Type II : If order of group is important : The number of ways in which 3p different things can be divided
equally into three distinct groups is (3 p)! t 3! = (3 p)!
3!(p!)3 (p!)3
Note : If order of group is not important : The number of ways in which mn different things can be
divided equally into m groups is mn!
(n!)m m!
If order of group is important: The number of ways in which mn different things can be divided
equally into m distinct groups is (mn)! × m! = (mn)! .
(n!)m m! (n!)m
Example: 37 In how many ways can 5 prizes be distributed among four students when every student can take one or more prizes
Solution: (a) [BIT Ranchi 1990; Rajasthan PET 1988, 97]
Example: 38
Solution: (d) (a) 1024 (b) 625 (c) 120 (d) 60
Example: 39
The required number of ways = 45 = 1024 [since each prize can be distributed by 4 ways]
Solution: (a)
Example: 40 The number of ways in which 9 persons can be divided into three equal groups is [Orissa JEE 2003]
Solution: (c) (a) 1680 (b) 840 (c) 560 (d) 280
Total ways = 9! = 9×8× 7× 6×5× 4 = 280.
(3! )3 3× 2× 3× 2×3× 2
The number of ways dividing 52 cards amongst four players equally, are [IIT 1979]
(a) 52! (b) 52! (c) 52! (d) None of these
(13! )4 (13!)2 4! (12!)4 4 !
Required number of ways = 52 C13 × 39C13 × 26C13 × 13C13 = 52! × 39! × 26! × 13! = 52! .
39!13! 26!13! 13!13! 13! (13! )4
A question paper is divided into two parts A and B and each part contains 5 questions. The number of ways in which a
candidate can answer 6 questions selecting at least two questions from each part is [Roorkee 1980]
(a) 80 (b) 100 (c) 200 (d) None of these
The number of ways that the candidate may select
2 questions from A and 4 from B = 5 C2 ×5C4 ; 3 questions form A and 3 from B = 5 C3 ×5C3
4 questions from A and 2 from B = 5 C4 × 5C2 . Hence total number of ways are 200.
Derangement.
Any change in the given order of the things is called a derangement.
If n things form an arrangement in a row, the number of ways in which they can be deranged so that no one
of them occupies its original place is n ! 1 − 1 + 1 − 1 + ...... + (−1)n . 1 .
1! 2! 3! n!
Example: 41 There are four balls of different colours and four boxes of colurs same as those of the balls. The number of ways in which
the balls, one in each box, could be placed such that a ball doesn't go to box of its own colour is [IIT 1992]
(a) 8 (b) 7 (c) 9 (d) None of these
Solution: (c) Number of derangement are = 4 ! 1 − 1 + 1 = 12 −4 +1 = 9 .
3!
2 ! 4 !
(Since number of derangements in such a problem is given by n ! 1 − 1 + 1 − 1 + 1 ...... + (−1)n 1 .
1! 2! 3! 4! n!
Some Important Results for Geometrical Problems.
(1) Number of total different straight lines formed by joining the n points on a plane of which m (< n) are
collinear is n C2 −m C2 + 1 .
(2) Number of total triangles formed by joining the n points on a plane of which m (< n) are collinear is
n C3 −mC3 .
(3) Number of diagonals in a polygon of n sides is n C2 − n .
(4) If m parallel lines in a plane are intersected by a family of other n parallel lines. Then total number of
parallelograms so formed is m C2 ×nC2 i.e mn(m − 1)(n − 1)
4
(5) Given n points on the circumference of a circle, then
(i) Number of straight lines = n C2 (ii) Number of triangles = n C3 (iii) Number of quadrilaterals = n C4 .
(6) If n straight lines are drawn in the plane such that no two lines are parallel and no three lines are
concurrent. Then the number of part into which these lines divide the plane is = 1 + Σn .
nn
∑ ∑(7) Number of rectangles of any size in a square of n × n is r 3 and number of squares of any size is r 2 .
r =1 r =1
(8) In a rectangle of n × p (n < p) number of rectangles of any size is np (n + 1)(p + 1) and number of squares
4
n
∑of any size is (n + 1 − r)(p + 1 − r) .
r =1
Example: 42 The number of diagonals in a octagon will be [MP PET 1984; Pb. CET 1989, 2000]
(a) 28 (b) 20 (c) 10 (d) 16
Solution: (b) Number of diagonals = 8 C2 − 8 = 28 − 8 = 20.
Example: 43
The number of straight lines joining 8 points on a circle is [MP PET 1984]
(a) 8 (b) 16 (c) 24 (d) 28
Solution: (d) Number of straight line = 8 C2 = 28.
Example: 44
The number of triangles that can be formed by choosing the vertices from a set of 12 points, seven of which lie on the
same straight line, is [Roorkee 1989, 2000; BIT Ranchi 1989; MP PET 1995; Pb. CET 1997; DCE 2002]
(a) 185 (b) 175 (c) 115 (d) 105
Solution: (a) Required number of ways = 12 C3 −7C3 = 220 – 35 = 185.
Example: 45
Out of 18 points in a plane, no three are in the same straight line except five points which are collinear. The number of
(i) straight lines (ii) triangles which can be formed by joining them [WB JEE 1992]
(i) (a) 140 (b) 142 (c) 144 (d) 146
(ii) (a) 816 (b) 806 (c) 800 (d) 750
Solution: (c, b) Out of 18 points, 5 are collinear
(i) Number of straight lines = 18 C2 −5C2 + 1 = 153 − 10 + 1 = 144
(ii) Number of triangles = 18C3 −5C3 = 816 − 10 = 806 .
Multinomial Theorem. .....(i)
Let x1, x 2 ,......., xm be integers. Then number of solutions to the equation x1 + x 2 + ...... + xm = n .....(ii)
Subject to the condition a1 ≤ x1 ≤ b1, a2 ≤ x 2 ≤ b2 ,......., am ≤ xm ≤ bm
is equal to the coefficient of xn in
(x a1 + x a1+1 + ...... + x b1 )(x a2 + x a2 +1 + ..... + x b2 )......(x am + x am+1 + ..... + x bm ) ......(iii)
This is because the number of ways, in which sum of m integers in (i) equals n, is the same as the number of
times xn comes in (iii).
(1) Use of solution of linear equation and coefficient of a power in expansions to find the number
of ways of distribution : (i) The number of integral solutions of x1 + x 2 + x3 + ...... + xr = n where
x1 ≥ 0, x 2 ≥ 0,......xr ≥ 0 is the same as the number of ways to distribute n identical things among r persons.
This is also equal to the coefficient of xn in the expansion of (x 0 + x1 + x 2 + x 3 + ......)r
= coefficient of xn in 1 1 x r = coefficient of xn in (1 − x)−r
−
= coefficient of xn in 1 + rx + r(r + 1) x 2 + ...... + r(r + 1)(r + 2)......(r + n − 1) x n + .......
2! n!
= r(r + 1)(r + 2)....(r + n − 1) = (r + n − 1)! = n+r −1Cr −1
n! n!(r − 1)!
(ii) The number of integral solutions of x1 + x2 + x3 + ..... + xr = n where x1 ≥ 1, x2 ≥ 1,.......xr ≥ 1 is same as
the number of ways to distribute n identical things among r persons each getting at least 1. This also equal to the
coefficient of xn in the expansion of (x1 + x 2 + x 3 + ......)r
= coefficient of xn in 1 x x r = coefficient of xn in xr (1 − x)−r
−
= coefficient of xn in x r 1 + rx + r (r + 1) x 2 + ..... + r (r + 1)(r + 2).....(r + n − 1) x n + .....
2! n!
= coefficient of xn−r in 1 + rx + r(r + 1) x 2 + ..... + r(r + 1)(r + 2).....(r + n − 1) x n + .....
2! n!
= r(r + 1)(r + 2)......(r + n − r − 1) = r(r + 1)(r + 2).....(n − 1) = (n (n − 1)! =n−1Cr −1 .
(n − r)! (n − r)! − r)!(r − 1)!
Example: 46 A student is allowed to select utmost n books from a collection of (2n + 1) books. If the total number of ways in which he
Solution: (b)
can select one book is 63, then the value of n is [IIT 1987; Rajasthan PET 1999]
(a) 2 (b) 3 (c) 4 (d) None of these
Since the student is allowed to select utmost n books out of (2n + 1) books. Therefore in order to select one book he has
the choice to select one, two, three,......., n books.
Thus, if T is the total number of ways of selecting one book then T = 2n+1C1 + 2n+1C2 + ..... + 2n+1Cn = 63 .
Again the sum of binomial coefficients
2n+1 C0 + 2n+1 C1 + 2n+1C2 + ...... + 2n+1Cn + 2n+1Cn+1 + 2n+1 Cn+2 + ..... + 2n+1C2n+1 = (1 + 1)2n+1 = 22n+1
or, 2n+1C0 + 2(2n−1C1 + 2n+1C2 + ..... + 2n+1Cn ) + 2n+1 C2n+1 = 22n+1
⇒ 1 + 2(T) + 1 = 22n+1 ⇒ 1+T = 2 2n+1 = 22n ⇒ 1 + 63 = 22n ⇒ 26 = 22n ⇒ n = 3.
2
Example: 47 If x, y and r are positive integers, then x Cr +xCr−1yC1 +xCr−2 yC2 + ..... +yCr =
[Karnataka CET 1993; Rajasthan PET 2001]
(a) x! y! (b) (x + y)! (c) x+y Cr (d) xy Cr
r! r!
Solution: (c) The result x+y Cr is trivially true for r = 1, 2 it can be easily proved by the principle of mathematical induction that the
result is true for r also.
Number of Divisors.
Let N = p1α1 .pα22 .pα33 ......pkαk , where p1, p2 , p3 ,......pk are different primes and α1,α 2 ,α 3 ,......,α k are natural
numbers then :
(1) The total number of divisors of N including 1 and N is = (α1 + 1)(α 2 + 1)(α 3 + 1)....(α k + 1)
(2) The total number of divisors of N excluding 1 and N is = (α1 + 1)(α 2 + 1)(α 3 + 1).....(α k + 1) − 2
(3) The total number of divisors of N excluding 1 or N is = (α1 + 1)(α2 + 1)(α3 + 1).....(αk + 1) − 1
(4) The sum of these divisors is = (p10 + p12 + p32 + ...... + p1α1 )(p20 + p12 + p22 + ... + pα2 2 ).....(pk0 + pk1 + pk2 + .... + pkαk )
(5) The number of ways in which N can be resolved as a product of two factors is
1 (α1 + 1)(α 2 + 1)....(αk + 1),If N is not a perfect square
2 + 1)(α 2 + 1).....(αk + 1) + 1], If N is a perfect square
1
2 [(α1
(6) The number of ways in which a composite number N can be resolved into two factors which are relatively
prime (or co-prime) to each other is equal to 2n−1 where n is the number of different factors in N.
Important Tips
• All the numbers whose last digit is an even number 0, 2, 4, 6 or 8 are divisible by 2.
• All the numbers sum of whose digits are divisible by 3,is divisible by 3 e.g. 534. Sum of the digits is 12, which are divisible by 3, and
hence 534 is also divisible by 3.
• All those numbers whose last two-digit number is divisible by 4 are divisible by 4 e.g. 7312, 8936, are such that 12, 36 are divisible by
4 and hence the given numbers are also divisible by 4.
• All those numbers, which have either 0 or 5 as the last digit, are divisible by 5.
• All those numbers, which are divisible by 2 and 3 simultaneously, are divisible by 6. e.g., 108, 756 etc.
• All those numbers whose last three-digit number is divisible by 8 are divisible by 8.
• All those numbers sum of whose digit is divisible by 9 are divisible by 9.
• All those numbers whose last two digits are divisible by 25 are divisible by 25 e.g., 73125, 2400 etc.
Example: 48 The number of divisors of 9600 including 1 and 9600 are [IIT 1993]
Solution: (c) [Rajasthan PET 2000]
Example: 49 (a) 60 (b) 58 (c) 48 (d) 46
Solution: (a) (d) 74
Since 9600 = 27 × 31 × 52
Hence number of divisors = (7 + 1)(1 + 1)(2 + 1) = 48 .
Number of divisors of n = 38808 (except 1 and n) is
(a) 70 (b) 68 (c) 72
Since 38808 = 8 × 4851 = 8 × 9 × 539 = 8 × 9 × 7 × 7 × 11 = 23 × 32 × 72 × 11
So, number of divisors = (3 + 1) (2 + 1) (2 + 1) (1 + 2) – 2 = 72 – 2 = 70.
Binomial theorem and mathematical induction
Binomial Expression.
An algebraic expression consisting of two terms with +ve or – ve sign between them is called a binomial
expression.
For example : (a + b), (2x − 3y), p − q , 1 + 4 etc.
x2 x4 x y3
Binomial Theorem for Positive Integral Index .
The rule by which any power of binomial can be expanded is called the binomial theorem.
If n is a positive integer and x, y ∈ C then
(x + y)n = nC0 x n−0 y 0 + nC1 x n−1y1 + nC2 x n−2 y 2 + ........ + nCr x n−r y r + ...... + nCn−1 xy n−1 + nCn x 0 y n
n .....(i)
∑i.e., (x + y)n = nCr .xn−r .yr
r=0
Here nC0 , nC1 , nC2 ,......nCn are called binomial coefficients and nCr = n! for 0 ≤r ≤ n.
r !(n − r)!
