a1,
34,4
limit
-1
35. 100
43. log3
45.
46. 0
47.-1
50. e
51.
S4
e/3
53.e2
52.
respectively
56. 48,-24,-48
respectively
55. 4
57. 4
58.24
59.5
61. (c)
62.((a)
63. (b)
64.(b)
65. (b)
66. (b)
67. (d)
68. (d)
70.(b)
71.
69. (d)
1)
1
49. e
b=-1
33.a=1,b=
2.
18.
to
)
32.
Choose the correct optiols).
where p, 4 #0 then Im is
21.e
39.2log 2
44. 15. e2 19.e
a
41.
-
12.8
18.1
a cos
+a'cos
40. 0
11.
31. a=-2,
68.
38. 2asin a
3
1
17.
26. 9sin
If fo)=2,ft)=1,ste)=-1,&to)=2 then
)-glaf)
10. co
9.,32
(b)-2
v2-cos-sin eaual
9/4
X-1
.2 14. 8 2(log 3)2
Vf)-3
Gr)s-V25-then lim G-G)
lim
60.
(d)-3p
37.12
36. then
equal to
is
n/2 Cos x
hmVr-3 Vr-3 r9
when (x, y)-> (0,0) along the curve yi=r
Answers
(a) 5 57. If f9)=9,f19)=4 then
=0
40 3-1log3
F)-3f(2)+f(4)
m X
na
y+0
(d) none of these
(a) 2p
X-2
I2
x-0
cot a
then
fr) is twice differentiable and f"0)=p
lim
75. lim
log(1 +)1
x-
1
(d) none of these
statements are true or false.
Xcos a
6)a+b-0
a=b
(c)2a=b 65. If
55.
cos )+bSin x
(1+a 64. If lim
Cot
lim
74.
(a)1(b)-1()0
t dr"+ex9-24h
ra
is equal to
im
(e) none of these x+1
73. lim S0S
3. true
74. false
.
(d)
true
60. (a)
72.
true
4.
Continuity,
Function of Graph Differentiability and
of Facts and Formulae
Recap
.Left-hand derivative off(x) 1.
Existence of limit
.lim h0 is
fla +h) is the
Sa -0) or f(a
and it right-hand limit off(x) at x =a
denoted byfla + 0) orf(a
+) or lim
a+0
r=a if
and it is
fa
fla-his orlim flx). -0 denoted by fla-0) orf(a-) a-0 im
exists iffla im fx) the value of lim
+ 0) =f(a - 0)
f(r) is
ie,
atx =a iff
Clearly, f(a) = lim
cannot does not exist then f(x)
Continuous at x = a
be
i
-
as from the'right). from the left as well
some standard Continuity and. differentiability of functions a), Polynomial functions (i.e., continuous and differentiable at sin x, cos x and é"are 5.
Differentiability of a function at a point
denoted by of flx) at x =a, lim
-a
and examining the continuity point at a differentiability of a function f(x) and diferentiability you start with the *=a, if conclude differentiable then you can you find find that f(x) is also continuous. But if is function the that have to differentiable at =a, you will also not is flx) separately. check the continuity continuity and find that the Instead, if you start with then you can conclude that function is not continuous nondifferentiable. But if you find the function is also check the you will also have to flx) is continuous, differentiability separately.
of
+), is the 0
fa)-fa)
fx)
0) does not arise. In So the question of getting f(a continuous at x=a it this case f) will be right-end a+0)=fla). Similarly, if x =a is the will be Point of the domain of definition then f(*) Continuous atr=a provided fla -0) =f@).
a9orfa
derivative offlx) at
While
domain the left-end point of the =ais then If definition from the left and cannot tend to a
h)-
finite
x=a
x = a.
Right-uandderiva
0
B=fa. -h
continuous at fx) is not differentiable (finitely) at x = a. is not
thenf(x) is three are unequal t any two of the above
dlim.f()
h
(finitely) at
continuity and differentiability Relation between 4. x differentiable (finitely) at =a is f) x=a fx) is continuous at
fla+0)=fla -0) =f(a), f(a-h)=fla). ie, lim fla+h)=lim discontinuous at
lim Tat)=f@h
r >a
Continuity of a function at a point
Note
lim
=a,
value of equal to the common
continuous A functionf(x) is
h0
is called the and the common limit denoted byf "(a).
fla+0) andfta -0). 2.
Tla=h-f(a)
= finite, + 0) =f'la-0)
h0
i
and
-), is the lim
differentiable said to be A function f(r) is
f(x).
atx =a the left-hand limit off(x)
at x =a, denoted by
a"+ax"+.+
a) D-41
Problems Plus in IIT Mathematics
D-42
all points of the set R of real numbers.
is continuous and differentiable at all points of
log
,
+oc).
Continuity, Diferentiability and Graph of Function
all points of R -
10,
,continuous 3,...
and differentiable at
27,
is continuous everywhere and differentiable at all points except atr = a.
Ix-al
.Ixl
21
diocuso the existence of limit at r =1,-1.
Fromthe
=
or r2 Fromtheslgn-scheme, x S-1 Thus,the function is f(x) = x, xs-1
x
f) is continuous (or differentiable) and 4() is Ifdiscontinuous (or nondifferentiable) then f) + 4() and flr)x o(*) are discontinuous (or nondifferen tiable) at x =a.
Fig.
1
Fig. 2
f(x)=
X0. V16+ Vx -4 If possible, find the value of a so that the function may be continuous at x =0.
(1+h} =1
lim f(1 h) = lim
exists and it is equal to
lim()
fris
is continuous in an interval if it is continuous at each point of the interval.
i)
3. Let
VX
= lim
h0
f)
lim
a)
Ifa functionf) is differentiable at x = a, the graph of flx) will be such that there is only one tangent to the graph at the corresponding point. But if nondifferentiable at x=a, there will not be unique tangent at the corresponding point of the graph
fr)
Gi)
=
Whon discontinuouo alx
f(1+h)
f)=lim
202 (Whon contlnuous at x a)
¢(x) are continuous
r
f)
yX-1
x,*21
Algebraic property of continuity and differentiability and 4(x) are both continuous (or differentiable)
7. Continuity and
1.
'-10
fU)
y=f)=cos x-sin x, 0Sxs4
(1+gth) G(h), using given relation
sin xcosX
7
D-46
Problems Plus in
IT Mathematics 11.
*-o)-in
Continuity, Diferentiability and Graph of Function
The functionf) is defined as follows:
D-47
f)-ul+11-xl,-1sxs53
where ll is the greatest integer function. Det the points, if any, where fa) is not differe ntiable. Draw a graph of the function.
h
0
y=5 atr =3.
So, the graph is as below:
Hence, the required graph is as below:
3,5)
(1.2)
Here f(x) is the sum of two functions wh: re themselves piecewisely defined. In cases, at first break the domain of definition: subintervals keeping an eye on the turning o definition of the constituent functions. Here, the domain is the interval -1,3] which is equalto -1,0)u 0, 1)u [1, 2)u [2, 3) u 13).
lim h0
TIP
h
0
2 sin
lim h0
-h
l-N2
f() = r*-x*+x+1 and 8t)= max {f)0Stsxl, 0Srs1, 3-x, 1 0 in b) thenf'(x) is m.i. in (a, n
f)0
he
in (a, b) thenf(x)
.fis
sketch of monotonic functions are as follows:
y-txlis m. in f a.b]
a y
f(x) is m.d.
thatf is
Note A function may be invertible though it is nonmonotonic.
For example y = x,x is rational -x, x is irrational.
f(x) O
invertible is
a sufficient condition forf to be strictly m.i. (or m.d.) over the interval la, bl
is m.d. in (a, b)
yf(
Invertibility of a function
The inverse function is x = y, y is rational y is irrational.
-
in [ a,b]
D-83
Problems Plus in IIT Mathematics
D-84
are real such thatf(@) and f(P) are of opposite then ftr)=0 has a real root lying veen signs
6.
Use of Rolle's and Lagrange's theorems in equations
Iffx)=0 has two real roots a, a (repeated root) then fx)=0 willhave a root a.
functionfr) = 2sin x+ tan r-3r*
(-T/2, m/2).
-3
2cos x +
2cos-3cosr+1 cos x (cos-1x2cos-cosx-1)
t)(-)(+
So,fx)
1
>0for all x e (-z/2, a/2).
>0 for x e (- /2,0) and r e (0, 7/2)
f)=0
forr=0
Hence,fr) is increasing in (-/2, /2. 2. Find the interval of monotonicity of the function
fr)=2-1oglrl,
x*0.
Ifx0
and
(2r+1)(2x-1)>0
cos
cOS 1)*(2cos
"-o=1
d1+24x-3x2, x>0.
(-1/2,0)
0
x-sin x>0
axe,s0
f-" We
is monotonic increasing in fr) monotonic decreasing in (--1/2). If x>0,
or
f)>
So.,ftr) is differentiable at x = 0 also.
when-1/2 0
cos
(cos -1XN2c0s I +1Xcos x-1)
is a non-negative integer) As" (where n difterentiable everywhere,
f(0+0) [1 +2ax- 3
Butr 0 for all r>0, x * mi.
fr)is
*so
constant, Find the interval in where a is a positive which f' () is increasing.
(+)
fx)=x-
x>x-for all x >0.
sin x
flr)=1-cos
Letfr) =re
D-85
5. Prove that x> sin
Let
fa)
ae
is increasing in the interval (- T/2, Td2)
Here, f(x) =2sin x +tan x-3x. Clearly, the function flx) is defined at all points in
t
continuous and except perhaps at is differentiable everywhere
sign-scheme
from
R.
bel,) 4.
and B.
Selected Solved Examples 1. Prove that the
orcos -bs0,i.e, greatest value of cos x =1 b2 the
a
where a 2.
3
fO0,
ie.ft) is mi
f0,ie,f) 1
is
md
ismd
[leastf(), greatestf(7)]
Find the maximum and the minimum values of
-6-7x+10)-6-5-2
The sign-scheme for
=
the greatest and the least values of the
fo -1 rol215r-211 Hence find the range of fT,
unction
ru*
on is continuous and differentiable everywhere, Tis continuous and differentiable in 0 S rs2. Wenow that the maximum (or minimum) value or = the greatest (or least) among values at the critical points and the values at end pointsof definition.
='
,b-b+b-1,0sx ), xe (a, b) then prove thatf(r) > 0 ),xe (a, b).
37.
.
an
[a, b]
If flr) and
gtx) are differentiable functions 0SxS1 such that f(0) =2, g(0) =0, f(1) = 6, 8) then show that there exists c satisfying 00
1except (3)
Now, lim
=
0,i.e.,
a(x'+xi-I+2
stds
perhaps at the turning point of
-1+h-lim-1+)*+(1+b+1 h
.
andf-1) = ji so,f(r) is continuous at r y is an increasing function ofz.
+S g(x)dx increasesaasb-a
Again, lim
increases.
-f-) h
(-1
+h)+1-2
m 0
Let
f)=J
It+1|dt.
Examine the continuity and
diferentiability of f() in the interval
lim
-h1h+1
f-2sxg-1then
to-f-14-
polynomial
f-1-i)=lim-1-h3-(-1--
for all x, implies that g is
gla). dz
are
and
definition, i.e., x = -1.
h0
g(4-)> (3)
f-0
ax*+ax-ax + 2a
bothr
functions which are continuous and differentiable everywhere, flr) is continuous and differentiable in
-2.
-sb) _s(4-a)-gla)
aa.
3a-2b+c=0
f)=
+X+1,-1Sxs1
db
da
an increasing function.
(2)+(3) = Letf:0.-Rsuch that f0) 3 and
Differentiating (2) wrt.a,
As
f-1)=0
and
=s) +stb)
da
relative maximum/minumum).
and
dy
wrt.a. d
dydas(a)
-2SxS-1
fo)-
0.
As
+;
=0
+
Thus, the definition of f(r) is as follows
.(2)
so) +s) d4)=g(0)-gtb).
or
3ax*+2bx + c.
-8a+4b-2c
(1)
=4- 2a
(4-a)-a
dy
thenfr) From the question we havef(-2) =0
(3)
has
increases.
strndr
3+
function f(r) vanishes at r=-2
If>0 dr
a
for all x, prove
increases as (b-)
We have to prove that
Letfx)=ax*+bx*+cx
f
std+stde
=b-a=
26. A cubic
and let g(x) be
a+b=4 where a*as
intersects
y+)SOdt = 2.
l'uttingy= mr iny
J
as |*
o*
f(U)Jdt => **
and
inl is continuous and diferentiable
1f f(1) is
,
= 0
=
mx*
Provethat
xlog sin x dx =
m**+|
syA.
14.
ie.,
o(r)=2,
a +J
fNdt
+ Vcot x)Jdr
(Vtan x
Prove that
log sin x dr =
15.
Prove that
16.
Provethat
17.
Prove
=
18.
Prove that
19.
Prove that
20.
EvaluateJ
n/2
Ifx>
Evaluate tlhe following
log(tan x + cot x}dx = rlog
1 cos
2.
n
o
Odr +x)
=
sin -10n
0.
(1
, 1+ c .
2.
3.
r "a-x) "ur
P'rove that]2
Cos 0-sin 6
1+
sin
+
r"a-a) "dx.
cos d6
Evaluatef2sin COS
=
(2+
22. Prove
8E
24.
Prove that
y=|
dr=[ og0 +*).
X0,,is
sin Vtdt+J
cos
sin x
=
0.
Prove that
2x dr(2sin o Cos 'r+ sin "r
Evaluate the following:
Xsin a
1+cos
dr.
dx
r
4.
Prove that
5.
Prve that (
6.
fa f()+f(21-
Evaluate
log(l
tan a Nix.
t
1+cos a
25.
Nt
dt,
the equation of a straight line
39.
1+ cos
*x
sina,
J
Is
fr-chda.
sin
fxydx
r
cos "xdx = 0.
=J fdx
*
Jf2a- 1)dx.
function, prove that
f)it is odd.
fthdt, a 20 odd? Justify your answer.
1f-x)+fa)
= 0 in
and f)
the interval
periodic with period à, prove that
.11fft)=
periodic function of the period
-x|+ -2|.evaluate fmd.
If [xl is the greate
integer function then evaluate
42. If
S
Prove
thata]+ [-x]]dx =a -b,
nere (x] is the greatest integer S r
.
f(t)dt is
is a
fla + x) =f(x), prove that
fo= 43. Prove
xdr
Xdr
Show that
40. 16f0) is an even
41.
cos x-sin x|dr
26.(lx+ 1x-1|\dx
sin x
a cos *+b"sin *r
= 2b +
Evaluate the following
28.
xdr log(1+ cos xhds = -tlog2
x| dx
ftrdr=
Hence, show that 0
1+x
cos
Where
cos
where be Nand ae 38. Prove that
Prove that
|cos x |}dx
37. Show that
that
23.
+
1)xdx,
bos6 d0-a-Va*-b*).
Prove that
x|
cos x|dr
36.
cos
being an integer.
problem. will have real solutions. Hence, the
7.
2x dr
35.sinxdr
dx
Vtanx 12
parallel to the x-axis. Find its equation.
I'rove that
prove that
0
Exercises
1.
0,
log cos xdx=-log 2.
J
= 2,
ndt =2
lz]dx then define f(x) as polynomials in
x) =J
the interval [0, 3).
22
Ttlog 2.
thatj
+ that[Vxldr ="- Ddn 1)
sSr. where n e N and [r] is the greatest integer
2
32.
21. nm
-log
30. Prove
31.I
when |x| ->
o for all real values of r between -« and «, (r) atains values between 0 and « [Note If ¢ (a) = p, d (b) = q and d (x) is continuous then by evcry value between p and q will be attained from he (a) for x between a and b-this is clear graph of a continuous function] all x will also attain the value 2 for some real and (x)
ie. =
because
realm.
sndt we get m'r?.J fodt.
Consider the function o(1)
13,
discontinuous function cannot be continuous.So
-b
Iim
So, fta) is differentiable at x
xtan Provethat o tan X+ sec r
dx=m(T-2)
0
function. As the suma continuous, being a polynomial continuous functions is also continuous, o(r) is ale Continuous.
= lim
12.
Sndt=ft)
is continuousand
Asf)
(n-1»
thatf (x) =
andevaluate
-
four.
sin2x
is
sin 'r+ cos "x
sin 2x
sin'r+ cos'
a periodic
funetion
D-152
Problems
in l1T
Mathematcs
Prgertis
b
l,
nt
where n e N.
rdx=J ftrdr
that
Show
=J cos "r dx, n e Z, show that l,
Hence, evaluate
x-a
56. 1in
=l-
dt
sin
If ne
46.
sin"
2
dr zx =
is the middle term in the expansion
47. If
,
=}
tm
where t,
of4
s. 59.
tan "r da, n e N, show that 60.
48. If
-Hene, evaluatetan r
-*
l
-
(G
1)
-
in : +) 1"I10
62.
76 ForFor nEe Show that
63. 1
=
(1-x)".
prove that I,
g
Hence, evaluate
s
fy-
51.
f
Find
lag-z.
sin
yy
and
=0,
fo-)
=0,
12-
-
55.
f() at x
in!
0.
penodis
id)}-
=
)
-fia}
.
V
which
a, cOs kz
then find/r)
where 2-1>0.
sin
s an )fz)-žiz.
0. E fi)
kz
is
fnctior. F
henl:i seal
V)"dz
=
t
a
en e
even ctien
ot
where à is acorstart. is
(a)4
loglx\ds =
x,0sr a orr 0and the
28.
cu by Sketch the region enclosed the find
yla+**)=x'a2-r) and
byr--2,y
region. Find the area of this
ard bouni
its
the reg curves and identify 9. Sketch the 2 y= = logx and
area
the
Find
36.
bounded by the parabola
and the chord joining the points and (4,6).
=6+4x-x
cune
Also find its the xy area of the region in the Determine z5. a'y^=x(a*-x') enclosed by the curve 4y =
12.
15.
the
and y axes in the frs +y*s =a *, the quadrant. Find its area. area enclosed by the curve x=acos (Or Find the y= asin'0 in the first quadrant.) a 21. Find the area bounded by the curve y=sin 1 x-axis in the interval 0Sxs2T. boundel 22. Find the ratio of the areas of two regions 7-axis n te by the curve y = *"sin 2x and the interval 0 srsT. bounded by the cut 23. Determine the area y=x(X- 1), the y-axis and the line y=2 by the cu Sketch the region bounded
Sxs
and the xX-axis.
Find the area bounded by the normal at (1, 2) to the parabola y '=4x (y20), the curve itself and the axis of the parabola.
-6)
2
24.
144.
1
11.
.
Sketch
x
Find the area or the region bounded by the curves y= logX, y= Sin zx and the line x = 0. area of the region bounded by the curve 4. Find the - tan x, the tangent to itatx = q and the r-axis.
area
4r-a
=X*-x?
46. Find the area of the region in the Argand plane in which the point z will be located if
33.
35.
37.
V5s Iz| S2v3 and
Fill in the blanks.
38.
Find the area enclosed by the parabola y* =2x and two tangents to the parabola from the point (-2, 0).
39.
the Find the area bounded by the curve y x-axis and the two ordinates corresponding to the minima of the function.
=2x-x,
48.
The total
ordinate r
X-axis,
part of the curve y=1+and the ordinates
|X|
=
y=4r",x=0, y =1 and y=4 50. The area
is
the ordinate x = e
bounded by y=log,
and the x-axis is The area bounded by
51.
y=
X0 and the lines
yx(x-1)07-3) =0 is area bounded by the curve x lines |z] = 1 is-
52. The
= cos
y
and the
area of the region bounded by r* =y, y=X*+2 and the r-axis is
53. The
into two equal parts then find a.
Find the continuous function ftr) such that the a bounded by the x-axis, the curve y=f(7) and the ordinates x =1, x =a is equal to V1 +a* -v2 tor all
Choose the correct option(s). 54.
>1
ne
area bounded by the continuous curve y=Ju) and the linesy= 2, x =1 and = a, a> 1is equal to
x
l(20)3-3a+3 -212).
Find f)
if f) >2
for all
R, so that the area bounded by the curves
X-bx2 4. The (0,1) *na
the maximum.
r2is (0, 0), (2, 0), (2, 1) and s of a rectangle arehaving vertices at thne and y =
two parabolas
The area bounded by y= V4- x",y20 I-axis is (6) 27 (a) 4t
()T 55.
The area bounded and the x-axis is
(a)4
[1, a].
DE
area enclosed by the lines y= |7
The area in the region x>0, y>0 bounded by
49.
2
=2,x=4
1yl =1 is
and y =0 is
a divides the area bounded by
If the
=
The area bounded by the lines |x]+
47.
.
Curves.
z+z-227+8z +82>0.
Objective Questions
Find the area enclosed by the curves x*+y'= 4, + 8 =0 and a common tangent to the
x+y-6r
40.
-+1 and
Prove that the area bounded by y=
45.
eog
56.
and the
(d) none of these by the curve y= Sin x,
x E
[0, 27]
(6)0
(d) none of these c)2 The ratio in which the area bounded by the curves y=12r and x*=12y is divided by the line x = 3 is
a) 15:16 ()1:2
(b) 15:419
(d) none of these
Problems Plus in IIT
D-172
Mathematics
Answe7s 28. 1. 4
2aV1-e(ev1
3.
-
e2 + sin
area=a(T-21
'e)
s171.sin71 6
6.
8.-
43
10.
9. 12 12.
2
11.
log,2
areaog
13.(8t-V3)
V2
143 16.