Important Tips
• The number of terms in the expansion of (x + y)n are (n + 1).
• The expansion contains decreasing power of x and increasing power of y. The sum of the powers of x and y in each term is equal to n.
• The binomial coefficients nC0, nC1, nC2........ equidistant from beginning and end are equal i.e., nCr =nCn−r .
• (x + y)n = Sum of odd terms + sum of even terms.
Some Important Expansions . by – y in (i), we get,
(1) Replacing y
(x − y)n = nC0 x n−0 .y 0 − nC1 x n−1 .y1 + nC2 x n−2 . y 2 .... + (−1)r nCr x n−r .y r + .... + (−1)n nCn x 0 .yη
n .....(ii)
∑i.e., (x − y)n = (−1)r nCr x n−r .y r
r=0
The terms in the expansion of (x − y)n are alternatively positive and negative, the last term is positive or
negative according as n is even or odd.
(2) Replacing x by 1 and y by x in equation (i) we get,
n
∑(1 + x)n =nC0 x 0 + nC1 x1 +nC2 x 2 + ...... + nCr x r + ...... +nCn x n i.e., (1 + x)n = nCr xr
r=0
This is expansion of (1 + x)n in ascending power of x.
(3) Replacing x by 1 and y by – x in (i) we get,
n
∑(1 − x)n = nC0 x 0 − nC1 x1 + nC2 x 2 − ...... + (−1)r nCr x r + .... + (−1)n nCn x n i.e., (1 − x)n = (−1)r nCr x r
r=0
(4) (x + y)n + (x − y)n = 2[nC0 x ny 0 + nC2 x n−2y 2 + nC4 x n−4 y 4 + .......] and
(x + y)n − (x − y)n = 2[nC1 x n−1y1 + nC3 x n−3 y 3 + nC5 x n−5 y 5 + .......]
(5) The coefficient of (r + 1)th term in the expansion of (1 + x)n is nCr .
(6) The coefficient of x r in the expansion of (1 + x)n is nCr .
Note : If n is odd, then (x + y)n + (x − y)n and (x + y)n − (x − y)n , both have the same number of terms
equal to n + 1 .
2
If n is even, then (x + y)n + (x − y)n has n + 1 terms and (x + y)n − (x − y)n has n terms.
2 2
Example: 1 x 5 + 10x 4 a + 40x 3a 2 + 80x 2a 3 + 80xa 4 + 32a 5 =
Solution: (c)
Example: 2 (a) (x + a)5 (b) (3x + a)5 (c) (x + 2a)5 (d) (x + 2a)3
Solution: (b)
Example: 3 Conversely (x + y)n =nC0 +nC1xn−1y1 +nC2xn−2y2 + .... + nCnx0yn
Solution: (c)
(x + 2a)5 = 5C0 x5 +5C1x4 (2a)1 +5C2x3(2a)2 +5 C3 x 2(2a)3 +5C4 x1(2a)4 +5C5 x0(2a)5
Example: 4
= x5 + 10x4a + 40x3a2 + 80x2a3 + 80xa4 + 32a5 .
Solution: (c)
The value of ( 2 + 1)6 + ( 2 − 1)6 will be
(a) – 198 (b) 198 (c) 98 (d) – 99
We know that, (x + y)n + (x − y)n = 2[xn +nC2xn−2y2 +nC4 xn−4y4 + .....]
( 2 + 1)6 + ( 2 − 1)6 = 2[( 2)6 +6C2( 2)4(1)2 +6C4( 2)2(1)4 +6C6( 2)0(1)6] = 2[8 + 15 × 4 + 30 + 1] = 198
The larger of 9950 + 10050 and 10150 is [IIT 1980]
(a) 9950 + 10050 (b) Both are equal (c) 10150 (d) None of these
We have, 10150 = (100 + 1)50 = 10050 + 50.10049 + 50.49 100 48 + .....(i)
2.1
and 9950 = (100 − 1)50 = 10050 − 50.10049 + 50.49 10048 − ....... ..... (ii)
2.1
Subtracting, 10150 − 9950 = 10050 + 2. 50.49.48 10047 + ..... > 10050 . Hence 10150 > 10050 + 9950 .
3.2.1
Sum of odd terms is A and sum of even terms is B in the expansion of (x + a)n , then [Rajasthan PET 1987]
(a) AB = 1 (x − a)2n − (x + a)2n (b) 2AB = (x + a)2n − (x − a)2n
4
(c) 4 AB = (x + a)2n − (x − a)2n (d) None of these
(x + a)n =nC0 xn +nC1xn−1a1 +nC2xn−2a2 + ... +nCnxn−n.an = (xn +nC2xn−2a2 + ..) + (nC1xn−1a1 +nC3 xn−3a3 + ....) = A + B .....(i)
Similarly, (x − a)n = A − B .....(ii)
From (i) and (ii), we get 4 AB = (x + a)2n − (x − a)2n
Trick: Put n = 1 in (x + a)n . Then, x + a = A + B . Comparing both sides A = x, B = a .
Option (c) L.H.S. 4 AB = 4 xa , R.H.S. (x + a)2 − (x − a)2 = 4ax . i.e., L.H.S. = R.H.S
General Term .
(x + y)n = nC0 x n y 0 + nC1 x n−1y1 + nC2 x n−2 y 2 + ..... + nCr x n−r y r + .... + nCn x 0 y n
The first term = nC0 x n y 0
The second term = nC1xn−1y1 . The third term = nC2 x n−2 y 2 and so on
The term nCr x n−r y r is the (r + 1)th term from beginning in the expansion of (x + y)n .
Let Tr +1 denote the (r + 1)th term ∴ Tr+1 = nCr x n−r y r
This is called general term, because by giving different values to r, we can determine all terms of the
expansion.
In the binomial expansion of (x − y)n , Tr+1 = (−1)r nCr x n−r y r
In the binomial expansion of (1 + x)n, Tr +1 =nCr xr
In the binomial expansion of (1 − x)n, Tr+1 = (−1)r nCr xr
Note : In the binomial expansion of (x + y)n , the pth term from the end is (n − p + 2)th term from
beginning.
Important Tips
• In the expansion of (x + y)n,n ∈N
Tr +1 = n − r + 1 y
Tr r x
• The coefficient of x n−1 in the expansion of (x − 1)(x − 2).......(x − n) = − n(n + 1)
2
• The coefficient of x n−1 in the expansion of (x + 1)(x + 2).....(x + n) = n(n + 1)
2
Example: 5 If the 4th term in the expansion of (px + x −1)m is 2.5 for all x ∈ R then
Solution: (b)
Example: 6 (a) p = 5 / 2,m = 3 (b) p = 1 , m = 6 (c) p = − 1 ,m = 6 (d) None of these
Solution: (d) 2 2
Example: 7 We have T4 = 5 ⇒ T3+1 = 5 ⇒ mC3 (px)m− 3 1 3 = 5 ⇒mC3 pm−3 x m−6 = 5 .......(i)
Solution: (b) 2 2 x 2 2
Example: 8
Solution: (c) Clearly, R.H.S. of the above equality is independent of x
∴m−6=0, m=6
Putting m=6 in (i) we get 6C3 p3 = 5 p= 1 . Hence p = 1 / 2,m = 6 .
2⇒ 2
If the second, third and fourth term in the expansion of (x + a)n are 240, 720 and 1080 respectively, then the value of n is
(a) 15 (b) 20 (c) 10 [Kurukshetra CEE 1991; DCE 1995, 2001]
(d) 5
It is given that T2 = 240,T3 = 720,T4 = 1080
Now, T2 = 240 ⇒ T2 =nC1 x n−1a1 = 240 .....(i) and T3 = 720 ⇒ T3 =nC2xn−2a2 = 720 .....(ii)
T4 = 1080 ⇒ T4 =nC3 xn−3a3 = 1080 .....(iii)
To eliminate x, T2 .T4 = 240.1080 = 1 ⇒ T2 . T4 = 1 .
T32 720.720 2 T3 T3 2
Now Tr +1 = nCr = n − r + 1 . Putting r=3 and 2 in above expression, we get n − 2 . 2 = 1 ⇒ n=5
Tr n Cr −1 r 3 − 2
n 1
The 5th term from the end in the expansion of x3 − 2 9 is
2 x3
(a) 63x 3 (b) 252 (c) 672 (d) None of these
− x3 x18
5th term from the end = (9 − 5 + 2)th term from the beginning in the expansion of x3 − 2 9 = T6
2 x3
⇒ T6 = T5 +1 =9C5 x3 4 − 2 5 = −9C4 .2. 1 = − 252 .
2 x3 x3 x3
If T2 in the expansion of (a + b)n and T3 in the expansion of (a + b)n+3 are equal, then n =
T3 T4
[Rajasthan PET 1987, 96]
(a) 3 (b) 4 (c) 5 (d) 6
T2 = n − 2 + 1 . b = n 2 1 b and T3 = n + 3 3 3 + 1 . b = n 3 1 b
T3 2 a − a T4 − a + a
T2 = T3 (given) ; ∴ n 2 1 b = n 3 1 b ⇒ 2n + 2 = 3n − 3 ⇒ n=5
T3 T4 − a + a
Independent Term or Constant Term.
Independent term or constant term of a binomial expansion is the term in which exponent of the variable is zero.
Condition : (n − r) [Power of x] + r . [Power of y] = 0, in the expansion of [x + y]n .
Example: 9 The term independent of x in the expansion of x + 3 10 will be
3 2x 2
Solution: (b)
Example: 10 [IIT 1965; BIT Ranchi 1993; Karnataka CET 2000; UPSEAT 2001]
Solution: (c)
(a) 3 (b) 5 (c) 5 (d) None of these
Example: 11 2 4 2
Solution: (b)
Example: 12 − r) 1 =10 1 8 / 2 3 2 5
Solution: (b) 2 3 2 4
(10 + r(−2) = 0 ⇒ r = 2 ∴ T3 C2 =
Example: 13
Solution: (b) (1 + x)n 1 + 1 n
Example: 14 x
The term independent of x in the expansion of is [EAMCET 1989]
(a) C02 + 2C12 + ...... + (n + 1)Cn2 (b) (C0 + C1 + ...... + Cn)2 (c) C02 + C12 + ...... + Cn2 (d) None of these
We know that, (1 + x)n =nC0 +nC1x1 +nC2x 2 + .......... +nCnxn
1 + 1 n =nC0 +nC1 1 +nC2 1 + ..... + nCn 1
x x1 x2 xn
Obviously, the term independent of x will be nC0.nC0 +nC1nC1 + ..... + nCn.nCn = C02 + C12 + ....... + Cn2
Trick : Put n=1 in the expansion of (1 + x)1 1 + 1 1 =1+ x+ 1 +1= 2+ x + 1 .....(i)
x x x
We want coefficient of x 0 . Comparing to equation (i). Then, we get 2 i.e., independent of x.
Option (c) : C02 + C12 + .....Cn2 ; Put n = 1 ; Then 1 C02 +1C12 = 1 + 1 = 2 .
The coefficient of x −7 in the expansion of ax − 1 11 will be [IIT 1967; Rajasthan PET 1996]
bx 2
(a) 462a6 (b) 462 a 5 (c) − 462a5 (d) 462 a 6
b5 b6 b6 − b5
For coefficient of x−7 , (11 − r)(1) + (−2).r = −7 ⇒ 11 − r − 2r = −7 ⇒ r =6; T7 =11C6 (a)5 − 1 6 = 462 a5
b b6
If the coefficients of second, third and fourth term in the expansion of (1 + x)2n are in A.P., then 2n2 − 9n + 7 is equal to
[AMU 2001]
(a) –1 (b) 0 (c) 1 (d) 3/2
T2 =2nC1 , T3 =2nC2, T4 =2nC3 are in A.P. then, 2.2nC2 =2nC1 +2nC3
2. 2n(.2n − 1) = 2n + 2n(2n − 1)(2n − 2)
2.1 1 3.2.1
On solving, 2n2 − 9n + 7 = 0
The coefficient of x 5 in the expansion of (1 + x 2 )5 (1 + x)4 is [EAMCET 1996; UPSEAT 2001; Pb. CET 2002]
(a) 30 (b) 60 (c) 40 (d) None of these
We have (1 + x 2 )5 (1 + x)4 = (5C0 +5C1 x 2 +5C2 x 4 + .....) (4C0 + 4C1 x1 + 4C2 x 2 + .......)
So coefficient of x 5 in [(1 + x 2 )5 (1 + x)4 ] = 5 C2.4 C1 +4C3 .5 C1 = 60
If A and B are the coefficient of xn in the expansions of (1 + x)2n and (1 + x)2n−1 respectively, then [MP PET 1999]
(a) A = B (b) A = 2B (c) 2A = B (d) None of these
Solution: (b) A = coefficient of xn in (1 + x)2n = 2nCn = (2n)! = 2.(2n − 1)! .....(i)
n!n! (n − 1)!n!