29.
y
15. T+
V3 7+ 8n
17.3v3-)
V3
1.
19.121:4
20.
l(o.a)
34.log2 32
ra
(0-a)
35.
37.3/2+ sin
36. 36
39.2 120
38 22.
24
3
32. 1og2
area
21. 4t
logeX
30. 2
(a.0
(-a,0)
=
457-4
40. 22
41.1+X)
22x, x21 44. 32y9x, 4x*=
43. t1
42.
45.
4y 3(x-1)
area:
0-6)
d
9y
2-log3 +tan3 12sin
x»l
4v3
+5sinE+
64
49.
47. 2
48.
51. 2
52.2sin 153
55. (a)
56. (b)
1
50.1 54.(b)
26.5
27.a(2-
13.DifferentialEquation of the First Order
Recap ofFacts and Formulae .Differential equation, its degree and order An
equation involving derivatives (i.e.,
,
The differential equation of 2% dx
differential equation. etc.) in x and y is a the highest order derivative in the The order of equation is the order of the equation. ,Thedegree of the highest order derivative (when put degree of the equation. in rational form) is the For example
dy
is a differential equation of the first
y+=2
order and the first degree (linear).
=0
G
is a differential equation of the
dx2 second order and the first degree.
mv-(
being order and the second degree, it =x in therationalised form.
first 1
=x3 is a differential equation of the
ax
differential equation is the We know x2+y2=a2 is a circle whose centre will it parameter, Ongin anda is the radius. If a is a epresent a family of concentric circles with the 2 Formation of
common
centre (0, 0).
Dilferentiating
2+y?=a dy 2x+2y=0 dx
equation of the second order, obtained by eliminating the parameters by differentiating the algebraic equation twice. Similar procedure is used to find differential equation of a family of curves of three or more parameters. differential equation the The general solution of a differential equation is integrating relation in the variables x, y obtained by (removing derivatives) where the relation contains as many arbitrary constants as the order of the equation. the The general solution of a differential equation of first order contains one arbitrary constant while that constants. of the second order contains two arbitrary of equation differential So, the general solution of a curves. the first order gives one-parameter family of of the In the general solution, if particular values solution arbitrary constants are put, we get a particular which will give one member of the family of curves. 3. Solution of
differential equation of the first order and the first degree Simple standard forms of differential equation of the first order and the first degree are as follows. 4. To solve
=
Method Make a suitable substitution so that the
equation becomes variable separable in the new variables.
sis and constant).
Forexample
curves of one h tferential equation of a family ofthe first order, is a differential equation. of eliminating
1) of two parameters is a differential
.Reducible into variable separable.
a differential equation for all the members of the it does not contain any parameter (arbitrary
ned by differentiation.
family of curves (like
.Variable separable. Form: flr)dx + oy)dy 0 Method Integrate it, i.e., find J ftx]dx +] d(yldy =c
1e,xdx + ydy = 0 (in differential form).
eter
=
a
the
parameter
Take the.equation (r
-y)2=1
This is not in the form of variable separable because we cannot write it in the formf(x)dx + ¢(y)dy = 0.
by
D-175
Problems Plus in lT
D-176
Dferential Equation of the First Orde
Mathenatics
xdy ydr
T'utr-y
1
1-
1
(r-y)
ax
To
Homogeneous equation. Form: d variablesX,
=
vx and solve he equation
the new
in
U.
Nonhonmogeneous equations. OV *
dy Fom
Solvable forp.
..
*
*.
*2 put r=X
If 2
a,h+b,k +C
0,
+
h, y= Y+k such that
+ c2 =0. In this way, the homogeneous in X, Y. Then use the =v
ar
or
+
by = v. The
variable separable equation changes in the form of
in
Solvable for x.
*
Q)
called by e", Method Multiply the equation becomes integratin8 factor. Then the equation
P
Q).
integrating, y
PirMix
e'rN = } Q(x) e"aM
dx
pc
Method Put S(y)
y = z;
or fp)
+
x
cx +f©) which
y
dx
length of the cartesian subnormal
=
AD =y =
1.
dtan dtan
yus y
tan)u
in the equation. The resulting
equation
y-x
is
the equation of the orthogonal
trajectories.
Find the differential equation of the family of curves arbltrary A,B
equation has two arbitrary constants. So its diferential equation will be of the second order. Here, the
We
.
have y me"(Acos x+ Bsin x)
Differentiating () wrt.
(1)
X,
+
orJedy=Je"dx;
.ca. 3.
Determine the equation of the curve passing through the origin, in the form y=f®), which satisfies the
sin(10x
differential equation
z; then
10
+6
10
Differentiating (2) w.t.t. X,
+e(-Asin
x + Bcos
the +e(-Acos
-y,using
or
This is
2.
'
X
tangent at P(x, y) he equation of the
y-f) is Y-y-(X-).
to to the
curv
6y).
+e'(-Asin x+ Bcos x)
or
(2) and (1)
X
- Bsin X
or
10
equation becomes
6sin z + 10;
6sinz+ 10
or3sinz+5"d
1
O
+
x +Bsin x)+ e"(-Asinx + Bcos x)
- (x)
4y =ydx xdy
place of in differential
are Acosx+ Bsin ) where ye constants. Also write its order and degree.
3
the differential equation.
1+tan"
xdy-
-
y.--0
Method To find the orthogonal trajectories of a lamily of curves whose differential equation is known, put
normais Some results on tangents and
which is in the linear form. Form 2flotx, y)hdo=0. Exact differential equations. cquation in this Method In order to put a differential following ditferentials. form, one has to remember the ydx d(ry) » xdy + ydx dlog dlog
=y
Selected Solved Examples
a
singular This solution is called the
Qr):
P(rz
The equation becomes
1s
+X =0 all the u andf'P) any arbitrary constant) touching (without = y cr +f(C) given by solution
Sy) d5(y) dy dy
0
curveswhose differential equation isflx.
P'ut 10x +6y
=0.
which we get a solution
dx
1
CP =y
or the cartesian subtangentt=CA
e(Acos
solution is When p= the general lines, gives a family of straight = p from y=pxI When f'p) +x 0, eliminating cu
R(y) hen
wrt
c,
P) Sty)-Q)
such that
orthogonal trajectories of a family of curves form another family of curves such that each cunve ot one tamily cuts all the curves of the other family at nge angles. differential cquation of the orthogonal The is the family of trajectories of the curvesf|x, y.
.The initial ordinate of the tangent -OB
x+fP)-0.
Reducible into linear equation. Form R(y)
Form x =f(y.P)
y which gives Method Differentiate x =f(y. P) in p and y. Solve first order and first degree equation p from x =f(y, P) and it to get vy. p, c) = 0. Eliminate VyP, C)=0. +f P) Clairaut'sequation. Form y=px x, we get Method Differentiating w.r.t.
X, v.
y= Linear cquations. Form+P(X)
=
7. Orthogonal trajectory
The
pP)
a Method Differentiatey=f%.p) W.rt. which gives in p and x. Solve cquation degree first and order first 0. it to get é(,p, c) = ©)=0. Eliminate p from y =f(X, P) and ¢(z, P.
azlh + bak
put a,r+ b,y
If
the curve
length ot the normal= PD
.The length . The
to
Tu,y)
the solution.
Solvable for y. Form y=f(x,
equation becomes method tor homogeneous equations.
R)
. The
x
p-Snlx. y) =0 Form p-fX, yNip -fT, order and first deg Method Solve the rst etc. If d,x, y ,c)=0, ete equations p -fixy) =0, same arbitrary constant c for the (taking solutions y, C) 'nz, y. ¢) =0 i each) then 4(X, y, e) **, y))
at
x).
ot the tangent The length
*
dx a+by
Method
dx
equation ot the first order bus solve differential
equao Is -yd(X-
-f()
v
equation of the f Standard forms of dilterentual (here wed follows are as and higher degree
separable. which is in the form of variable
Method 1lut y
The
"(ay)=-
higher degree
dz
=
dx
=z
-l
dsin
az
then 1-
2;
Solve
log,
Here=e at
see
d
ar+by. +
by.
edy=e
5
ax
5tan+6tan5+5
sin2
.
D-178
or
5t2+6t+5
Mathematics Problems Plus in IIT
x+C
puting
tan
ifferential Equatiom the of First Order
-
tfc)-y= z; thenf'()-d f'lt)
becomes,f'(r) -
(1)
esin
=X*
d(1-2f
v do
"sin
=
=0
v dv = c
I=Je "sin
-Ir
do
vdv = 0
"sin
or
Now,
or
v+
=
sin
2) = cC fo)+log(1 +y-f(a)) f)+ logil
=-sin
-
5.
tan
or
It is a
5t+3
4 or
an
5tan
Solve xdy-ydx
5+3=4tan 4(r
5tan(5x
+
3y)
+ 3
Je"cos
v dv
+c)
.. (1) =
y
UX.
=v+x. dx Vr+y
(sin v
I=e
"(sin v+cos
or
or
=x v1+v
x
Putx
or
cos v) =
c
+
Vo+
1)= log x+ log
y
e
+y*=
0; or
v+ Vo*+1 = cX
y+
log I+log(v* + 2o +2)
or
los+2yx+ 2)-an
8. Solve
-
tan
(o+
1)
=c
1
=.
(y-3+3)y--4
42y-X-4 iet dx y-3x+3 Putx = X+h, y=Y+k; then dx = dx, dy = dY the
equation becomes
a_20*-X+h)-4
dY.
or
Y+k-3(X
+ h)+3
X+(2*--4) dx Y-3X+(k-3h+3)
Select h, k such that 2-h-4=0,k-3h +3 0.
Vx* + y
=
Solve
xe-
to solve the equation,
xe"-
.
or
ysin
equation becomes,
y°+I+y°=0
ysind+rsinž dy=0
nenU+1So, the equation
fl)
v+x
cx,
puty Ths is a homogeneous equation. So,
where fl) is a given function ofx.
Y+X+Y=0
+ xsin
xe"-vxsin v + xsin
on
=0
v-+
becomes
substitution of r
DEcomes dY
=
X+2,y=Y+3,
2Y-X
dxY-3X hich
which is a homogenous equation
*he
25tan 4x 4x)
tan4-3tan 4x")
Here-f)-y)f'lu)
or
dX
Then dx = dX, 2ydy =
put y'= vx; then 2y
5(4-3tan4r)55(4-3tan 5tan 4r
f)=fr)
log
lere
ay. dY Using this in the given equation,
So, in order
tan 4x
4(3+4tan 4x)3
4do x5Í (+ 1)2+1 2+20+2 -fd=o
or
Solving these, h = 2,k=3.
54tan 47+tan
1-
Solve
**2 * (sin +
***
3y)=
0
2+2+2
v)
xc+s sincos
X, y*=Y.
+2
2+2)-2 du=
v)
*2 dx
tan 4r
4.
od
20
r
or
solve it.
1x
loglv
y
+
n
that the differential equation y 'dy + (* + y)dr = 0 can be reduced to a homogeneous equation. Hence,
5tan(5x+3y) +3 - 4tan |4r + tan" tant5r
v.
7. Show
Putting in (),
or
log
log
Putting in the equation from (1), we get
v1+
+
cos
21-e
trom(),
-y=
4tan 4c
tan-4c
4c;
vx* +y* dr.
x
v + cos
=-* "(sin v+ cos v)-I
equation. I'ut
We have the equation
- 4tan 4(x+c)
t passes through (0, 0). So, 5tan 0+3
tan tan
homogeneous
Then Then
4(7 +c)
=
=C.
v do
v-Jcos
or-og(1-2)=f)-c 5
D179
e"+x'sin v 0
is
PutY- VX; thenxV+XK
2X-X . V+X xVX-3XK V+X or
2V-1 orXXV-3
v-+3V-1 -5V-4V
Integrating).5V-dV
Xlv+*+x+dx
+x *1++
pX =0
1+0 2 dr
0
the equation
homgeneous.
= log
X+ log
-2V+5)av =log excx 0F-2-+5V-14log V or og -V 5V-1)*zv-5/2(21/2 =
log cX
Probems Plus
in iIT
that Select h, ksuch
+21 or
oF-io
X
. =
or
5-2
=2
log ex
+dg. 9. Solve (r +2yNdr-dy)=dr
Putx-2y=;
the
=0
-
14 + 24
nes 3u + 2
+ =
c(r
Solve
11.
or
=0
2-1+2y+1
or
Integrating factor=
euation becomes
y=-
()
uls
or
becomes-5*r=r
in the linear form. So,
integrating factor =
e"
Multiplying by it
+log =log u- log
c
or
-lo lo 4 u-v
or
X-2+=1 lo x-2-(-1) -log2+Y-1XX-2-Y-D
2ez=-2
sin
y-Isin Solve (1+y
2
Here.
(1y
dr= (tany-1dy. tan "y-
a--2. or
(putting*=9
J:*#
=log(X-2)+kg
orlogy31. (r-2i-8)y
-2 log(z*-2)+hg
or
-( =
I+y
s
is in
1+3
the linear form.
Cleari
grating factor= e
1
yi=1
Maxltiplying
žy=
dz, dt
-yi-3a*g
dy.
Put= X. =Y; thern 2zdr= X,2y theeuaion becomes (3X -2Y-3dr 2X 3Y-7K =
thea l= PutX=-k,Y=t-c
4
then
-2=-2x.
or
log u + log
Itis
-2or6-1) Here (23y-7)ziz
=;
Put
(X-2)
=
Hereyx'y
cos
or
3-1d
is in the linear form.
e
rr log
+c= (tan"y-1)
factor, Nfultiplying by the integrating
oF3osl--og
then 1 +2
=
r'y-ry.
13. Solve
**" du
=
xean
.y-costhis
3+2
11-- :
where tan "y
=- cos
Here,
where X=ut2,Y=+1
3u+2uz 3+2
" du
1+y
eds,
=y-cos
1.
-2)
- 4log c = log c°=log k (say)
--jeid=-1
**)
= 1.
becone du
Integrating we get
.2)
3h+2k-8 =0
x(5)., -5h
and so k
c
-3=k{x* -yi-1)3.
2k-8)
2h+ 3k-7
Put r= u; then
d
(7 -2¥* 1M
(1)
+
+
2-y2-1)
or
+ 3k- 72
-1/2
2NI2-3)2-3)-(5-121 XF -2))
=
From2 x (2)-3
o2Y-5+ 121X
2N-3)-(1=--3)-
(a-2y-1ir
*k)-8 3(4 h)+2("
(i 2+3+ 3u+2+ (3/h
log ex
er
2(u+)* 5("+ )-7
ddu
(-v5V- 1).5-
logty"-)-žbst-y-1) log
becomes,
(1)
oF-o
Diferential Equation of the irst Order
NMatkemaiths
bogy-9-ost-1) log(-y-1)+
by it
-1
Here-2-1-2
logta*-
loglz*-2)kg
=
+y
z
roblems Plus in irT
Ditferential Equation of the First Order
Mathemath
D182
dy dr sin
2
or
the
xdy
dy
-2lincarfom log
integrating factor =
rcos Bd0) or
,
-yax = TcOs
the
=r'd6 equation becomes
-
y Now, y
rdr
+
rdr + de
0;
or
yax
ydx = xy °(1 +log x)dx
=xy(1+ log )4x
xy(1+ "(1+ log xjdx -d
We know, d(x
"
+
xuy
ydy +-
=0.
+y
or
+y) =2(xdx + ydy)
theequation becomes
dxy+
dtan
0
+
log
+P)-y
or
y=
curve passing throug Find the equation of the
whose differential equation is yr+y ydr=zy-x)dy.
xtan -
P(x)
42 = QP)
v(X)
+p(x) v(x) = g(7) +p(xMu(x)- v(x)) =f)-g{r)
(1) dx .
- otr)-e'PMJ= {fx)-s(»)) e'Pa
orlu(x)
2
(it
. 8)
or
=0 y(ydx -xdy) + x(ydx + zdy)
or
xdxy)=0
)+
dx
P(x)
y-y)=0
.
Let
(4)
(2)-(3)
-x" **
+r(-sin 8jd0 and
in
linear torm)
d
-2)+
From (4)
and (5),
P)
(y1-y»)=0
.
,
(u(t)-
(5)
are positive. v{T)}%
-
: u(x)-v(X)}A
y-
{u(z;)-vtz,)la
> 0
f)>gla) for * > > {u(71)
x,
and eJFaM*
> 0}
- v{T)})u> 0 because u(X,)> Ar,)
u(r)-v(r) -1
(1)
(eod,eo
Clearly
:
tan 0
was
F)-s(r) e""dx .
P--y)-0
Here, (ry+y ydr (iy3-x dy
ie,-x+yi, dr=
=Q)
..
+p(x) u(x) =fr)
=
= rcos 8 and y = rsin 6 Second Method Put x
drcos 6
1y2
and
solutions of the differential equation
Pr)y-Q)
g -3
17.
.
Here,
From (1)- (2),
Integrating.a+y)+ tan"
=c
then prove that y=y4 +Ch -y)
general solution
y, are two
rom
+y+ 2tan
respectively
ux)-v(x)Me'PRM
logš-*
(exact equation)
satisfy the differential equations
and glr) are continuous functions. If where p)f() u(z)> v{z) for some x, and f) > gt*) for all x>ru prove that any point (, Y), where x>*y does ot satisfy the equations y=u(7) and y =v(r),
of the equation where c is any constant. For what relation between the constants Bya also be a a,P will the linear combination ay+ solution? is the
xd
a + B =1.
d+PM=f andp)p =g)
are two solutions of the differential equation
+Pry=Q)
and
xdy-ydx xdy-yax +y
dtan
-14-/r*a
y +2x-5x y= 0.
curve is
If yu ya
As y
Q), using (2) and (3)
19. Let u(r) and v()
he
log r)dx
15. Solve xdx
if
Qr) = Q).
+B)
Hence
c
tan"
2)} dx=0. Solve xdy-ly +xy°(1+ log -
o
(ay* Py») =Qr)
aQt)+p-Q)-
(a
2+2+2 2=0
2+0=c
18.
-2log y-
or
y+2x +2cx 'y = 0
= 0
(1, 2). So tpasses through
Here, xdy
y+Cy -a) ay+ By2 will bea solution
ay+By)+P)-
y= an 16.
*
or
Jrdr+J de =G
y
y)
=log c(y1 -y)
log-V)
rsin B(cos bdr- rsin0 de)
e
ze*y) or
+
integrating, log (y-yi)= log(Y1 or
b{Sin dar + rcos 6d0) -
equation becomes
Multiplying by
or
rdr
then 2xd
d
rsin B(sin bdr
dividing by x'y
0
rcos b(cOs Bdr- rsin Bde) t
yi--
Put-z;
0 +
xdx + ydy=
2y d-x--(d+2)
or
rcos Bde
u(r)
* D(X)
> 0
for x >xi
when
x > X1.
Hience the problem.
Mathematics Pnvlems Plus in IIT
2py-c= x
20. Solve
+(y
dr
1)+ yp
pp-
2v
1) 0
-
1Mxp + y)=0
(
=
4xp*- 3p*x
=
xp*
y Au= -2x* +X
x(y- 2xp)
=
y = 2xp
+p*
.the
methods
equation
to nna
+ 4y(y+)= 4x(Ty + ©XY+*)+ (ry c).
***"
Put
dy
xdy + ydx
is (y
-
X -
c\ay
-
d
dy
Where p
Here, xp=l+p"
C
general solution
dy.
2. Solve 1
=0
=0
d(ry)
xy .the
4y+x*=0.
a tamily of straight lines
But the
singular ssolution 4y tr*=0. vith a of
tne
solving first order differential general solution of
2y+22y+2
Now, p-1 =0
the temperature of the substance at time r anu :beT.The rate of cooling of the substance kT-30)C/minute (from the question).
Let
solutior tions are y=cr +C
+ 2x
Putting this in
or P*y*.
PIO
Dyferential Equation the o First Order
p)-3p * +2p)
P(2y+2x )=xy +c Xy* p =
yp - y=0
-ap+
or
=0.
-xp - y =0where p dx
Here ap(y or
x)y
-
+
29|2xp
= P p. Then
ddx
dx
equation is
the
-
3 dp
=
rate of cooling rate of decrease of temperature-d
from the question,-=k(T
Hog(T-30))=kll
logk15;
:which is of the formr =f0.
y=2" dr *dx
Solve
Here, y 2xp +p* where p
p 2p+
dr
dp
=
solvable
or
dy
0
Here, y= xp
px)= or
lo%P
byp',
+2px= dp
aax,
=
+
++ b)
Thus, we have y
and 3p 'r+ 2pC
=
2xp + p*
+2p=0
Odx
0
The ra
?A3*
y=xp +p*gives y = cx
0
+c
p=-
nce
ha
15 cooling of a substance in moving air to the difference of temperatures or tue and the air. A substance cools irom minutes. Find when the substance wi"
the it being known that of air is 30°C
emperature 32°C,
constant temperature
Solving
V (from
(1),-k-dt;
=e**i t=
0, u = 2V
(1)
... (2)
kp
Att= 0, u = 2V and v=
When
.
-ku
=
r
onal
x+2p
and B are two separate reservoirs of water. The capacity of A is double that of B. Both the reservoirs are filled completely with water. Water is released simultaneously from both the reservoirs. For each of the reservoirs, the rate of flow out at any instant is proportional to the quantity of water left in the reservoit. After one hour, the quantity of water in A is 1.5 times the quantity of water in B. After how many A hours from the time of release of water, do both and B have the same quantity of water? B u Let at timet hours, the volume of water in A and be and v respectively. From the question,
and
d(p r)= -2p dp
ior
2
Y-2ax
0;
2p)
minutes.
t
-20
2p
Pp
p'r-2pdp--
dx
y=
this is of the form y = p +fP). Differentiating w.rt. x,
(x+
+b
P=V-2arh
+ b)=p
p p° where
p?.