B= coefficient of xn in (1 + x)2n−1= 2n−1Cn = (2n − 1)! .....(ii)
n!(n − 1)!
Example: 15 By (i) and (ii) we get, A = 2B [AIEEE 2004]
The coefficient of xn in the expansion of (1 + x)(1 − x)n is
(a) (−1)n−1n (b) (−1)n(1 − n) (c) (−1)n−1(n − 1)2 (d) (n − 1)
Solution: (b) Coefficient of xn in (1 + x)(1 − x)n = Coefficient of xn in (1 − x)n + coefficient of xn−1 in (1 − x)n
= Coefficient of xn in [nCn(−x)n + x.nCn−1(−x)n−1] = (−1)n nCn + (−1)n−1.nC1 = (−1)n + (−1)n.(−n) = (−1)n[1 − n] .
Number of Terms in the Expansion of (a + b + c)n and (a + b + c + d)n .
(a + b + c)n can be expanded as : (a + b + c)n = {(a + b) + c}n
= (a + b)n +nC1(a + b)n−1(c)1 +nC2 (a + b)n−2 (c)2 + ..... + nCn c n = (n + 1) term + n term + (n − 1)term + ... + 1term
∴ Total number of terms = (n + 1) + (n) + (n − 1) + ...... + 1 = (n + 1)(n + 2) .
2
Similarly, Number of terms in the expansion of (a + b + c + d)n = (n + 1)(n + 2)(n + 3) .
6
Example: 16 If the number of terms in the expansion of (x − 2y + 3z)n is 45, then n =
(a) 7 (b) 8 (c) 9 (d) None of these
Solution: (b) Given, total number of terms = (n + 1)(n + 2) = 45 ⇒ (n + 1)(n + 2) = 90 ⇒ n=8.
2
Example: 17 The number of terms in the expansion of [(x + 3y)2 (3x − y)2 ]3 is [Rajasthan PET 1986]
(a) 14 (b) 28 (c) 32 (d) 56
Solution: (b) We have [(x + 3y)(3x − y)]6 = [3x 2 + 8xy − 3y 2 ]6 ; Number of terms = (6 + 1)(6 + 2) = 28
2
Middle Term.
The middle term depends upon the value of n.
(1) When n is even, then total number of terms in the expansion of (x + y)n is n + 1 (odd). So there is only
n 1 th
2
one middle term i.e., + term is the middle term. T n +1 =nCn / 2 x n / 2yn / 2
2
(2) When n is odd, then total number of terms in the expansion of (x + y)n is n + 1 (even). So, there are two
n + 1 th n + 3 th n+1 n−1 n−1 n+1
2 2
middle terms i.e., and are two middle terms. T n+1 =nCn−1 x 2 y 2 and T n+3 =nCn+1 x 2 y 2
2 2
2 2
Note : When there are two middle terms in the expansion then their binomial coefficients are equal.
Binomial coefficient of middle term is the greatest binomial coefficient.
x + 1 10
x
Example: 18 The middle term in the expansion of is [BIT Ranchi 1991; Rajasthan PET 2002; Pb. CET 1991]
Solution: (b) (a) 10 C4 1 (b) 10 C5 (c) 10 C5 x (d) 10 C7 x 4
Example: 19 x
∵ n is even so middle term T 10 = T6 ⇒ T6 = T5+1 =10 C5 x 5 . 1 =10 C5
2 x5
+1
The middle term in the expansion of (1 + x)2n is [Pb. CET 1998]
(a) 1.3.5......(2n − 1) x 2n+1 (b) 2.4.6......2n x 2n+1 (c) 1.3.5......(2n − 1) xn (d) 1.3.5......( 2n − 1) x n. 2n
n! n! n! n!
Solution: (d) Since 2n is even, so middle term = T2n = Tn+1 ⇒ Tn+1 =2nCnxn = (2n)! xn = 1.3.5........( 2n − 1) . 2n x n .
n!. n! n!
2 +1
To Determine a Particular Term in the Expansion.
In the expansion of xα ± 1 n , if xm occurs in Tr+1 , then r is given by nα − r(α + β)= m ⇒ r = nα − m
xβ α+β
Thus in above expansion if constant term which is independent of x, occurs in Tr+1 then r is determined by
nα − r(α + β ) = 0 ⇒ r = nα
α+β
Example: 20 The term independent of x in the expansion of 3x 2 1 9 is
Solution: (b) 2 3x
Example: 21 − [Rajasthan PET 1996; AMU 1998]
Solution: (a) (a) 7/12 (b) 7/18 (c) – 7/12 (d) – 7/16
n =9, α =2, β = 1 . Then r 9(2) = 6 . Hence, T7 =9C6 3 3 − 1 6 = 9 × 8 × 7 . 1 = 7 .
= 1+2 2 3 3 × 2 × 1 23.33 18
If the coefficient of x7 in ax 2 + 1 11 is equal to the coefficient of x −7 in ax − 1 11 then ab =
bx bx 2
(a) 1 (b) 1/2 (c) 2 [MP PET 1999; AMU 2001; Pb. CET 2002]
(d) 3
For coefficient of x7 in ax 2 + 1 11 ; n = 11, α = 2, β =1, m=7
bx
r = 11.2 − 7 = 15 = 5
2+1 3
Coefficient of x7 in T6 =11C5 a6. 1 ......(i)
b5
and for coefficient of x −7 in ax − 1 11 ; n = 11,α = 1, β = 2, m = −7 ; r = 11.1 + 7 =6
bx 2 3
Coefficient of x −7 in T7 = 11 C6 .a 5 . 1 .....(ii)
b6
From equation (i) and (ii), we get ab = 1
Greatest Term and Greatest Coefficient.
(1) Greatest term : If Tr and Tr+1 be the rth and (r + 1)th terms in the expansion of (1 + x)n , then
Tr +1 = nCr x r = n − r + 1 x
Tr nCr−1 x r−1 r
Let numerically, Tr +1 be the greatest term in the above expansion. Then Tr+1 ≥ Tr or Tr +1 ≥1
Tr
∴ n − r + 1| x |≥ 1 or r ≤ (n |+x1|) )| x | ......(i)
r (1+
Now substituting values of n and x in (i), we get r ≤ m + f or r ≤ m
where m is a positive integer and f is a fraction such that 0 < f < 1 .
When n is even Tm+1 is the greatest term, when n is odd Tm and Tm+1 are the greatest terms and both are equal.
Short cut method : To find the greatest term (numerically) in the expansion of (1 + x)n .
(i) Calculate m = x(n + 1)
x +1
(ii) If m is integer, then Tm and Tm+1 are equal and both are greatest term.
(iii) If m is not integer, there T[m]+1 is the greatest term, where [.] denotes the greatest integral part.
(2) Greatest coefficient
(i) If n is even, then greatest coefficient is nCn / 2
(ii) If n is odd, then greatest coefficient are nCn+1 and n Cn+3
22
Important Tips
y)n . we xn 1 + y n
x
• For finding the greatest term in the expansion of (x + rewrite the expansion in this form (x + y)n = .
Greatest term in (x + y)n = xn . Greatest term in 1 + y n
x
Example: 22 The largest term in the expansion of (3 + 2x)50 , where x = 1 is
Solution: (c,d) 5
Example: 23 (a) 5th (b) 8th (c) 7th (d) 6th
Solution: (b)
2x)50 350 1 + 2x 50 1 + 2x 50
3 3
(3 + = , Now greatest term in
2x (50 1) 2. 1 (51)
3 5 +1
x(n + 1) + 3
r= 1+ x 6 (an integer)
= 2x = 2 =
3
+1 15
∴ Tr and T[r]+1 = T6 and T[6]+1 = T6 and T7 are numerically greatest terms
The greatest coefficient in the expansion of (1 + x)2n+2 is [BIT Ranchi 1992]
(a) (2n)! (b) (2n + 2)! (c) (2n + 2)! (d) (2n)!
n!2 [(n + 1)!]2 n!(n + 1)! n!.(n + 1)!
n is even so greatest coefficient in (1 + x)2n+2 is = 2n+ 2Cn+1 = (2n + 2)!
[(n + 1)!]2
Example: 24 The interval in which x must lie so that the greatest term in the expansion of (1 + x)2n has the greatest coefficient is
(a) n − 1 , n (b) n , n + 1 (c) n , n + 2 (d) None of these
n − + n + n
n 1 n 1 n 2
Solution: (b) Here the greatest coefficient is 2nCn
∴ 2nCn xn > 2nCn +1 x n −1 ⇒ x> n and 2n Cn xn > 2nCn −1 x n +1 ⇒ x< n + 1 . Hence the result is (b)
n+1 n
Properties of Binomial Coefficients.
In the binomial expansion of (1 + x)n, (1 + x)n =nC0 + nC1 x + nC2 x 2 + ..... + nCr x r + .... + nCn x n .
where nC0 , nC1, nC2 ,......, nCn are the coefficients of various powers of x and called binomial coefficients,
and they are written as C0 , C1, C2,.....Cn .
Hence, (1 + x)n = C0 + C1 x + C2 x 2 + ..... + Cr x r + ..... + Cn x n .....(i)
(1) The sum of binomial coefficients in the expansion of (1 + x)n is 2n .
Putting x = 1 in (i), we get 2n = C0 + C1 + C2 + ..... + Cn .....(ii)
(2) Sum of binomial coefficients with alternate signs : Putting x = −1 in (i)
We get, 0 = C0 − C1 + C2 − C3 + ...... …..(iii)
(3) Sum of the coefficients of the odd terms in the expansion of (1 + x)n is equal to sum of the coefficients of
even terms and each is equal to 2n−1 .
From (iii), we have C0 + C2 + C4 + ..... = C1 + C3 + C5 + ........ ......(iv)
i.e., sum of coefficients of even and odd terms are equal.
From (ii) and (iv), C0 + C2 + C4 + ..... = C1 + C3 + C5 + ..... = 2n−1 ......(v)
(4) nCr = n n−1Cr −1 = n . n −1 n−2Cr −2 and so on.
r r r −1
(5) Sum of product of coefficients : Replacing x by 1 in (i) we get 1 + 1 n = C0 + C1 + C2 + ... Cn + .... (vi)
x x x x2 xn
Multiplying (i) by (vi), we get (1 + x)2n = (C0 + C1x + C2x 2 + ......) C0 + C1 + C2 + .....
xn x x2
Now comparing coefficient of x r on both sides. We get, 2nCn+r = C0Cr + C1Cr +1 + ......Cn−r .Cn .....(vii)
(6) Sum of squares of coefficients : Putting r = 0 in (vii), we get 2nCn = C02 + C12 + ......Cn2
(7) nCr +nCr −1 =n+1Cr
Example: 25 The value of C1 + C3 + C5 + ..... is equal to [Karnataka CET 2000]
Solution: (a) 2 4 6
2n + 1
Example: 26 (a) 2n − 1 (b) n.2n (c) 2n (d) n+1
Solution: (c) n+1 n
Example: 27 We have C1 + C3 + C5 + ...... = n + n(n − 1)(n − 2) + n(n − 1)(n − 2)(n − 3)(n − 4) + .....
2 4 6 2.1 4.3.2.1 6.5.4.3.2.1
= 1 (n + 1)n + (n + 1)(n)(n − 1)(n − 2) + ..... = 1 [2(n+1)−1 −1] = 2n − 1
+ 2! 4! + n+1
n 1 n 1
Trick: For n =1, = C1 = 1 C1 = 1
2 2 2
Which is given by option (a) 2n − 1 = 21 − 1 = 1 .
n+1 1+1 2
The value of C0 + 3C1 + 5C2 + ..... + (2n + 1)Cn is equal to [IIT 1993]
(a) 2n (b) 2n + n. 2n−1 (c) 2n(n + 1) (d) None of these
nn nn
∑ ∑ ∑ ∑We have C0 + 3C1 + 5C2 + ..... + (2n + 1) Cn =
(2r + 1)Cr = (2r + 1) nCr = 2r nCr + n Cr
r=0 r=0 r=0 r=0
∑ ∑ ∑ ∑= n n n n n = 2n[(1 + 1)n−1] + [1 + 1]n 2n. 2n−1 + 2n = 2n.[n + 1] .
r. r . n−1Cr −1 + r=0 nCr = n−1 Cr −1 + r=0 n Cr =
2. 2n
r =1 r =1
Trick: Put n = 1 in given expansion 1C0 + 3.1C1 = 1 + 3 = 4 .
Which is given by option (c) 2n.(n + 1) = 21(1 + 1) = 4 .