-2p
2
4x
2p
20r
log3
26. A
integrating
y=*
Hog(T-30),,lo
e °%(Op= ap*
from ()
.(1)
"=e
log a =
=3logP*
eliminating p the p-eliminant, obtained by and (2), is the general solution.
The integrating factor
(1)
a being an arbitrary constant
y+c=5-logP
23. Solve
integrating, we get
dp= 3log p + log a.
p1-d
yse--
dxx-2 dpp 2log
dp
dlos log
for y)
2x+2p=0
Multiplying
wheret minutes is the required timne.
or
Differentiating (1) w.r.t. y.
or
nn"E+p
,
isolvable for x)
dx
4*P dx *P Or
p)
=
It is in the form y =f, P) Differentiating w.r.t. X,
klog
dx
c)= 0.
dy
2x
30)
T-30dt
.(1)
21.
-
the question).
logu-k,t +c
. (3) 2V=e
Trovlems 1lus in
)gives,
u
=
Simiiariy, from (2) we get From the question,
if p
r-
..
Ve m I
then
u =
z
u
(p
4
Using (o),2-G1.ie
Tos-log 2 ie.
and
be
los2
T-
same will have the
The
quation
curve
(*
+
- 1). So it
Now, the line is 2x -y-4
y
cx
t2V-.
(1,
1
1
is a
Hence it
=0,
or
ie,+1.
dy+(-yMr-0.
Puty vx; thenr**
this in td
.
()
or
y-
ie,
But in the first
quadrant x, y are positive.
ay1.
.
Solving y 2= 4(r+y)-8
of the equation and
*y* (or lines) are tne requíred curves
ay Y = 0
and
X
46. A
(X- x), i.e., X=- " OA)
Curve passing
point(
through
y) equal to
r(OB)
(1, 2)
Find
has
Here, siope* dx
y-2
its Sio
the area
the line Dounded by the curve and dy
-
and 2r-y-4=0, we get
e ta
o
2r-y-4
or
24)4y-8 -6y=0, ie. y=0,b Then y = 6, * =
dx
ie, Y=y
*
l.
=0 successively, we get
2
1-0
12
1y- 1,-3.
solution This is the singular (l, a curve and it passes through
-n,
2xy
It is a homogeneous equation.
pomt (r, y) of t..e of the tangent at the
Solving (1) with
a
place of Puttingin dy or
p-.Putting
=0
0.
2r+
75S,6)
So r+y quadrant x, y are positive.
*
r=t
=0
4r y This is the differential equation of the circles The equation of orthogonal trajectories is
-2 0.
lne is y = -r t2,ie, r *y=t2
But in the tirst
y
passes through A(2,0) and (0,-4).
=
the
d
+y-ay
parabola.
vertex C is (1,2) and the axis is y
Its
dy
2x+2
y-4y +4= 4r-4
y)-8
12 =9.
af().
T-y(X-r)
and Y-y
=
(-2)=4(*
or
-p
y*
-
0, differentiating w.rt. x,
-8
the curve does not cut the y-axis.
= 21
Find the orthogonal trajectories of the circles x"+y°-ay=0where a is a parameter.
29.
Here,y2-ay=
its roots are imaginary.
ct 2V-¢, ie, (l-c)*-4e. cm-1 (1+c) -0
then
NOw,
is
y
y-4y+8=0;
or
in (2, If it passes through This gives a tamily of lines.
ditferential possible curves
0)-8, i.e., x=2
When x= 0,y'= 4(0 +y)
i,
Pc
d putting
0 = 4(T *
*4y
urve dy = f
(1), the required area
from
A(2,0) only. the curve cuts the x-axis at
or
O
,
When y =
Again,
=0
ar(BCAODB)-S
2+c,ie., c=-4
find the area, we have to draw a rough sketch of the
To
y-x=t2V-p
-4
thecquation of the curve is-2y 2-4 y-4y= 4x - 8, ie, y '=4(x + y) -8.
curve
-P
of water
y
=p
taking
p
2
kSheurs the nservoirs
curve be
apP),
-
Pp*dt2
(, 1) such that the A cure (or line) passes through coondinate aves and the triangle tormed by the curve is in the first tangent at any point of the area equal to 2. Form the quadrant and has its ot the equation and find the equations
Let the
y)g
or y=ap+2N-p Differentiating w.rt.
q =1
2e
i.e.,
A
B
-
=- (y-ap);
(6)
27,
So passes through (1,2).
2
or
of water in Let, atter 7 hours, the volume yual. 2V¢*= Ve ** » (5) )and
241-21
-2y 2c.
S-2dy2J dr;
Vo
=c-i **
z
=*,
-2)ady = 2dx
.
OB
or
Dividing these,32
qantity
5OA
%= Ve
and (5)
atter
= 2 =
dy
t, whent
=
the first quadrant
the axes in
(5)
D17
Diferential Equation of the First Order
NMathematics
by the tangent a rea of the triangle formed nd
.. (9
2Ve *
In
log*log(l+*)-bg al+r)-c n5. So B= (5,
or
learly, the required area
ar{OABDO)-ar(BCAODB)
arDABDO)
-
d=0: antegrating we get
dy-
(1)
1-c.
ie,a+y*-a
orm the ame Note It the orthogonat trajectories amy ot Curves tamily ot curves as the given curves is caled o then the gIven system seforthago.
IIT ProbiemshPlus in
42.
Exercises to e
equation corresponding 1. Find the differential arbitrary -c), where c is an tamily of curves y = dr constant. conic sections difterential equation of al 2 Obtain the a, b are parameters. where ax by*1, family of circles differential equation of the 3. Find the origin. touching the I-axis at the 4. 5.
Solve
xdy +y*dx= zy{zdy yáx).
Solve1+r*tan "x +y=0. sec
6. Solve sec'r. tan ydx 7.
Solve
8.
Solve
9.
10.
Solve
Solve
-
13. Solve
yMy =
+
26.
cos Solve the equaion
+ dx
27. Solve
sin
yCos(x dx
16.
.
+|1-je
it. homogeneous form. Hence, solve = 0. Solve (4r-y+3)dy + (2r -3y-1dx
in the
21. Solve
dy 3r-4y +2
22. Solve a7 23. Solve
1dx.
(x-1)dy
- y)dx. =
36.
y
-.
(1+y)+(r-e-an
+
ydy +
(r*+y)dy =0.
1x
sin =cos x sinx-
:
56.
ve where v is a function of
equation when y
= 1
and=0, dr
= 0.
smallest integer bigger than or equal to
log 10- Iog9 log (1.04) -0.03 47. Show that the equation of the curve passing throug (,0) and satisfying the differential equation *o*
(1+y jdx- xy dy = 0 will be x"-y'=1. whose slope at TOVe that the equation of a curve (,y) is
ify=1 when x =2
24y_ +2 d =**
and
58.
60.
t
curves for which the cartesian equation of the constant lengn. ength of the tangent is of portion ot the tangen Find the curves for which the between the axes o
xy
aany
p=
point intercepted
rectangular hyperbola. Find the equation of curves for which the cartesian subtangent varies as the abscissa. Find the equation of curves for which the cartesian subnormal is constant Find the orthogonal trajectory of the family of parabolas y°s 4ax where
a is a
parameter
Prove that the family of curves A
Ab+a
isa parameter, is self orthogonal.
objective Questions
Fill
in the blanks.
62. For
the differential equation
and
degree 3.
65.
order
Vis,
the
=
equation is general solution of a differential then (y+c)° cx, where c is an arbitrary constant the order of the differential equation is it the
:64. The general
(
solution of
The particular solution where y(0) = 1, 15 =
of
lE
66.
Let
67.
If y() is a solution of (1+)
>0.
=0is_
+
log y)dy
sin -dx
=
+
ydx = 0
F(t)- F(1)
then one of the possible values of kis-
equal to the radius vector
*nd
Find the curves for which the distances of the origin from the tangent and the normal at any point (¥, y) e equal. ne normal at any pont P(z, y) of a curve meets the be X-axis at G. It the distance of G from the origin a twice the abscissa of P prove that the curve is
where
which passes through the point
(2,1), is r3+21y =8. on the square of the intercept cut by any tangent y-axis is equal to the product of the coordinates he suchn of the point of contact then find the equation of curves. normal is the curves for which the length of the Find
*".
-+e)+1-0.
y=2px + yp°where
the point ot contact.
-2+y=xe",
the average food requirement per person remains constant, prove that the country will become self-sufficient in food after n years where n is the
differential equation
solution of the equaton Find the particular
41. Solve
55.
4%61.
40. Solve
=0.
.
the59.
y(1)=5
ylr+y*+1)
y
A radioactive Substance decays with time such that at any moment the rate or decay of volume is proportional to the volume at that time. The half-life of the substance is the time it takes for half substance to disappear. Calculate the half-life of the substance if 20%% ot it disappears in 15 years. Its (b) A country has a food deficit of 10%. S7% per population grows contunuously at a rate ot year. Its annual food production every year is more than that of the previous year. Assuming that
dx.
Solve
39. Solve
order equations,
du
x+20.
Reduce the differential equation
forr
+xy)dy =0. + 1)dx.
38. Solve
Using methods ot sOlving first
46. (a)
x(t -1)
- x*)+x log x.
y-=r+ Vi-s
reference, is of constant length. Find the equation of the curve for which the length or the normal is constant (a) and the curve passes through the point (1, 0) 54. Find the family of curves which are such that the area of the rectangle constructed on the abscissa of any point and the initial ordinate of the tangent at that point 1s a constant. Find the curves for which the portion of the tangent included between the coordinate axes is bisected at 53.
Obtain all the solutions of (y+ 1)+2=z
x. Hence, solve the
cos (T-y)-sin x sin 2
y(zy°
+3
x(1+x)y(1 (1-x
+ y) +
xdy =
37. Solve
=(2y +X19. Solve (2r+4y+ 1dy 20. Solve
45.
ax
d y-3y+2y
4x5y
44.
nas a solution which is a
solve the differential equation
dy=0
ž+ tan
dx
cos'
ycos(xdy-ydx)+ Xsin (xdy + ydr) =0
put the equation By making suitable substifufions,
18. 5ove
43.
dy
34.
ay2x*2ry-4 17.
2r)
1t
= using the substitution y
Solve (1+*°4y y(T
when 15. Solve
Td-(tan
3x where y(0)=4
vx +4
33. Solve xdx
-3y
"M) dr 14. Solve (1+e
+2* y -y-5r dr =0.
Solve
35. Solve the
dr
straight lhnes ana parabola.
ytr).
and y
where Ix
32. Solve
z
x(x
Solve x(1 -x"dy
+ 31. Solve ydx x(1
vr+y+1 dx =r+ y-1. xdy= Vr^-y dx. +
25.
+ 30. Solve ydx
dy
Solve y'dx
Ify+ay)= x{sinx* log X), find
29.
(r+y*+ 2ry)a I+y= sin
24.
tan xdy =0.
+y)+1 =0.
+rtan(r
11. Solve ydr 12.
"y
Prove that tne geriera 50uton of the differential form equation y= (r+ of
1)z- dx
.
129
Diferential Equation of the First Orde
Mathematics
D-188
then y1)-
y=1 and y(0) =-1
Problems Plus in l1T Mathena
D-190
AuswersS
dy
(dy
dy
2.y dx2
28. cos
29. 3.
(-y)2xy
5.
X= tan
7. log
4.log
C
6.
sin(r+y)+r=c
9. tan(x
ay
tan r tan y
8.r+y=
atan
2vr +y+1+log(v
+
y+1 -1)
13.
ye
sin=c =
y+1+2)
12. kxy= x+2y
kr2
14. x + yeY = c
15. cx = sin =
X+1,y2= Y+1;
6Vy*-1+vr-1 tan 23 V-1
23
log 12x+3y3-5+Vr-1-1))=c 17. (x +y+2)*=cy-2x-1)2
33. 2y +
34. sin
19. 20.
y 22.
+ceN1
1-
y=c+ tan"y
32.3x4
(2x2 +2y2-1) 2tan
=z
(y-e+ c\y+e+c)
25.
26. ycos 2x 27.
= sin
X
cos°x
yr+ Nx2+4) =r*+ (x2+4)32
+c
38.
(2y-x-c)2y +3x-
39.
(xy-c(x-y2-c) =0 eliminant of xp
x
= 0
c) = 0
= ce P
and y=x(p +p^)
y?= 2cx +c 43. (y +1)c+2 =cx, (y + 1)2 + 8x=0
41.
44. xy + c= kx
y=e
1
46. nearly 46.6 years
49.
x= ce2 Ny/*
51.
X=alog
52.
x+y
53.
y2+(x-1 ta)2 =a254.y=cx*
55. xy
50. xty2=a2
a-va-y+ Na-y +c
2 =
a2
=c
58. cx=y
y 5x + cx v1 -x
y
36.y sin
37.
x =1+ y*+cv1 +y
23. xetan
1) +c(r -1)-2a
log(r*+y*)=c
35. 2xy sin
18. (y-x-1)(5y
-3x -1) = c 3log(4x +8y-1)= 4(x-2y +C) y(1 +x)=xlog x-x*+ cx
+**)+ cy
xY
40. p
16. Substitutions: x2
V1
y
-
-log( log x+
V1+x =ylog(x +
y+C
+ y) sec(x + y) = x +C
10. x+c=
11.
C
-
20--1)+ 31. log y=+c 30.
=
sin 'x sin x ++ce sin
y
56.
x+y=e
59.y
-2tan
2ax +c
60.
2x+y2=c2
62.1,2 respectively
63.
one
64. yx*= o(y + 2)
65.
y(x- 1+log y) + 1 =0
66. 16
67
Probability
1. Elementary
Probability
Recap of Facts and Formulae
.
1.
Sample space and event
Here n(S)
of all possible outcomes of a random is experiment called the sample space or probability space. space is an event. Every subset of a sample example In throwing a dice, the sanmple space
setS
The
n(E)
S={1,2,3, 4, 5, 6} and n(S) = 6. = E 1,3,5} c S. So E, is an event and n(E,) 3. The event E = {1,3, 5} is also expressed as the event of getting an odd number in throwing a
the number of numbers of four distinct digits beginning with 4 that can be made with 1,2, 3 and 4
=
=
dice.
1x°P =3!
required probability =
and S c
cS
P=4!.
=
For
S. So o and S are also events for the S. ¢ is the impossible event (null event)
total length
Probability of an event is finite) then
Ifthe sample space S is discrete (i.e. n(S) by the probability P(E) of the event E is given P(E)
or
nE)
or
=*
n(S)
number of favourable outcomes total number of outcomes permutation or The problems of restricted problems of combination are convertible into work out such probability. Students are advised to
favourable area total areaa
favourablevolume total volume
according as the sample points (outcomes) are distributed over a length (one dimension) or an area dimensions). (two dimensions) or a volume (three from 4 PM to For example A football match is played match (not before 6 PM. A boy arrives to see the probability that he the match starts). What is the which takes will miss the only goal of the match match? place at the 15th minute of the has to reach between To arrive at the match, the boy total possible length of time 4 PM and 6 PM. So, the will miss the goal if he = 120 minutes. He hours 2 6 PM. So, the favourable arrives between 4.15 PM and = 105 minutes length of time = 1 hour 45 minutes goal probability of his missing the
corresponding chapter roblems by using exercises in the in Algebra.
=
probability can If n(S) and n(E) are both infinite, the below: be estimated by geometrical method as PE)=avourable length
sample space and S is the certain event.
2.
total number of numbers of four distinct 4 digits that can be made with 1,2, 3 and (without restriction)
=
For such problems,
n(E) number of ways under the restriction n(S) number of ways without restriction distinct For example How many numbers of four 1, 2, 3 and 4 digits can be made with the digits which begin with 4? tis a problem on restricted permutation. A similar Sum in probability will be as the following: P(E) =-
the
A four-digit number of distinct digits is written 1, 2,3 and 4. What is Wn at random by using digits number begins with the
105. 7 120
Probability (chance) that
8
0sP(E)S1.
the digit 4?"
F-3
Problcms Plus in
Multiplication theorem) P(E,) P(E = P(E;): P(E;) P(E3). PE, events E, Ez Cannot be utually .Tiwo non-null at the same time. exclusive and independent independent events then E, E, :Ei. If E, E are also independent. and E,, Ez' are
PEnE)=
nE E)
Complementary event
3.
erent E ' (or E or E") of The complementary happening and E is
the event
the event of not + P(E') = 1 = 1- PE), i.e., PE) E
P(E )
The The
If
odds in favour ot the event
P(E) PE')
E
PE)
odds against the event E
E
=
a
:
b then P(E) =
Union and intersection of events E, is the event of at E E, The union E, u of events and
given by P(E2/l
-
LAddition theorem)
u E,) =E P(E)
PE EuE,U...
-
2sitjsn
PE;n E)
.the number of favourable selections = r{E) =7x
(5, 6), (6, 1), (6, 2),
..,
..,
..
x
31 50
determinant of the second order is made with the elements 0 and 1. What is the probability that the determinant made is non-negative?
The number of determinants of the second order that canbe made with 0,1
each
Repeated trial is p and that o the probability of success in one trial p then failure is q so that +g =1 trials = "C, p9 the probability of successes inn expansion o binomial the in term 1e., (r+1)th
(q+P)
PE,)+ P(E>) + UEuE) = PCE;) + PE) P(E)
successes in n trials the probability of at least r +"CGP
+..
="Cp'q"-+"C,.1p*'q"-ve
.Two events E, and E, are independent iff = P(E,n E) P(E,) P(E). Thus, for independent events:
successes in n trlais the probability of at most r
="C9+"C, pq"-l+... +"C,p'4
n(S)
=
16
ofthefour places ofelements can
16.
=the event of gettingnon-negative determinants. Then E' = the event of getting negative determinants. Clearly, negative determinants of the second order that can be made with 0, 1 are
1-19-11
10) with replacement, determine the probability that the 0ots of the 11, 2, 3, 4, 5, 6, 7, 8, 9,
equationx+px +q=0 are real.
:
Roots of x?+ px +g=0
or
pz 44,i.e.,
where 1 Sx, s6,1
and x, then
sx
6,.1
Sr,
$ 6.
The number of solutions of this equation
in (r+x*+r'+r'+*+)
coefficient of x =coefficient of
coeficient of
r" in (1+x+x+r*+x'+r555 x "
coefficient of x
"
in in
(1-x°)-1-x)*
tC-cr*+C"..
x('C+Cr+ C?..+SCa"... to «)
required probability=
13
16
1514 2413:12.9-8-7.6 15 7 13
are real if p-4q2
Xy, Xy
+X+Is =16
PE)=1-P(E') =1-;
Tw
set
If the numbers shown be integers r, X
=coefficient of x " in
Selected Solved Examples If two numbers p and q are chosen at random from the
Five ordinary dice are rolled at random and the sum of the numbers shown on them is 16. What is the probability that the numbers shown on each is any one from 2,3, 4 or 5?
=
"(E) 3
P(E)=
.
+
Let E
Hence, the
1.
c64x6314
(10, 10)
2. A
2 x2x2 x 2
r
PEE)=
n(S)
(8, 10),
to have the
9.
if
nE)7x147x14x2.
(6,9),
be filled in 2 ways-by 0 or 1).
If
14.
the required probability
1+2+4 +6 +9+ 10 10+10+10 62.
i=1
Mutually exclusive and independent events
PE
corner = 2x7=14.
10.
62 required probabinty(S) 100
the
x;Pi
E(x)=
PEnE,nEx). 1siejsksn .+(-1)"- P(E, n E2... En) exclusive .Two events E and E are mutually events PE, nE)= 0. So, for mutually exclusive
10 ,
2), (4, 1), (4, 2), (4,3), (4,4), ,
(10, 1), (10, 2),
+
6.
10
7,1), (7, 2), ..., 7, 10), (8, 1), (8, 2), (9, 1), (9,2),.. 9, 10),
n(E)=
z, isp value x is pPix if the probability of the value variable r Xn of the random Iz, values x, the X,... If Pn respectively have the probabilities pi,P2» P3 E(x), is given by then the expectation of x, denoted by
PE3)-P(E, nE) PEnEs)- P(Es n E) + P(E, nEz n E)
5, 1), (5, 2),..
PEE)
.The expectation of the random variable
+
..,
= 10 x 10 = 100
{(2, 1), (3, 1), (3,
and E which
E)PE
are selected (or The number of ways to select two consecutive rows columns) =7, because there are 8 rows (or columns) of small squares. For each pair of two consecutive rows (or columns), the number of pairs of squares having exactly one common
ifp=6thenq=1,2,3,..9 ifp 7 thenq=1,2, 3,.., 10
Expectation
8.
P(E, n Ez)
PE,) + PE)
common if they Two squares selected can have a comer from two consecutive rows (or columns).
ifp=3
ifp= 10 then q= 1,2, 3,..
g
least one of the events E, E, happening. E, is the event E, The intersection E, n E, of events and of both the events E,, E, happening.