∑ ∑If n 1 and tn n r . Then tn is equal to
Sn = r=0 nCr = r=0 n Cr Sn [AIEEE 2004]
(a) 2n − 1 (b) 1 n − 1 (c) n − 1 (d) n
2 2 2
Solution: (d) Take n = 2m , then, Sn = 1 + 1 + ...... + 1 = 2 1 + 1 + ...... + 1 + 1
2m C0 2m C1 2m C2m 2m C0 2m C1 2mCm
Example: 28 2m Cm−1
Solution: (a)
∑ ∑n r 2m r = 1 2 + ...... + 2m
Example: 29 nCr 2mCr 2m C1 + 2m C2 2m C2m
Solution: (c) tn = =
Example: 30 r=0 r=0
Solution: (c)
Example: 31 tn = 1 + 2m − 1 + 2 + 2m − 2 + ...... + m−1 + m+1 + m + 2m
Solution: (c) 2m C1 2m C2m−1 2m C2 2m C2m−2 2m Cm−1 2m Cm+1 2m Cm 2m C2m
Example: 32
= 2m 1 + 1 + ..... + m + 2m = 2m 1 + 1 + ... + 1 + m = mSn − 1 + m = mSn
2m C1 2m C2 2m Cm 2m C0 2m C1 2m Cm 2m Cm
2m Cm−1 2m Cm
tn = mSn ⇒ tn =m= n
Sn 2
If (1 − x + x2)n = a0 + a1x + a2x2 + ...... + a2nx2n . Then a0 + a2 + a4 + ...... + a2n =
[MNR 1992; DCE 1996; AMU 1998; Rajasthan PET 1999; Karnataka CET 1999; UPSEAT 1999]
(a) 3n + 1 (b) 3n −1 (c) 1 − 3n (d) 3n + 1
2 2 2 2
(1 − x + x 2 )n = a0 + a1 x + a2 x 2 + ...... + a2n x 2n
Putting x = 1, we get (1 − 1 + 1)n = a0 + a1 + a2 + .... + a2n ; 1 = a0 + a1 + a2 + ..... + a2n .....(i)
Again putting x = −1 , we get 3n = a0 − a1 + a2 − ...... + a2n ......(ii)
Adding (i) and (ii), we get, 3n + 1 = 2[a0 + a2 + a4 + ...... + a2n ]
3n + 1 = a0 + a2 + a4 + ....... + a2n
2
∑If n 1 + C1 1 + C2 ......1 + Cn
(1 + x)n = r=0 Cr x r , then C0 C1 Cn−1 = [BIT Ranchi 1987]
(a) nn−1 (b) (n + 1)n−1 (c) (n + 1)n (d) (n + 1)n+1
(n − 1)! (n − 1)! n! n!
We have 1 + C1 1 + C2 ........1 + Cn = 1 + n 1 + n(n − 1) / 2! ......1 + 1
C0 C1 Cn−1 1 n n
= 1 + n 1 + n 1 + n ....... 1 + n = (n + 1)n
1 2 3 n n!
Trick : Put n = 1, 2, 3......., S1 = 1+ 1 C1 = 2, S2 = 1 + 2 C1 1 + 2 C2 = 9
1 C0 2 C1 2 C1 2
Which is given by option (c) n =1, (1 + 1)1 = 2 ; For n= 2, (2 + 1)2 9
1! 2! =2
In the expansion of (1 + x)5 , the sum of the coefficient of the terms is [Rajasthan PET 1992, 97; Kurukshetra CEE 2000]
(a) 80 (b) 16 (c) 32 (d) 64
Putting x = 1 in (1 + x)5 , the required sum of coefficient = (1 + 1)5 = 25 = 32
If the sum of coefficient in the expansion of (α 2x2 − 2α x + 1)51 vanishes, then the value of α is [IIT 1991; Pb. CET 1988]
(a) 2 (b) –1 (c) 1 (d) – 2
The sum of coefficient of polynomial (α 2x2 − 2α x + 1)51 is obtained by putting x = 1 in (α 2x2 − 2α x + 1)51 . Therefore by
hypothesis (α 2 − 2α + 1)51 = 0 ⇒ α = 1
If Cr stands for nCr , the sum of given series 2(n / 2)! (n / 2)! [C02 − 2C12 + 3C22 − ...... + (−1)n(n + 1)Cn2] where n is an even
n!
positive integer, is [IIT 1986]
(a) 0 (b) (−1)n / 2 (n + 1) (c) (−1)n(n + 2) (d) (−1)n / 2(n + 2)
Solution: (d) We have C02 − 2C12 + 3C22 − ........ + (−1)n(n + 1)Cn2 = [C02 − C12 + C22 − ....... + (−1)nCn2] − [C12 − 2C22 + 3C32...... + (−1)nn.Cn2]
Example: 33 = (−1)n / 2.nCn / 2 − (−1)n / 2 −1. 1 n.n Cn / 2 = (−1)n / 2 1 + n nCn / 2
Solution: (a) 2 2
2 . n ! n !
2 2 (−1)n n n!
Therefore the value of given expression = / 2 . 1 + 2 = (−1)n / 2 (n + 2)
!
n! n ! n
2 2
If (1 + x)n = C0 + C1x + C2x2 + ....... + Cnxn , then the value of C0 + 2C1 + 3C2 + ...... + (n + 1)Cn will be
[MP PET 1996; Rajasthan PET 1997; DCE 1995; IIT 1971; AMU 1995; EAMCET 2001]
(a) (n + 2) 2n−1 (b) (n + 1)2n (c) (n + 1) 2n−1 (d) (n + 2) 2n
Trick: Put n = 1 the expansion is equivalent to 1C0 + 2.1C1 = 1 + 2 = 3 .
Which is given by option (a) = (n + 2)2n−1 = (1 + 2) 20 = 3
(1) Use of Differentiation : This method applied only when the numericals occur as the product of binomial
coefficients.
Solution process : (i) If last term of the series leaving the plus or minus sign be m, then divide m by n if q
be the quotient and r be the remainder. i.e., m = nq + r
Then replace x by x q in the given series and multiplying both sides of expansion by x r .
(ii) After process (i), differentiate both sides, w.r.t. x and put x = 1 or − 1 or i or – i etc. according to given series.
(iii) If product of two numericals (or square of numericals) or three numericals (or cube of numerical) then
differentiate twice or thrice.
Example: 34 C1 + 2C2 + 3C3 + .....n Cn = [Rajasthan PET 1995; MP PET 2002]
(a) 2n
(b) n. 2n (c) n. 2n−1 (d) n.2n+1
Solution: (c) We know that, (1 + x)n = C0 + C1x + C2x2 + ...... + Cnxn ......(i)
Differentiating both sides w.r.t. x, we get n(1 + x)n−1 = 0 + C1 + 2 . C2x + 3C3x2 + ....... + nCnxn−1
Putting x = 1, we get, n . 2n−1 = C1 + 2C2 + 3C3 + ...... + n Cn .
Example: 35 If n is an integer greater than 1, then a −nC1(a − 1) +nC2(a − 2) − ........ + (−1)n(a − n) = [IIT 1972]
(a) a (b) 0 (c) a 2 (d) 2n
Solution: (b) We have a[C0 − C1 + C2 ......] + [C1 − 2C2 + 3C3 .....] = a[C0 − C1 + C2 ......] − [−C1 + 2C2 − 3C3 + ......]
We know that (1 − x)n = C0 − C1x + C2x2...... + (−1)nCnxn ; Put x = 1 , 0 = C0 − C1 + C2 − .......
Then differentiating both sides w.r.t. to x , we get n(1 − x)n−1 = 0 − C1 + 2C2 x − 3C3 x 2 + .....
Put x = 1 , 0 = −C1 + 2C2 − 3C3 + .... = a[0] − [0] = 0 .
(2) Use of Integration : This method is applied only when the numericals occur as the denominator of the
binomial coefficients.
Solution process : If (1 + x)n = C0 + C1 x + C2 x 2 + ..... + Cn x n , then we integrate both sides between the
suitable limits which gives the required series.
(i) If the sum contains C0 , C1, C2,....... Cn with all positive signs, then integrate between limit 0 to 1.
(ii) If the sum contains alternate signs (i.e. +, –) then integrate between limit – 1 to 0.
(iii) If the sum contains odd coefficients i.e., (C0, C2, C4.....) then integrate between –1 to 1.
(iv) If the sum contains even coefficients (i.e., C1,C3,C5 .....) then subtracting (ii) from (i) and then dividing by 2.
(v) If in denominator of binomial coefficients is product of two numericals then integrate two times, first taking
limit between 0 to x and second time take suitable limits.
Example: 36 C0 + C1 + C2 + ..... + Cn = [Rajasthan PET 1996]
Solution: (c) 1 2 3 n+1
Example: 37 (a) 2n (b) 2n − 1 (c) 2n+1 − 1 (d) None of these
Solution: (a) n+1 n+1 n+1
Example: 38
Solution: (d) Consider the expansion (1 + x)n = C0 + C1x + C2x2 + .... + Cnxn .....(i)
Integrating both sides of (i) within limits 0 to 1 we get, 1 x)ndx 1 1 1 C2x2 + ....... + 1 xndx
(1 + C0 + C1x + Cn
∫ ∫ ∫ ∫ ∫0 =
00 0 0
(1 + x)n+1 1 = C0[x]10 + C1 x2 1 + ........ + Cn xn+1 1
2 n+1
n+1 0 0 0
2n+1 − 1 = C0[1] + C1 1 + C2 1 + ....... + Cn. 1 ; C0 + C1 + C2 + .... + Cn = 2n+1 − 1 .
n+1 n+1 2 3 n+1 2 3 n+1 n+1
2C0 + 22 C1 + 23 C2 + ...... + 211 C10 = [MP PET 1999; EAMCET 1992]
2 3 11
(a) 311 − 1 (b) 211 − 1 (c) 113 − 1 (d) 112 − 1
11 11 11 11
It is clear that it is a expansion of (1 + x)10 = C0 + C1x + C2x2 + ..... + C10x10
Integrating w.r.t. x both sides between the limit 0 to 2.
(1 + x)11 2 = C0[x]02 + C1 x2 2 + C2 x3 2 + ...... + C10 x11 2 ⇒ 311 − 1 = 2C0 + 22 .C1 + 23 .C2 + ...... + 211 C10 .
2 3 11 11 2 2 11
11 0 0 0 0
The sum to (n + 1) terms of the following series C0 − C1 + C2 − C3 + ..... is
2 3 4 5
(a) 1 (b) 1 (c) 1 (d) None of these
n+1 n+2 n(n + 1)
(1 − x)n = C0 − C1x + C2 x 2 − C3 x 3 + .......
∫⇒ x(1 − x)n = C0 x − C1x 2 + C2 x 3 − C3 x 4 + ..... ⇒ 1 − x)n dx = C0 x2 1 − C1 x3 1 + C2 x4 1 − ....... (i)
2 3 4
x(1 0 0
0 0
0 n by putting 1− x =t, 1 n t n+1)dt 1 1
∫ ∫The integral on L.H.S. of (i) = t) t n+1 n+
(1 − (−dt) ⇒ (t − = − 2
1 0
Whereas the integral on the R.H.S. of (i)
= C0 1 − C1 1 + C2 − ....... = C0 − C1 + C2 − ....... to (n + 1) terms = 1 − n 1 2 = (n 1 + 2)
2 3 4 2 3 4 n+1 + + 1)(n
Trick : Put n = 1 in given series = 1 C0 − 1 C1 = 1 . Which is given by option (d).
2 3 6
An Important Theorem.
If ( A + B)n = I + f where I and n are positive integers, n being odd and 0 ≤ f < 1 then (I + f). f = Kn where
A − B2 = K > 0 and A − B < 1.
Note : If n is even integer then ( A + B)n + ( A − B)n = I + f + f ′
Hence L.H.S. and I are integers.
∴ f + f ′ is also integer; ⇒ f + f ′ = 1 ; ∴ f ′ = (1 − f)
Hence (I + f)(1 − f) = (I + f)f ′ = ( A + B)n ( A − B)n = (A − B 2 )n = K n .
Example: 39 Let R = (5 5 + 11)2n+1 and f = R − [R] where [.] denotes the greatest integer function. The value of R.f is [IIT 1988]
Solution: (a)
(a) 4 2n+1 (b) 4 2n (c) 4 2n−1 (d) 4 −2n
Since f = R − [R] , R = f + [R]
[5 5 + 11]2n+1 = f + [R] , where [R] is integer
Now let f ' = [5 5 − 11]2n+1,0 < f ' < 1
[ ]f + [R] − f ' = [5 5 + 11]2n+1 − [5 5 − 11]2n+1 = 2 2n+1 C1(5 5 )2n(11)1 + 2n+1C3 (5 5 )2n−2 (11)3 + .......
= 2.(Integer) = 2K (K ∈ N) = Even integer
Hence f − f ' = even integer – [R], but −1 < f − f ' < 1 . Therefore, f − f ' = 0 ∴ f = f '
Hence R.f = R.f ' = (5 5 + 11)2n+1(5 5 − 11)2n+1 = 4 2n+1 .
Multinomial Theorem (For positive integral index).
∑If n is positive integer and a1, a2, a3 ,....an ∈ C then (a1 + a2 + a3 + ... + am )n = n1 ! n ! n! !...nm ! a1n1 a n2 ...a mnm
n3 2
2
Where n1,n2 ,n3 ,.....nm are all non-negative integers subject to the condition, n1 + n2 + n3 + .....nm = n .
(1) The coefficient of a n1 .a n2 .....amnm in the expansion of (a1 + a2 + a3 + ....am)n is n!
1 2 n1! n2! n3!....nm!
(2) The greatest coefficient in the expansion of (a1 + a2 + a3 + ....am )n is n! + 1)!]r
(q!)m−r [(q
Where q is the quotient and r is the remainder when n is divided by m.