PE uEE)=
ifp =4 thenq=1,2,3, 4 3,4,5,6 ifp 5 then q=1,2,
nS)="C
ifp=9 then q= 1,2, 3,.
nB).
total number of ways to select 2 squares
the
occur in conjunction place (i.e, E, i and E, takes place after E has taken E takine dependent on E,) then the probability of denoted by P(E2/ E,), is place after E, has taken place,
E, E are two events
1
ifp =8 then q= 1,2,3,
we have
Conditional probability
7.
5.
PE uE)= P(E,) + P(E)-
PMA) =P(AoB)+ P(A
B
q=
ifp =2 then thenq= 1,2
Now, (S)
= a:b then
. If
.If odds against the event
-
-
any two events A and
P(E)
odds in favour of the event
E
then are independent events P(E)N1 P(Ez).
PE UE={1
For
PEa+b
.
E, E
If
Odds in favour, odds against
4.
e
Tt
are chosen at random from the small C chance drawn on a chessboard. What is the two squares chosen have exactly one corner in common?
uares
the
otal number of
= small squares on a chessboard 64
10.9.7= 1365
- 630
=
735
n(S)=735. Now, n(E) = the number of integral solutions of
16 where X X Ss5 2xS5, 2s2s5, 2
+2+=
Problems Plus in IT
cocificient ofr
"
in
+X*+x*+r*'
coefficient of r " in coefficient of r ° in
(1
+
x
in which the remaining n 1 The number of ways two places on two Sides can take places keeping ne of
xP'Co+C,x+°C^x*+.. +Csx") 10 x 10
+
n(S)"Cm-1
3)
n(E) 15249
quired probaDintyn(S)
(m-1)! (n-m -2) !
735
-
consecutively, three Out of (2n + 1) tickets numbered that the are drawn at random. Find the chance numbers on them are in AP + , Let the tickets be numbered 1, 2, 3, ... 2n 1 m(S)= the number of ways to select three numbers 2n
Now, possible Al's with common ditference 4), ...,(2n-1, 2n, 2n + 1),
1
possibilities. 1.e, Possible Al's with common difference 2 are (1,3, 5), (2,4,6), .,(2n-3, 2n-1, 21+ i.e., 2n -3 possibilities.
2-
(n -1)(n-2)
divisible by 10? x, y. Let the two non-negative integers be if the sum of digits Nowx+y' will be divisible by 10 10. y is 0 or x in the units places of and x 10 ways units place of can be filled in
The
1
1, 2,3,.. ,9 can be used). in 10 ways. Also, the units place of y can be filled x, y to have a the number of ways for the numbers
(:any one of 0,
:
1),
places= 10 x 10 = 100. digit each in their units can be x* as well as y digits in the units place of can square of any number from 1,4,5,6,9 because the in the units place. digits one of these have only be places of x' and yshould Sum of the digits in units
***********************************************************
The
Possible AP with common difference n is 1), ie., possibility (1,n+ 1, 1
2+
n{E)
=
(27
-
1) +
(2n
1+3+5t...
-
3)
+
(Zn - 5) +
..+1
+(2n--1)
-12x1(n- 1)2)
=
(S)
= C,
or
:
n?
So,
number of ways of drawing 2n cards of which are white and n are black; n = 1, 2,... 26
"C,"C,
if x
have 0 in unt has 0 in units place, y must
1)
in a row, his man parks his car among n cars standing On his return he finds car not being parked at an end. cars are still there. What is the that exactly m of the n on two sides of probability that both the cars parked
6. A
his car, have left? crosses (x). Clearly, his car is at one of the
Xxx *
X
*
*
5
un must have 5 in in units place, y
x has 4 or 6 in units in units place
place, y must have
the
1+2x2+2* 1x1+2 x2+2x 2+1x
3-2
2
+7.6+7
A
(
number is divisible by
11
if the difference of the sum in even
of the digits in odd places and that of the digits
places is divisible by 11. As the number is of seven digits we must have (for favourable cases), sum of four digits in odd places- sum of three digits in even places 33, 44, 55.
X+y=* X -y=ll
X+y=59
"Co+°C X+ "C2*r'+...+Czsx
or
X-y
-y=
+y=5*
x+y=5
or
33
or
I-y=44
(Gii)
55
I-y=
(vi)
(
(iv
:Clearly, ),
+y= 59 X-y=2 (ii)
or
and Gv) do not give integral values of
x
and y.
constant term in
(1
+X)
(i)
But
(vi)
+"C
c+c}+*c+. of x
(1+)
in ln25
Coetficient of x
in (1
y =24;
»
iv)
46, y = 13;
=57,y=2
.
+x)="C26
C=C3+*c}+"C}.*C nE)="C2»-"C6= 52 (26!3 therequired probaolty
I= 35,
=
Obviously, from (a), (b), (C), (d) and (e) we get, sum of three digitsy cannot be 2 or 13. Hence, only favourable case takes place when the sum of four digits in the odd places = 35 and the sum of the three aigits in even places= 24. in the favourable numbers we will get, 9,9,9,8 in odd places and 8, 8, 8 in even places places 9,9,9, 8 in odd places and 9, 8,7 in even or even places or 9,9,9,8 in odd places and ,9,6 in by 11 the number of numbers divisible
constant term in (1+
coefficient
3'3*31*3*2
52 (26 ) .
-4+24+12
5
. )
= 40
from (1) and (2), *
place, y musthave it x has 3 or 7 in units in units place to have numbers T, number of ways for the sum is 0 or whose places digit in their units
02,
X+y=
ne
or or9
5!2! 5! 6
4!2!
62 210
or
s place, y must have 0r if x has or 9 in units in units place 0r place, y musthave 4 if x has 2 or 8 in units
has place
7
3!4!
seven-dig
7.7,7
:If the two sums are denoted by x and y respectively then
in units place
(21+1)X2n-1) 4n-1
7
0,11,22,
1
(2m+1)2n(2n
total number of ways to form a whose sum of digits is 59
the
number
E("C,
Multiplying these and equating the constant terms on both sides,
,
10.
=
n
-c+C+*c3+... "c Now, (1 +)"=
place
the required probability
nE)
nE) =
-2
-1 (using property of binomial coefficients)
(-1)1
Two non-negative integers are chosen at random from replacement. the set of non-negative integers with What is the probability that the sum of their squares is
7.
are
(1,2, 3), (2,3,
(n-m)!
C+"C6t+"Ca=2*'-1
-2
11
+
m-1)
C
7- m- 1)
5.
,
total numberof ways of drawing even numberof cards
S)=
135.
5 x5=
from 1,2,3,..
n ordinary pack of 2 cards an even number of are drawn at random. Find the probability of getting equal numoe Or DIack and red cards. an ordinary pack ot 2 cards, 26 are black and 26 are
red.
-
n(E)
:CC,+C2 C2+ *C°C
In
(d)9,9,9,9,9,8,6
c)9,9,9,9,9,7,7 (e) 9,9,9,9,9,9,5
8.ds
his car vacant=" Cm-1
therequired probability
Probubility
probability= 9 the required 0050
="-"Cm-1:
)°-(1+*)
rC+Cr+Cr*..+Cr 10+
Element
ways in which the remaini The number of the -1 (excluding the car of cars can take their places man) there are n -1 places for them-
+x*+x'+x*
in (x* (1
coefficient of
MatheaticS
sum of the digits of a seven-digit numbers is divisible by ne probability that this number is 59, clearly at
As 7 x least
digits and the sum of seven ast three of the digits must be number will be as the seven digits of the b
y,
1Ollows: ta)
9,9,9, 8, 8, 8, 8
(b)9,9,9,9,8,8,7
the required probability
210 21 0.3, PMB)=04
that PA)= 10. A, B, C are events such = PC)= 0.8, P{AB) = 0.08, PLAC) 0.25 and 0.75 then PMABC)=0.09. If
P{AuBU)2
0.23 Sx P{BC) lies in the interval
:We know, P{A
FEiiillItiE
U
BuC)S1.
s048.
show that
Problems Plus in
sP(A UBUC)Ss1 = But, PlAUBuC) PA) + P(B) +
075
0.4 +0.8 0.08-P(BC) 159-0.36 PBC)
-
P(Å
-
PLA
n B)
(HH..
m times) xx
T(HH..
m times) xx
...
028 +0.09
PA)=P(C) or P(B)=
xT(HH...
... x
times) xx ..
But0 ) P(D/E;) + P{E) PDE)
C
PE)
P(E/E)+ P(Ey) P(E/E,) +P(Ep) P(E/ED) + P(Ec) P(E/E-}
10 50
+ As PE)+ PEz) P(E;)= Ez, E
PEs) P(E/E;)
P(Es/E)=
9
20C1
above).
because after drawing 2 black balls in the first draw there are 4 black balls and 2 white balls for the next draw. (1), the required probability
B
spades.
x9,111 20 x 19 9190
10025
C
Let A,
Similarly, P(E/ED) =P(EEc)=78 1275
the last defective in the 12th drawv
P(D/E)
events E
1312.
150125
We have to find P(Es/E), ie., the probability of the missing card being spade when the two cards drawn are
= 2x 10!8I*20
PE)
C
PE/E,)= probability of getting two defectives and nondefectives in the first 11 draws and 2
from
draw
F-29
probability of drawing 2 spades when one heart is missing
PEE)= .. (1)
PE/E,)= Probability of getting one defective and 10 nondefectives in the first 11 draws and the last defective in the 12th
C
PB/E)
Conditional Probability and BayesS Thcorenm
the required probability + PE)- P(E)- P(E/E2) P(Eg) P(E/E,)
a
PE)21
PLE)P(B/E,)
+
Total
roduced event ot a product being producedat A event of product being produced at B event of a product being produced
=PC)
As the question is answered by either guessing or copying or knowing,
PU-1Now, PA/G) = probability of answering correctly by
8uessing
=:
there are 4 choices and only one of them is correct
Inlems
pobubility
A=
of
answering correctly
Plus n
In
ATatiNmaatics
otal Conditional Prabability and Bayes' Thevrem
1Le, the probabi have to tind rb/T), spoke truth. ball was drawn when they By Bayes' theorem,
We
by
copying
given) PA/K)
=
probability ot answering corrextiy
P(B/1)=
by
t
ab
PB/E) rEDP(B/E,)+ P¢B/E,)+ P(B/E,)+ P{B/E,) 35
he knows he certainiy
1.3Z5.13 10 6 64
21+15
36
4
8. An
10
times out of 6. A speaks truth 3 times out of 4 while 10. A ball is drawn at random from a bag containing one black ball and five other balls of different colours. Both A and B report that a black ball has been drawn from the bag Find the probability of their assertion B, 7
six bals
=
3
bag containing balls ot other colours
the event of a black ball drawn from the bag
Obviously, exclusive. Now, P(T/B)
speaking truth speaking lies.
Let
B
=
B
E
ball of another colour
a
= P(Ez) =
P(E) = P(Eg)=
P(L/B)
=
balls. Let B= the event of drawing 3 black 3 black bals Now, PIB/E,) = probability of drawing when E, happens
balls and 5 white balls the event of the urm containing 6 black E, are equiprobable and they are
exhaustive. Now, PW/E,) =
probability of 3 balls drawn being white in case of
11
2, 66 x 70 280
ever Bayes theorem for equiprobable
Let
B
909
909
303
=the event of drawing a black ball in the next draw.
Now, the ball in the next draw may be black after the 3 balls were dravn from the bag containirng 6 black white and 3 white balls. Its probability is PE/W) PIBAE,/W), etc. by the theorem on total conditional probability P(B) =
PE, /W)
PIBAE, /w
-
+P(E,/W)-PIBAE, /W}. Now, PLB/E,/W)}
already drawn
C =
=
C
C3 3
PIBAE,/W)l=77
Bayes' theorem for equiprobable events,
PIBAE/W
P(W/E;) P(W/E)+ P(W/E,)
+
P(W/ES) + P(W/E)
C
PEAE,/W))=
5
C using (,
3b
T12,1 84
x1
66x70 55
84
909
140
0 90**30
P(B) PB)-9051*07
30 33 909
(55132+210+280)=c0
909
(1)
1 the probability of drawing black in case ot E when 3 whites are
=
10C.
P(W/E) C 12c
after draws The bag will contain no black balls with By balls if B took place in conjunction
Dy
x70 420 14
66 11
3
PE,/W)=
So, we have to find P(E/B).
84 3033*
+PE, /W)-PLBAE, /W)
E
6
Similarly, P(WIEA)
by
C1. PB/E,)=g
11
PE/W)1.1
PEs/W)-PIBAE, /W)}
7
C Similarly, P(BJEg)= 6
=
probability of both telling lies when a black ball is not drawn
33
1,2+ 33 30
84
ball will be drawn in the
balls and 6 white balls.
(:their sum 1
10
10
black
the event of the um containing 6 black balls and 4 white balls
P(W/E)=
probability of both telling truth when black ball is drawn
probability of B telling truth
909
P(E,/W)=1,
contains at least 3 white balls. the event of the urn containing 6 black
Each of Ex Ey Es
are exhaustive and mutually
:probability of A telling truth
a
Es=the event of the urn containing 6 black
and P(B) 1- 6 and
unknown number
balls and 3 white balls
the event of the bag containing 4 black balls and 2 balls of other colours 1
Es
E
=
=
and
Clearly, the urn
3 black bals
E,= the event of the bag containing 6 black balls Clearly, they are equally probable and no cbhe are blaà possibilities exist as 3 balls drawn at random means the bag has at least 3 black balls.
PE,)
being true?
urn ontains 6 black balls
that the chance that
12
E=the event of the
and
Clearly, PB)
15
909
30
white balls. 1hree balls are drawn successively and not replaced and are all found to be white. Prove
Es=the event of the bag containing 5 black bals
B
b6 x 70
next draw is
there are
and
B
22 11
1
(S 6) of
or unknown colours; th a balls are drawn at random and found to be Find the probability that no black ball is left in theb (Or Find the probability that the bag contained erad 3 black balls.) ag
E
T- the event of both A and L= the event of both A and
35
10
Let
B =
1
20
I.
ansiwers cormxtiy) probability that he knew We have to find PiK/A,ie., the when he answered corrctly. By Baves' theorem, PK) PA/K) PK/4A) PMG)- PMAG)+ P(C)- PiA/C) P(K) P(A/K)
Let
0 =
to
+ P(B) P(L/B)
P(B) P(T/8)
nowing 1
imilarly, P(E, /W)
P(B)- PCIJ8)
677
PrWems Plas
contains
Apue
5 coins, each either a
in
ilT Mathematics Total Cenditional Probability anal Bayes' "Theorem
PE/E,)
titty-paisa coin
PE/E
and One-rupee coin. wo coins are drawn at random expected at found to be one-rupee coins. kind the coins. remaining of the total value pobable) Clearly. at least two of the coins are one-rupee coins. Let coin and f stand for tifty-paisa oneruper tor stand or
rNE/E)+P(E Eg)
* PEE)P(EE
I0
+
3
1
10
20
10
coin
following: he coins in the purse can be any one of the 27. 3 3. : ir, If 5. Let E, = the event ot the purse containing 2r and 3f E the event ot the purse containing šr and 25 etc. Clearly, the events are E;, Es. E, and Es, and they are
Similarly, P(E3/E) =
3
10 20
PE/E)=1
PE)=
Also E, E. E, and exclusive. Now, P(E/E;)
=
Es
-+ 0
are exhaustive and mutually
PEsE)1. +
probability of drawing wo one-rupee coins when E happens =
10
++1
20
(value
of Ir and 2f)
+
P(E,/E) (value of 2r and 1f)
+
P(Es/E) (value of 3r) -
20(Rs 1.50)2 Rs 2)10 (Rs 2.50) R
C P(E/ES)==1 probability of the purse containing 2r and 3f when 2r are drawn, etc. By Bayes theorem for equiprobable events, Now, PE/E)
3
P(E2/E) (value of 3f)
P(E,/E)
PE/E)C1 E)
0
expected total value of the remaining coins
the
Similarly,
+15 200
3
3
3
=
=
**|D'0*Í*Rs
2.625.
. There are
three bags each containing 5 white and 2 black balls. AlSo, there are 2 bags each containing 1 white and black balis.A ball is drawn at random from a bag and it is found to be black. What is the chance that the black ball came from the first group? 9. A can hit a target 4 times in 5 shots; B can hit 3 times in 4 shots and C twice in 3 shots. They fire once each. f two of them hit, what is the chance that C has missed it? 10. A bag contains 5 balls of unknown colours. A ball is drawn and replaced twice. On each occasion it is found to be red. Again, two balls are drawn at a time. What is the probability of both the balls being red ? 11. A letter is known to have come from either MAHARASTRA or MADRAS. On the postmark only consecutive letters RA can be read clearly. What is the chance that the letter came from MAHARASTRA? 12. A bag contains 10 coins of which at least 2 are onerupee coins. 1wo coins are drawn and both are Tound to be not one-rupee coins. Whatis the probabilhty ot the bag to contain exactly 2 one-rupee coins?
There are two bags, one of which contains three black and four white balls while the other contains four black and three white belis. A dice is cast. If the face 1: or 3 turns up, a ball is ta ken out from the first bag. But if any other face turns up, a ball is taken from the second bag. Find the probability of getting a black
3.
respectively. A ball is drawn from an urn chosen at random. What is the probability that a white ball is drawn if the choices of urns are equiprobable?
:
5.
e
not i probability that a certain eiec does n used is 0.10. If it o ais when first for lasts it that probability the pon Edately, a new co s U.9. What is the probability that one wll last year? A factory A produces 10% detecvealves defective another factory B produce 20% nd 5 valve ron bag contains 4 valves of factory A a randu at actory B. If two valves are drawn
4. The
ball. 3 2. Three urns contain 2 white and 3 black balls, white 1 black ball and 2 black balls; and 4 white and
An urn contains two balls each of which i5e um. Wnat A white ball is added to the tne u probability of drawing a white ball from mpone
1
2
. 0.29 3
0.891
train and the probabilities
of these are
late if he takes car, scooter, bus or train
Given
that he reached office in
:and is5
time,
fhnd the
the 22nd
century
probability that he travelled by a car
Objective Questions Fillin
the blanks
15. The probability that in a chosen
-
at random
there
year of will
be 53
Sundays,
is
There are two bags X and Y; X contains 3 white balls and 2 black balls, and Y contains 2 white balls and
16.
4 black balls. A bag and a ball out of that bag are picked at random. The probability that the ball is
white,
coupors each with a word written on them. On one of them the word PATNA is written and on the other PLATE. A coupon is taken at random and 3 letters are selected at random tem the letters of the word on the coupon probability that the selection contains two vowels, is
17. There are two
he
15.
1wo persons A and I throw two dice each. 1he probability that Y throws a sum greater than X s it is knoWn that X hrows a sum of at when
leasto.
13
12
1615
19
2
respectively:. The probability that he reaches office
6.
10.
C
A heads and 1 was chosen. 4. A person goes to ofice either by car, scooter, bus or
AnswersS
or black
coin
random and tossed 3 times giving ta'l Find the probability that the coin
man has three coins A, B, C. The coin A is unbiased. The probability that a head will show
Exercises
in case of the
A coin is chosen at
5. A
1.
.35
when B is tossed is while it is
o
20
equaly probable. PrE,) = P\E;) = PME) =
find the probability that at least valve is defective. unbiasecd coim is tossed. If the result is a 6. An head, a s pair ot unDiasea ace is rolled and the number btained by adding the numbers shown on them is noted. It the resuit is a tal, àa card trom a well-shuffled erea 2, 3, 4,., pack of eleven ca 12 is picked and the number ot the card is noted. What is the chance that the noted number is either 7 or 8? . A bolt factory has three machines A, B and manufactunng respectvely z7o, 35% and 40% of the total production. hese the machines produce 5%, 4% and 27o aerectve bolts respectively. A bolt is selected at random and it is tound to be defective. Find the probability that it was manutactured by the machine (a) A (b) B (c)C. bag,
18.
108
14 17.
Total Conditional Probabilitya
Bays Theovem
Chapter Test Time 75 reseztes Abag contains 3 white and 3 black balls. A person draws 3 ballsat rardom fon t He en drops 3 red balls into the bag and azain draws out 3 bals at rardom. What is te arce that the latter 3 balls will be of different colour? 2. Abag contains n coins of unknon values. A con is drawmat rardom and it is ford o a rupee. What is the chance that it is te the only rupee con in the baz? 3. In a city filmland there are 3 male superstars and 3 ferrale sperstarn Ad r make a film with 3 superstars. If the film deides has 2 rale saperstas yestz te chance of the film to be a hit iswhile the ae chance is here 1 mae zd 2 superstars. f the 3 superstars are from the same sex, he carce oé te begats the time of making the film, only 3 superstars arearalabe 1.
rd1eale
ai
At
toigtrte
te
director takes them all then what is the probablity of his flbeigak? 4. Three groups A, B and C are competing for positions an he Boardé Des cf a company. The probabilities of their winning are 05, 03 and 02 spesrey The probabilities of introducing a new product ifa particular gop ws 2219erey, 0.6 and 0.5. Find the probability that a new product will be intodacd 5. A purse containing 16 notes, four each of denominations Rs 10, Rs 20, Ps 50 znds 100, is left on a table. A maid servant steals two of the notes. Afer at ie za akas ot wo notes at random from the purse and finds that they anont to Rs 200 What is te probability that an amount of Rs 200 was stolen by the aid servart Écn is pse? 6. An um containing 6 white balls and 4 black balls. A ball is drawn at randoaadisp back into the un along with 5 balls of the same colour as that of the bal dza Aball is drawn again at random. What is the probability that the ball drawa now is white? 7. Abag contains p white and q red balls. m(m are