(3) If n is +ve integer and a1 , a2 ,.....am ∈ C, a1n1 . a n2 .........amnm then coefficient of xr in the expansion of
2
∑(a1 + a2x + .....amxm−1)n is n!
n1!n2!n3 !.....nm!
Where n1,n2.....nm are all non-negative integers subject to the condition: n1 + n2 + .....nm = n and
n2 + 2n3 + 3n4 + .... + (m − 1)nm = r .
(4) The number of distinct or dissimilar terms in the multinomial expansion (a1 + a2 + a3 + ....am )n is n+m−1 Cm−1 .
Example: 40 The coefficient of x5 in the expansion of (x 2 − x − 2)5 is [EAMCET 2003]
Solution: (c)
(a) – 83 (b) – 82 (c) – 81 (d) 0
Example: 41
Solution: (b) ∑Coefficient of x5 in the expansion of (x2 − x − 2)5 is n1 !. 5! ! (1)n1 (−1)n2 (−2)n3 .
n2 ! n3
where n1 + n2 + n3 = 5 and n2 + 2n3 = 5 . The possible value of n1, n2 and n3 are shown in margin
n1 n2 n3
13 1
21 2
05 0
∴ The coefficient of x5 = 1! 5! (1)1(−1)3 (−2)1 + 5! (1)2 (−1)1(−2)2 + 5! (1)0 (−1)5 (−2)0 = 40 − 120 − 1 = −81
3!1! 2!1!2! 0!5!0!
Find the coefficient of a3b4c5 in the expansion of (bc + ca + ab)6
(a) 0 (b) 60 (c) – 60 (d) None of these
In this case, a3b4c5 = (ab)x(bc)y(ca)z = ax+z.bx+y.cy+z
z + x = 3, x + y = 4, y + z = 5 ; 2(x + y + z) = 12 ; x + y + z = 6 . Then x = 1, y = 3, z = 2
Therefore the coefficient of a3b4c5 in the expansion of (bc + ca + ab)6 = 6! = 60 .
1!3! 2!
Binomial Theorem for any Index.
n(n − 1)x2
Statement : (1 + x)n = 1 + nx + 2! + n(n − 1)(n − 2) x3 + .... + n(n − 1)......(n − r + 1) xr + ...terms up to ∞
3! r!
When n is a negative integer or a fraction, where − 1 < x < 1 , otherwise expansion will not be possible.
If x < 1, the terms of the above expansion go on decreasing and if x be very small a stage may be reached
when we may neglect the terms containing higher power of x in the expansion, then (1 + x)n = 1 + nx .
Important Tips
• Expansion is valid only when −1 < x < 1 .
• nCr can not be used because it is defined only for natural number, so n Cr will be written as (n)(n − 1)......(n − r + 1)
r!
• The number of terms in the series is infinite.
(x + y)n = x n 1 + y n y
x x
• If first term is not 1, then make first term unity in the following way, , if <1.
General term : Tr +1 = n(n − 1)(n − 2)......(n − r + 1) x r
r!
Some important expansions:
(i) (1 + x)n = 1+ nx + n(n − 1) x2 + ....... + n(n − 1)(n − 2)......(n −r + 1) xr + ......
2! r!
(ii) (1 − x)n =1 − nx + n(n − 1) x2 − ....... + n(n − 1)(n − 2).....(n − r + 1) (−x)r + .......
2! r!
(iii) (1 − x)−n =1 + nx + n(n + 1) x2 + n(n + 1)(n + 2) x3 + ..... + n(n + 1)......(n + r − 1) xr + .....
2! 3! r!
(iv) (1 + x)−n =1 − nx + n(n + 1) x2 − n(n + 1)(n + 2) x3 + ..... + n(n + 1)......(n + r − 1) (−x)r + ......
2! 3! r!
(a) Replace n by 1 in (iii) : (1 − x)−1 = 1 + x + x 2 + ..... + x r + ......∞ , General term, Tr+1 = x r
(b) Replace n by 1 in (iv) : (1 + x)−1 = 1 − x + x 2 − x 3 + ..... + (−x)r + ......∞ , General term, Tr+1 = (−x)r .
(c) Replace n by 2 in (iii) : (1 − x)−2 = 1 + 2x + 3x 2 + ..... + (r + 1)x r + .....∞ , General term, Tr+1 = (r + 1)xr .
(d) Replace n by 2 in (iv) : (1 + x)−2 = 1 − 2x + 3x 2 − 4 x 3 + ...... + (r + 1)(−x)r + .....∞
General term, Tr+1 = (r + 1)(−x)r .
(e) Replace n by 3 in (iii) : (1 − x)−3 =1+ 3x + 6x 2 + 10x 3 + ..... + (r + 1)(r + 2) xr + ..........∞
2!
General term, Tr+1 = (r + 1)(r + 2) / 2!. x r
(f) Replace n by 3 in (iv) : (1 + x)−3 =1− 3x + 6x2 − 10x3 + ..... + (r + 1)(r + 2) (−x)r + .....∞
2!
General term, Tr +1 = (r + 1)(r + 2) (−x)r
2!
Example: 42 To expand (1 + 2x)−1/ 2 as an infinite series, the range of x should be [AMU 2002]
Solution: (b)
(a) − 1 , 1 (b) − 1 , 1 (c) [−2, 2] (d) (– 2, 2)
2 2 2 2
(1 + 2x)−1 / 2 can be expanded if |2x|< 1 i.e., if | x|< 1 i.e., if − 1 < x < 1 i.e., if x ∈ − 1 , 1 .
2 2 2 2 2
Example: 43 If the value of x is so small that x 2 and higher power can be neglected, then 1 + x + 3 (1 − x)2 is equal to
1+ x + 1+ x
Solution: (b)
[Roorkee 1962]
Example: 44
Solution: (a) (a) 1 + 5 x (b) 1− 5 x (c) 1+ 2 x (d) 1 − 2 x
6 6 3 3
Example: 45
Solution: (b) 1 + 1 x + − 1 x 2 + ..... + 1 − 2 x − 1 x2 − ....
Example: 46 2 8 3 9
Solution: (c) Given expression can be written as = (1 + x)1 / 2 + (1 − x)2 / 3 =
1 + x + (1 + x)1 / 2 1 + 1 1 + .....
Example: 47 1+ x + 2 x − 8 x2
Solution: (c)
Example: 48 1− 1 x − 1 x2 + ..... 5 5 x 2, x 3 ....
12 144 6 6
= 3 1 =1− x + ..... =1 − x , when are neglected.
4 16
1+ x − x2 + .....
If (1 + ax)n = 1 + 8x + 24 x 2 + ..... then the value of a and n is [IIT 1983; Pb. CET 1994, 99]
(a) 2, 4 (b) 2, 3 (c) 3, 6 (d) 1, 2
We know that (1 + x)n =1+ nx + n(n − 1)x 2 + .....
1! 2!
(1 + ax)n = 1 + n(ax) + n(n − 1)(ax)2 + ...... ⇒ 1+ 8x + 24 x 2 + ...... = 1+ n(ax) + n(n − 1)(ax)2 + ....
1! 2! 1! 2!
Comparing coefficients of both sides we get, na = 8, and n(n − 1)a 2 = 24 on solving, a = 2 , b = 4.
2!
Coefficient of xr in the expansion of (1 − 2x)−1/ 2 [Kurukshetra CEE 2001]
(a) (2r)! (b) (2r)! (c) (2r)! (d) (2r)!
(r! ) 2 2r.(r!)2 (r! )2.22r 2r.(r + 1)!(r − 1)!
− 1 − 1 − 1 − 1 2.... − 1 − r + 1 1.3.5...(2r − 1).(−1)r .(−1)r .2r 1.3.5...(2r − 1) (2r)!
2 2 2− 2 2r r! r! r! r!2r
Coefficient of xr = r! (−2)r = = =
The coefficient of x25 in (1 + x + x2 + x3 + x4 )−1 is
(a) 25 (b) – 25 (c) 1 (d) – 1
Coefficient of x25 in (1 + x + x2 + x3 + x4 )−1
= Coefficient of x 25 in 1(1 − x5 ) −1 = Coefficient of x 25 in (1 − x5)−1 .(1 − x)
− x
1
= Coefficient of x25 in [ (1 − x5)−1 − x(1 − x5)−1 ] = [1 + (x5)1 + (x5)2 + ......] − x[1 + (x5)1 + (x5)2] + ......]
= Coefficient of x25 in [1 + x5 + x10 + x15 + .....] – Coefficient of x24 in [1 + x5 + x10 + x15 + .....] = 1 − 0 = 1 .
1− 1 + 1 . 3 − ......... = [EAMCET 1990]
8 8 16
(a) 2 (b) 2 (c) 2 (d) None of these
5 5 5
We know that (1 + x)n = 1 + nx + n(n − 1) x 2 + ......∞
2!
Here nx = − 1 , n(n − 1) x2 = 3 ⇒ x= 1 ,n = − 1 ⇒ 1 − 1 + 1 . 3 − ...... = 1 + 1 −1 / 2 = 2.
8 2 8.16 4 2 8 8 16 4 5
If x is so small that its two and higher power can be neglected and (1 − 2x)−1/ 2 (1 − 4 x)−5 / 2 = 1 + kx then k =
(a) 1 (b) – 2 (c) 10 [Rajasthan PET 1993]
(d) 11
Solution: (d) (1 − 2x)−1 / 2 (1 − 4 x)−5 / 2 = 1 + kx
Example: 49 1 + (−1/ 2)(− 2x) + (− 1/ 2)(− 3 / 2)(−2x)2 + ...... 1 + (− 5 / 2)(− 4x) + (− 5 / 2)(− 7 / 2)(−4 x)2 + ...... = 1 + kx
Solution: (c)
Example: 50 1! 2! 1! 2!
Solution: (a) Higher power can be neglected. Then 1 + x 1 + 10 x = 1 + kx ; 1 + 10x + x = 1 + kx ; k = 11
1! 1!
The cube root of 1 + 3x + 6x 2 + 10x 3 + ..... is
(a) 1 − x + x 2 − x 3 + .....∞ (b) 1 + x 3 + x 6 + x 9 + ...... (c) 1 + x + x 2 + x 3 + .... (d) None of these
We have (1 + 3x + 6x2 + 10x3 + .....)1 / 3 = [(1 − x)−3]1 / 3; [ (1 − x)−3 = 1 + 3x + 6x2 + ...∞] ⇒ (1 − x)−1 = 1 + x + x2 + ...∞
The coefficient of xn in the expansion of 1 1 x 3 1 x is
− −
(a) 3n+1 − 1 (b) 3n+1 − 1 (c) 2 3 n+1 − 1 (d) None of these
2 . 3n+1 3 n+1 3 n+1
1 = (1 − x) −1 (3 − x)−1 = 3 −1 (1 − x)−1 1 − x −1 = 1 [1 + x + x2 + .....x n ] 1 + x + x2 + ..... + x n−1 + xn
x)(3 3 3 3 32 3 n−1
(1 − − x) 3n
Coefficient of xn = 1 + 1 + 1 + .....(n + 1) terms = 1 [3n+1 − 1] = 3n+1 − 1 .
3n+1 3n 3n−1 3n+1 3−1 2.3n+1
Trick: Put n = 1, 2, 3...... and find the coefficients of x, x2, x3...... and comparing with the given option as :
Coefficient of x 2 is = 111 = 1 [33 − 1] = 13 ; Which is given by option (a) 3n+1 − 1 = 33 −1 = 13 .
33 + 32 + 31 33 3−1 27 2.(3n+1 ) 2.3 3 27
Three / Four Consecutive terms or Coefficients .
(1) If consecutive coefficients are given: In this case divide consecutive coefficients pair wise. We get
equations and then solve them.
(2) If consecutive terms are given : In this case divide consecutive terms pair wise i.e. if four consecutive terms
be Tr , Tr+1 , Tr+2 , Tr+3 then find Tr , Tr +1 , Tr + 2 ⇒ λ1,λ2,λ3 (say) then divide λ1 by λ2 and λ2 by λ3 and solve.
Tr +1 Tr + 2 Tr + 3
Example: 51 If a1, a2 , a3 , a4 are the coefficients of any four consecutive terms in the expansion of (1 + x)n , then a1 + a3 =
a1 + a2 a3 + a4
[IIT 1975]
(a) a2 (b) 1 a2 (c) 2a 2 (d) 2a 3
a2 + a3 2 a2 + a3 a2 + a3 a2 + a3
Solution: (c) Let a1, a2 , a3 , a4 be respectively the coefficients of (r + 1)th ,(r + 2)th ,(r + 3)th ,(r + 4)th terms in the expansion of (1 + x)n .
Then a1 =nCr ,a2 = nCr+1,a3 = nCr+2 ,a4 =nCr+3 .
Now, a1 + a3 = nCr + n Cr+2 = nCr + nCr+2 = nCr + n Cr+2
a1 + a2 + a4 nCr + nCr+1 n Cr+2 +nCr+3 n+1Cr +1 n+1Cr +3 n+1 n+1
a3 r +1 nCr r+3 n Cr + 2
= r +1 + r+3 = 2(r + 2) = 2. nCr +1 = 2. nCr +1 = 2a2
n+1 n+1 n +1 n+1Cr + 2 n Cr +1 +nCr +2 a2 + a3
Some Important Points.
(1) Pascal's Triangle :
1 (x + y)0
11 (x + y)1
121 (x + y)2
1331 (x + y)3
14641 (x + y)4
1 5 10 10 5 1 (x + y)5
Pascal's triangle gives the direct binomial coefficients.
Example : (x + y)4 = 1x 4 + 4 x 3y + 6x 2y 2 + 4 xy 3 + y 4
(2) Method for finding terms free from radical or rational terms in the expansion of
N −r r
(a1 / p + b1 / q )N ∨ a, b ∈ prime numbers : Find the general term Tr+1 = N Cr (a1/ p )N−r (b1/ q )r = N Cr a p .b q
Putting the values of 0 ≤ r ≤ N , when indices of a and b are integers.
Note : Number of irrational terms = Total terms – Number of rational terms.
Example: 52 The number of integral terms in the expansion of ( 3 + 8 5)256 is [AIEEE 2003]
Solution: (b)
(a) 32 (b) 33 (c) 34 (d) 35
Example: 53
Solution: (a) 256−r r
Tr+1 = 256 Cr . 3 2 . 5 8
First term = 256 C0 3128 5 0 = integer and after eight terms, i.e., 9th term = 256C8 3124.51 = integer
Continuing like this, we get an A.P., 1st , 9 th....... 257 th ; Tn = a + (n − 1) d ⇒ 257 = 1 + (n − 1) 8 ⇒ n = 33
( )The number of irrational terms in the expansion of 8 5 + 6 2 100 is
(a) 97 (b) 98 (c) 96 (d) 99
100−r r
Tr+1 =100 Cr 5 8 .2 6
As 2 and 5 are co-prime. Tr+1 will be rational if 100 − r is multiple of 8 and r is multiple of 6 also 0 ≤ r ≤ 100
∴r = 0, 6,12.......96 ; ∴100 − r = 4,10,16.....100 ......(i)
But 100 − r is to be multiple of 8.
So, 100 − r = 0, 8, 16, 24,......96 .....(ii)
Common terms in (i) and (ii) are 16, 40, 64, 88.
∴ r = 84, 60, 36, 12 give rational terms ∴ The number of irrational terms = 101 – 4 = 97.
Binomial theorem and mathematical induction
First Principle of Mathematical Induction .
The proof of proposition by mathematical induction consists of the following three steps :
Step I : (Verification step) : Actual verification of the proposition for the starting value “i”
Step II : (Induction step) : Assuming the proposition to be true for “k”, k ≥ i and proving that it is true for the
value (k + 1) which is next higher integer.
Step III : (Generalization step) : To combine the above two steps
Let p(n) be a statement involving the natural number n such that
(i) p(1) is true i.e. p(n) is true for n = 1.
(ii) p(m + 1) is true, whenever p(m) is true i.e. p(m) is true ⇒ p(m + 1) is true.
Then p(n) is true for all natural numbers n.
Second Principle of Mathematical Induction .
The proof of proposition by mathematical induction consists of following steps :
Step I : (Verification step) : Actual verification of the proposition for the starting value i and (i + 1).
Step II : (Induction step) : Assuming the proposition to be true for k – 1 and k and then proving that it is true
for the value k + 1; k ≥ i + 1.
Step III : (Generalization step) : Combining the above two steps.
Let p(n) be a statement involving the natural number n such that
(i) p(1) is true i.e. p(n) is true for n = 1 and
(ii) p(m + 1) is true, whenever p(n) is true for all n, where i ≤ n ≤ m
Then p(n) is true for all natural numbers.
For a ≠ b, The expression an − bn is divisible by
(a) a + b if n is even. (b) a – b is n if odd or even.
Some Formulae based on Principle of Induction .
For any natural number n
∑(i) n = 1+ 2 + 3 + ....... + n = n(n + 1) ∑(ii) n2 = 12 + 22 + 32 + ....... + n2 = n(n + 1)(2n + 1)
2 6
n2 (n + 1)2 n2
4
∑ (∑ )(iii)
n3 = 13 + 23 + 33 + ...... + n3 = =
Example: 1 The smallest positive integer n, for which n!< n + 1 n hold is [Pb. CET 2001]
2
(a) 1 (b) 2 (c) 3 (d) 4
Solution: (b) Let P(n) : n ! < n + 1 n
2
Step I : For n = 2 ⇒ 2 ! < 2 + 1 2 ⇒ 2< 9 ⇒ 2 < 2.25 which is true. Therefore, P(2) is true.
2 4
Step II : Assume that P(k) is true, then p(k) : k ! < k + 1 k
2
Step III : For n = k + 1,
k + 2 k +1 k + 1 k (k + 1)k+1
2 2 2k
P(k + 1) : (k + 1) ! < ⇒ k ! < ⇒ (k + 1)k ! <
⇒ (k + 1)!< (k + 1)k+1 …..(i) and (k + 1)k+1 < k + 2 k +1 …..(ii)
2k 2k 2
k + 2 k +1 1 1 k+1 1 + k+1 1 2
k + 1 + 1 + +
⇒ >2 ⇒ + k >2 ⇒ 1 + (k + 1) k 1 C 2 k 1 + ........ > 2
1 + 1 + k+1 1 2
+
⇒ C 2 k 1 + ...... > 2
Which is true, hence (ii) is true.
From (i) and (ii), (k + 1)!< (k + 1)k+1 < k + 2 k +1 ⇒ (k + 1)! < k + 2 k +1
2k 2 2
Hence P(k + 1) is true. Hence by the principle of mathematical induction P(n) is true for all n ∈ N
Trick : By check option
(a) For n = 1, 1 ! < 1 + 1 1 ⇒ 1<1 which is wrong (b) For n = 2, 2 ! < 3 2 ⇒ 2< 9 which is correct
2 2 4
(c) For n = 3, 3 ! < 3 + 1 3 ⇒ 6 < 8 which is correct
2
(d) For n = 4, 4 ! < 4 + 1 4 ⇒ 24 < 5 4 ⇒ 24 < 39.0625 which is correct.
2 2
But smallest positive integer n is 2.
Example: 2 Let S(k) = 1 + 3 + 5 + ....... + (2k − 1) = 3 + k2 . Then which of the following is true. [AIEEE 2004]
Solution: (c)
Example: 3 (a) Principle of mathematical induction can be used to prove the formula
Solution: (c) (b) S(k) ⇒ S(k + 1)
Example: 4 (c) S(k) ⇒/ S(k + 1)
Solution: (b) (d) S(1) is correct
We have S(k) = 1 + 3 + 5 + ...... + (2k − 1) = 3 + k2 , S (1) ⇒ 1 = 4 , Which is not true and S(2) ⇒ 3 = 7 , Which is not true.
Hence induction cannot be applied and S(k) ⇒/ S(k + 1)
When P is a natural number, then P n+1 + (P + 1)2n−1 is divisible by [IIT 1994]
(a) P (b) P 2 + P (c) P 2 + P + 1 (d) P 2 − 1
For n =1, we get, P n+1 + (P + 1)2n−1 = P 2 + (P + 1)1 = P 2 + P + 1 ,
Which is divisible by P 2 + P + 1 , so result is true for n =1 …..(i)
Let us assume that the given result is true for n = m ∈ N
i.e. P m+1 + (P + 1)2m−1 is divisible by P 2 + P + 1 i.e. P m+1 + (P + 1)2m−1 = k(P 2 + P + 1) ∀ k ∈ N
Now, P (m+1)+1 + (P + 1)2(m+1)−1 = P m+2 + (P + 1)2m+1 = P m+2 + (P + 1)2 (P + 1)2m−1
= P m+2 + (P + 1)2 [k(P 2 + P + 1) − P m+1 ] by using (i)
= P m+2 + (P + 1)2 ⋅ k (P 2 + P + 1) − (P + 1)2 (P)m+1 = P m+1[P − (P + 1)2 ] + (P + 1)2 ⋅ k(P 2 + P + 1)
= P m+1[P − P 2 − 2P − 1] + (P + 1)2 ⋅ k(P 2 + P + 1) = −P m+1[P 2 + P + 1] + (P + 1)2 ⋅ k(P 2 + P + 1)
= (P 2 + P + 1)[k ⋅ (P + 1)2 − P m+1 ]
Which is divisible by p 2 + p + 1 , so the result is true for n = m + 1 . Therefore, the given result is true for all n ∈ N by
induction.
Trick : For n = 2, we get, P n+1 + (P + 1)2n−1 = P 3 + (P + 1)3 = P 3 + P 3 + 1 + 3P 2 + 3P = 2P 3 + 3P 2 + 3P + 1
Which is divisible by P 2 + P + 1 . Given result is true for all n ∈ N
Given Un+1 = 3 Un − 2Un−1 and U0 = 2 , U1 = 3 , the value of Un for all n ∈ N is
(a) 2n − 1 (b) 2n + 1 (c) 0 (d) None of these
Un+1 = 3 Un − 2 Un−1 …..(i)
Step I : Given U1 = 3
For n =1, U1+1 = 3 U1 − 2 U0 , U2 = 3.3 − 2.2 = 5
Option (b) Un = 2n + 1
For n = 1, U1 = 21 + 1 = 3 which is true. For n = 2, U2 = 22 + 1 = 5 which is true
Therefore, the result is true for n = 1 and n = 2
Step II : Assume it is true for n = k then it is also true for n = k – 1
Then Uk = 2k + 1 …..(ii) and Uk−1 = 2k−1 + 1 …..(iii)
Step III : Putting n = k in (i), we get
Uk+1 = 3 Uk − 2 Uk −1 = 3[2k + 1] − 2[2k−1 + 1] = 3.2k + 3 − 2.2k−1 − 2 = 3.2k + 1 − 2.2k−1
⇒ 3.2k − 2k + 1 = 2.2k + 1 = 2k+1 + 1 ⇒ Uk+1 = 2k+1 + 1
This shows that the result is true for n = k + 1 , by the principle of mathematical induction the result is true for all n ∈ N .
Divisibility Problems.
To show that an expression is divisible by an integer
(i) If a, p, n, r are positive integers, then first of all we write apn+r = apn. ar = (ap)n. ar.
(ii) If we have to show that the given expression is divisible by c.S
Then express, a p = [1 + (a p − 1], if some power of (a p − 1) has c as a factor.
a p = [2 + (a p − 2)] , if some power of (a p − 2) has c as a factor.
a p = [K + (a p − K)], if some power of (a p − K) has c as a factor.
Example: 5 (1 + x)n − nx − 1 is divisible by (where n ∈ N )
Solution: (b)
(a) 2x (b) x 2 (c) 2x 3 (d) All of these
Example: 6
Solution: (c) (1 + x)n = 1 + nx + n(n − 1) x2 + n(n − 1)(n − 2) x3 + ..... ⇒ (1 + x)n − nx −1 = x 2 n(n − 1) + n(n − 1)(n − 3) x + .....
2! 3! 2! 3!
From above it is clear that (1 + x)n − nx − 1 is divisible by x 2 .
Trick : (1 + x)n − nx − 1 . Put n = 2 and x = 3 ; Then 4 2 − 2.3 − 1 = 9
Is not divisible by 6, 54 but divisible by 9. Which is given by option (c) = x 2 = 9 .
The greatest integer which divides the number 101100 − 1 is [MP PET 1998]
(a) 100 (b) 1000 (c) 10000 (d) 100000
(1 + 100)100 = 1 + 100 .100 + 100.99 (100)2 + .... ⇒ 101100 − 1 = 100.1001 + 100.99 + 100.99.98 100 + .....
1.2 1.2 3.2.1
From above it is clear that, 101100 − 1 is divisible by (100)2 = 10000
Straight-line
Definition.
The straight line is a curve such that every point on the line segment joining any two points on it lies on it. The
simplest locus of a point in a plane is a straight line. A line is determined uniquely by any one of the following:
(1) Two different points (because we know the axiom that one and only one straight line passes through two
given points)
(2) A point and a given direction.
YY
Y
Unique line through O 45o X O XO 45º 45º 45º 45º X
two given points Unique line through a given
point with a given direction Infinite number of lines Infinite number of lines
through a given point with a given direction
Thus, to determine a line uniquely, two geometrical conditions are required.
Slope (Gradient) of a Line.
The trigonometrical tangent of the angle that a line makes with the positive direction of the x-axis in
anticlockwise sense is called the slope or gradient of the line. Y X X′ Y
The slope of a line is generally denoted by m. Thus, m = tan θ B
B θ
(1) Slope of line parallel to x – axis is m = tan 0o = 0 . θ
(2) Slope of line parallel to y – axis is m = tan 90o = ∞ . X′ A O O AX
(3) Slope of the line equally inclined with the axes is 1 or –1. Y′ Y′
(4) Slope of the line through the points A(x1, y1) and B(x2, y2) is y2 − y1 taken in the same order.
x2 − x1
(5) Slope of the line ax + by + c = 0,b ≠0 is − a .
b
(6) Slope of two parallel lines are equal.
(7) If m1 and m2 be the slopes of two perpendicular lines, then m1.m2 = −1 .
Note : m can be defined as tanθ for 0 < θ ≤ π and θ ≠ π
2
If three points A, B, C are collinear, then
Slope of AB = Slope of BC = Slope of AC
Example: 1 The gradient of the line joining the points on the curve y = x 2 + 2x , whose abscissae are 1 and 3, is [MP PET 1997]
(a) 6 (b) 5 (c) 4 (d) 3
Solution: (a) The points are (1, 3) and (3, 15)
Hence gradient is = y2 − y1 = 12 = 6
x2 − x1 2
Example: 2 Slope of a line which cuts intercepts of equal lengths on the axes is [MP PET 1986]
Solution: (a)
(a) – 1 (b) 0 (c) 2 (d) 3
Equation of line is x + y =1
a a
⇒ x + y = a ⇒ y = −x + a . Hence slope of the line is – 1.
Equations of Straight line in Different forms. Y
(1) Slope form : Equation of a line through the origin and having slope m is y = mx.
(2) One point form or Point slope form : Equation of a line through the X' A θc B
O X
point (x1, y1) and having slope m is y − y1 = m(x − x1) .
Y'
(3) Slope intercept form : Equation of a line (non-vertical) with slope m
and cutting off an intercept c on the y-axis is y = mx + c.
The equation of a line with slope m and the x-intercept d is y = m(x − d)
(4) Intercept form : If a straight line cuts x-axis at A and the y-axis at B then OA and OB are known as the
intercepts of the line on x-axis and y-axis respectively.
The intercepts are positive or negative according as the line meets with Y
positive or negative directions of the coordinate axes. B
In the figure, OA = x-intercept, OB = y-intercept. b
Equation of a straight line cutting off intercepts a and b on x–axis and y–axis X' A X
Oa
respectively is x y = 1.
a +b Y'
Note : If given line is parallel to X axis, then X-intercept is undefined.
If given line is parallel to Y axis, then Y-intercept is undefined.
(5)Two point form: Equation of the line through the points A(x1, y1) and B(x 2 , y2 ) is (y − y1) = y2 − y1 (x − x1) .
x2 − x1
In the determinant form it is gives as: Y
L
x y1
x1 y1 1 = 0 is the equation of line. B
(x2,
x2 y2 1 A(x1,Oy1 X
(6) Normal or perpendicular form : The equation of the straight line upon
which the length of the perpendicular from the origin is p and this perpendicular makes
Y
an angle α with x-axis is x cosα + y sinα = p .
Y B X
pP
α
OA
Y
(7) Symmetrical or parametric or distance form of the line : Equation of a line passing through (x1,y1)
and making an angle θ with the positive direction of x-axis is x − x1 = y − y1 =r, Y
cos θ sin θ (x1A,y1)θ r P(x, y)
where r is the distance between the point P (x, y) and A(x1, y1) .
The coordinates of any point on this line may be taken as X' θ X
(x1 + r cos θ , y1 + r sin θ ) , known as parametric co-ordinates, ‘r’ is called the O
parameter. Y'
Note : Equation of x-axis ⇒ y = 0
Equation a line parallel to x-axis (or perpendicular to y-axis) at a distance ‘b’ from it ⇒ y = b
Y
b
X' O X
Y′
Equation of y-axis ⇒ x = 0
Equation of a line parallel to y-axis (or perpendicular to x-axis) at a distance ‘a’ from it ⇒ x = a
Y
X' a X
O
Y’
Example: 3 Equation to the straight line cutting off an intercept 2 from the negative direction of the axis of y and inclined at 30° to the
Solution: (d)
positive direction of x, is [MP PET 2003]
Example: 4
Solution: (a) (a) y + x − 3 = 0 (b) y − x + 2 = 0 (c) y − 3x − 2 = 0 (d) 3y − x + 2 3 = 0
Example: 5
Let the equation of the straight line is y = mx + c .
Here m = tan 30° = 1 and c = – 2
3
Hence, the required equation is y = 1 x −2 ⇒ 3y − x + 2 3 = 0 .
3
The equation of a straight line passing through (– 3, 2) and cutting an intercept equal in magnitude but opposite in sign
from the axes is given by [Rajasthan PET 1984; MP PET 1993]
(a) x − y + 5 = 0 (b) x + y − 5 = 0 (c) x − y − 5 = 0 (d) x + y + 5 = 0
Let the equation be x + y = 1 ⇒ x −y = a
a −a
But it passes through (–3, 2), hence a = −3 − 2 = −5 . Hence the equation of straight line is x − y + 5 = 0 .
The equation of the straight line passing through the point (4, 3) and making intercept on the co-ordinates axes whose sum
is – 1, is [AIEEE 2004]
(a) x − y = −1 and x + y =1 (b) x − y = −1 and x + y = −1
2 3 2 1 2 3 −2 1
(c) x − y = 1 and x + y = 1 (d) x + y = −1 and x + y = −1
2 3 2 1 2 3 −2 1
Solution: (d) Let the equation of line is xy = 1 , which passes through (4, 3). Then 4 + 3 a = 1 ⇒ a = ±2
Example: 6 a + −1−a a −1−
Hence equation is x − y = 1 and x + y = 1 .
2 3 −2 1
Let PS be the median of the triangle with vertices P(2, 2), Q(6, − 1) and R(7, 3). The equation of the line passing through
(1, – 1) and parallel to PS is [IIT Screening 2000]
(a) 2x − 9y − 7 = 0 (b) 2x − 9y − 11 = 0 (c) 2x + 9y − 11 = 0 (d) 2x + 9y + 7 = 0
Solution: (d) S = mid point of QR = 6 + 7 , −1 + 3 = 13 ,1
2 2 2
∴ Slope (m) of PS = 2−1 = −2 ; ∴ The required equation is y +1 = −2 (x − 1) ⇒ 2x + 9y + 7 = 0
9 9
2 − 13
2
Equation of Parallel and Perpendicular lines to a given Line.
(1) Equation of a line which is parallel to ax + by + c = 0 is ax + by + λ = 0
(2) Equation of a line which is perpendicular to ax + by + c = 0 is bx − ay + λ = 0
The value of λ in both cases is obtained with the help of additional information given in the problem.
Example: 7 The equation of the line passes through (a, b) and parallel to the line x + y =1, is [Rajasthan PET 1986, 1995]
Solution: (b) a b
Example: 8 (a) x + y = 3 (b) x + y = 2 (c) x + y =0 (d) x + y +2=0
Solution: (d) a b a b a b a b
Example: 9
The equation of parallel line to given line is x + y = λ .
a b
This line passes through point (a, b).
∴ a + b = λ ⇒ λ=2
a b
Hence, required line is x + y = 2 .
a b
A line passes through (2, 2) and is perpendicular to the line 3x + y = 3 . Its y-intercept is [IIT 1992]
(a) 1 (b) 2 (c) 1 (d) 4
3 3 3
The equation of a line passing through (2, 2) and perpendicular to 3x + y = 3 is
y−2= 1 (x − 2) or x − 3y + 4 = 0 . Putting x=0 in this equation, we obtain y= 4 .
3 3
So y-intercept = 4 .
3
The equation of line passing through − 1, π and perpendicular to 3 sin θ + 2 cosθ = 4 is [EAMCET 2003]
2 r
(a) 2 = 3 r cosθ − 2r sinθ (b) 5 = −2 3 r sinθ + 4r cosθ
(c) 2 = 3 r cosθ + 2r sinθ (d) 5 = 2 3 r sinθ + 4r cosθ
Solution: (a) Equation of a line, perpendicular to 3 sinθ + 2cosθ 4 is 3 sin π + θ + 2 cos π +θ = k
=r 2 2 r
It is passing through − 1, π . Hence, 3 sinπ + 2cosπ = k /− 1 ⇒ k = 2
2
∴ 3 cosθ − 2 sinθ = 2 ⇒ 2= 3 r cosθ − 2r sinθ .
r
Example: 10 The equation of the line bisecting perpendicularly the segment joining the points (– 4, 6) and (8, 8) is [Karnataka CET 2003]
Solution: (a)
(a) 6x + y − 19 = 0 (b) y = 7 (c) 6x + 2y − 19 = 0 (d) x + 2y − 7 = 0
Equation of the line passing through (–4, 6) and (8, 8) is
y−6 = 8 − 6 (x + 4) ⇒ y−6 = 2 (x + 4) ⇒ 6y − x = 40 ......(i)
8 + 4 12
Now equation of any line ⊥ to it is 6x + y + λ = 0 .....(ii)
This line passes through the midpoint of (–4, 6) and (8, 8) i.e., (2, 7)
∴ From (ii) 12 + 7 + λ = 0 ⇒ λ = −19 , ∴ Equation of line is 6x + y − 19 = 0
General equation of a Straight line and its Transformation in Standard forms.
General form of equation of a line is ax + by + c = 0 , its
(1) Slope intercept form: y = − a x − c , slope m = − a and intercept on y-axis is, C = − c
b b b b
(2) Intercept form : x + y = 1, x intercept is = − c and y intercept is = − c
−c/a −c/b a b
(3) Normal form : To change the general form of a line into normal form, first take c to right hand side and
make it positive, then divide the whole equation by a 2 + b 2 like
ax by c , where cosα = − a , sinα = − b and p = c
−−=
a2 + b2 a2 + b2 a2 + b2 a2 + b2 a2 + b2 a2 + b2
Selection of Co-ordinate of a Point on a Straight line.
(1) If the equation of the straight line be ax + by + c = 0 , in order to select a point on it, take the x co-ordinate
according to your sweet will. Let x =λ ; then aλ + by + c =0 or y = − aλ + c ;
b
∴ λ,− aλ + c is a point on the line for any real value of λ. If λ = 0 is taken then the point will be 0,− c .
b b
Similarly a suitable point can be taken as − c ,0 .
a
(2) If the equation of the line be x = c then a point on it can be taken as (c,λ) where λ has any real value.
In particular (c, 0) is a convenient point on it when λ = 0 .
(3) If the equation of the line be y = c then a point on it can be taken as (λ,c) where λ has any real value.
In particular (0, c) is a convenient point on it when λ = 0 . [MP PET 2001]
Example: 11 If we reduce 3x + 3y + 7 = 0 to the form x cosα + y sinα = p , then the value of p is
(a) 7 (b) 7 (c) 37 (d) 7
23 3 2 32
Solution: (d) Given equation is 3x + 3y + 7 = 0 , Dividing both sides by 32 + 32
⇒ 3x + 3y + 7 = 0 ⇒ 3 x + 3 y = −7 , ∴ p= −7 = 7 .
32 + 32 32 + 32 32 + 32 32 32 32 32 32
Point of Intersection of Two lines.
Let a1x + b1y + c1 =0 and a2x + b2y + c2 = 0 be two non-parallel lines. If (x′, y′) be the co-ordinates of their
point of intersection, then a1x′ + b1y′ + c1 = 0 and a2 x′ + b2y′ + c2 = 0
b1 b2 c1 c2
c1
Solving these equation, we get (x′, y′) = b1c 2 − b2c1 , c1a2 − c2a1 = a1 c2 , a1 a2
a1b2 − a2b1 a1b2 − a2b1 a2 a1
a2
b1 b2 b1 b2
Note : Here lines are not parallel, they have unequal slopes, then a1b2 − a2b1 ≠ 0 .
In solving numerical questions, we should not remember the co-ordinates (x′, y′) given above, but
we solve the equations directly.
General equation of Lines through the Intersection of Two given Lines.
If equation of two lines P = a1x + b1y + c1 = 0 and Q = a2x + b2y + c2 = 0 , then the equation of the lines
passing through the point of intersection of these lines is P + λ Q = 0 or a1 x + b1y + c + λ(a2 x + b2y + c2 ) = 0 ;
Value of λ is obtained with the help of the additional information given in the problem.
Example: 12 Equation of a line passing through the point of intersection of lines 2x − 3y + 4 = 0 , 3x + 4y − 5 = 0 and perpendicular to
6x − 7y + 3 = 0 , then its equation is [Rajasthan PET 2000]
(a) 119x + 102y + 125 = 0 (b) 119x + 102y = 125 (c) 119x − 102y = 125 (d) None of these
Solution: (b) The point of intersection of the lines 2x − 3y + 4 = 0 and 3x + 4y − 5 = 0 is −1 , 22
17 17
Example: 13
Solution: (c) The slope of required line = −7 .
6
Hence, Equation of required line is, y− 22 = −7 x + 2 ⇒ 119x + 102y = 125 .
7 6 34
The equation of straight line passing through point of intersection of the straight lines 3x − y + 2 = 0 and 5x − 2y + 7 = 0
and having infinite slope is [UPSEAT 2001]
(a) x = 2 (b) x + y = 3 (c) x = 3 (d) x = 4
Required line should be, (3x − y + 2) + λ(5x − 2y + 7) = 0 .....(i)
⇒ (3 + 5λ)x − (2λ + 1)y + (2 + 7λ) = 0 ⇒ y = 3 + 5λ x + 2 + 7λ ......(ii)
2λ + 1 2λ + 1
As the equation (ii) has infinite slope, 2λ +1=0 ⇒ λ = −1
2
Putting λ = −1 in equation (i), We have (3x − y + 2) + −1 (5 x − 2y + 7) = 0 ⇒ x =3
2 2
Angle between Two non-parallel Lines.
Let θ be the angle between the lines y = m1 x + c1 and y = m2 x + c2 . Yy=m1x + c1
and intersecting at A.
where, m1 = tanα and m2 = tan β A y=m2x + c2
α =θ +β ⇒ θ =α −β
O βθ α X
C B
⇒ tanθ = tanα − tan β
1 + tanα tan β
∴ θ = tan −1 m1 − m2 .
1 + m1m2
(1) Angle between two straight lines when their equations are given : The angle θ between the lines
a1x + b1y + c1 =0 and a2x + b2y + c2 = 0 is given by, tanθ = a2b1 − a1b2 .
a1a2 + b1b2
(i) Condition for the lines to be parallel : If the lines a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 are
parallel then, m1 = m2 ⇒ a1 = a2 ⇒ a1 = b1 .
b1 b2 a2 b2
(ii) Condition for the lines to be perpendicular : If the lines a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0
are perpendicular then, m1m2 = −1 ⇒ a1 × a2 = −1 ⇒ a1a2 + b1b2 = 0.
b1 b2
(iii) Conditions for two lines to be coincident, parallel, perpendicular and intersecting : Two lines
a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 are,
(a) Coincident, if a1 = b1 = c1 (b) Parallel, if a1 = b1 ≠ c1 (c) Intersecting, if a1 ≠ b1
a2 b2 c2 a2 b2 c2 a2 b2
(d) Perpendicular, if a1a2 + b1b2 = 0
Example: 14 Angle between the lines 2x − y − 15 = 0 and 3x + y + 4 = 0 is [Rajasthan PET 2003]
(a) 90° (b) 45° (c) 180° (d) 60°
Solution: (b) tan θ = a2b1 − a1b2 = (3)(−1) − (2)(1) ⇒ tanθ = −3 − 2 = −5 =|− 1|
a1a2 + b1b2 (3)(2) + (−1)(1) 6 −1 5
θ = tan−1 | −1|= tan−1 1 = 45o .
Example: 15 To which of the following types the straight lines represented by 2x + 3y − 7 = 0 and 2x + 3y − 5 = 0 belongs [MP PET 1982]
(a) Parallel to each other (b) Perpendicular to each other
(c) Inclined at 45° to each other (d) Coincident pair of straight lines
Solution: (a) Here, a1 = b1 ≠ c1 ; 2 = 2 ≠ 7 . Hence, lines are parallel to each other.
a2 b2 c2 3 3 5
Equation of Straight line through a given point making a given Angle with a given Line.
Since straight line L makes an angle (θ + α) with x-axis, then equation of line L
L is y − y1 = tan(θ + α)(x − x1) and straight line L′ makes an angle (θ − α) with x-
axis, then equation of line L′ is α y=
⇒ y − y1 = tan(θ − α)(x − x1) α θ L′
where m = tanθ Oθ X
(θ –
Hence, the equation of the straight lines which pass through a given point
(x1, y1) and make a given angle α with given straight line y = mx + c are y − y1 = m ± tanα (x − x1)
1 m tanα
Example: 16 The equation of the lines which passes through the point (3, – 2) and are inclined at 60o to the line 3x + y = 1
Solution: (a)
[IIT 1974; MP PET 1996]
Example: 17
Solution: (b) (a) y + 2 = 0, 3x − y − 2 − 3 3 = 0 (b) x − 2 = 0, 3x − y + 2 + 3 3 = 0
(c) 3x − y − 2 − 3 3 = 0 (d) None of these
The equation of lines passing through (3, –2) is (y + 2) = m (x − 3) .....(i)
The slope of the given line is − 3 .
So, tan 60o = ± m − (− 3) . On solving, we get m = 0 or 3
1 + m(− 3)
Putting the values of m in (i), the required equation is y + 2 = 0 and 3x − y − 2 − 3 3 = 0 .
In an isosceles triangle ABC, the coordinates of the point B and C on the base BC are respectively (1, 2) and (2, 1). If the
equation of the line AB is y = 2x , then the equation of the line AC is [Roorkee 2000]
(a) y= 1 (x − 1) (b) y = x (c) y = x − 1 (d) 2y = x + 3
2 2
Slope of BC = 1− 2 = −1 A
2−1
AB = AC , ∴ ∠ABC = ∠ACB
2+1 m+1 m+1 m+1 1 y = 2x
1 + 2 (−1) + m (−1) 1−m 1−m 2
⇒ = 1 ⇒ =| −3| ⇒ = ± 3 ⇒ m = 2, .
But slope of AB is 2; ∴ m = 1 (Here m is the gradient of the line AC) B C
2 (1, 2) (2, 1)
Equation of the line AC is y−1 = 1 (x − 2) ⇒ x − 2y = 0 or y= x .
2 2
A Line equally inclined with Two lines.
Let the two lines with slopes m1 and m2 be equally inclined to a line with slope m
then , m1 − m = − m2 − m m2 m
1 + m1m 1 + m2m
θ
Note : Sign of m in both brackets is same. θ m1
Example: 18 If the lines y = 3x + 1 and 2y = x + 3 are equally inclined to the line y = mx + 4 , then m = [ISM Dhanbad 1976]
(a) 1+ 3 2 (b) 1−3 2 (c) 1+3 2 (d) 1±5 2
7 7 7 7
Solution: (d) If line y = mx + 4 are equally inclined to lines with slope and 1 ,then 3−m 1 −3 1±5 2
2 1 + 3m − 2 7
m1 = 3 m2 = = 1 ⇒ m =
1+ 2
m
Equations of the bisectors of the Angles between two Straight lines.
The equation of the bisectors of the angles between the lines a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 are
given by, a1x + b1y + c1 = ± a2x + b2y + c2 .....(i)
a12 + b12 a22 + b22
Algorithm to find the bisector of the angle containing the origin :
Let the equations of the two lines a1 x + b1y + c1 = 0 and a2 x + b2y + c2 = 0 . To find the bisector of the angle
containing the origin, we proceed as follows:
Step I : See whether the constant terms c1 and c2 in the equations of two lines positive or not. If not, then
multiply both the sides of the equation by –1 to make the constant term positive.
Step II : Now obtain the bisector corresponding to the positive sign i.e., a1 x + b1y + c1 = a2 x + b2y + c2 .
a12 + b12 a 2 b 2
2 + 2
This is the required bisector of the angle containing the origin.
Note : The bisector of the angle containing the origin means the bisector of the angle between the lines
which contains the origin within it.
(1) To find the acute and obtuse angle bisectors
Let θ be the angle between one of the lines and one of the bisectors given by (i). Find tanθ . If | tanθ |< 1,
then this bisector is the bisector of acute angle and the other one is the bisector of the obtuse angle.
If | tanθ |> 1, then this bisector is the bisector of obtuse angle and other one is the bisector of the acute angle.
(2) Method to find acute angle bisector and obtuse angle bisector
(i) Make the constant term positive, if not. (ii) Now determine the sign of the expression a1a2 + b1b2 .
(iii) If a1a2 + b1b2 > 0 , then the bisector corresponding to “+” sign gives the obtuse angle bisector and the
bisector corresponding to “–” sign is the bisector of acute angle between the lines. L1
(iv) If a1a2 + b1b2 < 0 , then the bisector corresponding to “+” and “–” sign
given the acute and obtuse angle bisectors respectively. Acute bisector
Note : Bisectors are perpendicular to each other. P(x, y)
If a1a2 + b1b2 > 0 , then the origin lies in obtuse angle and if L2
a1a2 + b1b2 < 0 , then the origin lies in acute angle. Obtuse bisector
Example: 19 The equation of the bisectors of the angles between the lines | x |=| y| are [Orissa JEE 2002]
Solution: (c)
Example: 20 (a) y = ±x and x = 0 (b) x= 1 and y= 1 (c) y = 0 and x = 0 (d) None of these
Solution: (b) 2 2
The equation of lines are x + y = 0 and x − y = 0 .
∴ The equation of bisectors of the angles between these lines are x+y = ± x−y ⇒ x + y = ±(x − y)
1+1 1+1
Taking +ve sign, we get y = 0 ; Taking –ve sign, we get x = 0 . Hence, the equation of bisectors are x = 0, y = 0 .
The equation of the bisector of the acute angle between the lines 3x − 4y + 7 = 0 and 12x + 5y − 2 = 0 is
[IIT 1975, 1983; Rajasthan PET 2003]
(a) 21x + 77y − 101 = 0 (b) 11x − 3y + 9 = 0 (c) 31x + 77y + 101 = 0 (d) 11x − 3y − 9 = 0
Bisector of the angles is given by 3x − 4y + 7 = ± 12x + 5y − 2
5 13
⇒ 11x − 3y + 9 = 0 .....(i) and 21x + 77y − 101 = 0 ......(ii)
3 11
m1 − m2 4− 3 35
Let the angle between the line 3x − 4y + 7 = 0 and (i) is α , then tanα = 1 + m1m2 = 3 11 = 45 < 1 ⇒ α < 45 o
4 3
1+ ×
Hence 11x − 3y + 9 = 0 is the bisector of the acute angle between the given lines.
Length of Perpendicular.
(1) Distance of a point from a line : The length p of the perpendicular from the point (x1, y1) to the line
ax + by + c = 0 is given by p = |ax1 + by1 + c | .
a2 + b2
Note : Length of perpendicular from origin to the line ax + by + c = 0 is c.
a2 + b2
Length of perpendicular from the point (x1, y1) to the line x cosα + y sinα = p is
x1 cosα + y1 sinα − p .
(2) Distance between two parallel lines : Let the two parallel lines be ax + by + c1 = 0 and ax + by + c2 = 0 .
First Method : The distance between the lines is d = |c1 − c2 | . ax + by + c1 = 0
(a2 + b2)
d
ax + by + c2 = 0
Second Method : The distance between the lines is d = λ , where ax + by + c1 = 0
(a2 + b2)
(i) λ =|c1 − c2 | if they be on the same side of origin. ax + by + c2 = 0
(ii) λ =|c1 |+|c2 | if the origin O lies between them.
.O (0, 0)
Third method : Find the coordinates of any point on one of the given line, preferably putting x = 0 or
y = 0 . Then the perpendicular distance of this point from the other line is the ax + by + c1 = 0
required distance between the lines.
Note : Distance between two parallel lines ax + by + c1 = 0 and .O (0, 0)
c1 − c2 ax + by + c2 = 0
k
kax + kby + c2 = 0 is
a2 + b2
Distance between two non parallel lines is always zero.
Position of a Point with respect to a Line.
Let the given line be ax + by + c = 0 and observing point is (x1, y1), then
(i) If the same sign is found by putting in equation of line x = x1, y = y1 and x = 0 , y = 0 then the point
(x1, y1) is situated on the side of origin.
(ii) If the opposite sign is found by putting in equation of line x = x1, y = y1 and x = 0 , y = 0 then the point
(x1, y1) is situated opposite side to origin.
Position of Two points with respect to a Line.
Two points (x1, y1) and (x 2, y2 ) are on the same side or on the opposite side of the straight line
ax + by + c = 0 according as the values of ax1 + by1 + c and ax 2 + by2 + c are of the same sign or opposite sign.
Example: 21 The distance of the point (–2, 3) from the line x − y = 5 is [MP PET 2001]
Solution: (a)
Example: 22 (a) 5 2 (b) 2 5 (c) 3 5 (d) 5 3
Solution: (c)
p = x1 − y1 − 5 = − 2 − 3 − 5 = − 10 = 5 2
Example: 23 12 + 12 12 + 12 2
Solution: (d) The distance between the lines 4 x + 3y = 11 and 8x + 6y = 15 is [AMU 1979; MNR 1987; UPSEAT 2000; DCE 1999]
Example: 24 (a) 7 (b) 4 (c) 7 (d) None of these
Solution: (a) 2 10
Given lines 4 x + 3y = 11 and 8x + 6y = 15 , distance from the origin to both the lines are − 11 and − 15 ⇒ 11 , 15
25 100 5 10
Clearly both lines are on the same side of the origin.
Hence, distance between both the lines are, 11 − 15 = 7 .
5 10 10
If the length of the perpendicular drawn from origin to the line whose intercepts on the axes are a and b be p, then
[Karnataka CET 2003]
(a) a 2 + b 2 = p 2 (b) a2 + b2 = 1 (c) 1 + 1 = 2 (d) 1 + 1 = 1
p2 a2 b2 p2 a2 b2 p2
Equation of line is x + y =1 ⇒ bx + ay − ab = 0
a b
Perpendicular distance from origin to given line is p = − ab ⇒ a2 + b2 1 ⇒ a2 + b2 = 1 ⇒ 11 1
a2 + b2 ab =p a2b2 p2 a2 + b2 = p2
The point on the x-axis whose perpendicular distance from the line x + y =1 is a, is[Rajasthan PET 2001; MP PET 2003]
a b
(a) a (b ± a2 + b2 ), 0 (b) b (b ± a 2 + b 2 ),0 (c) a (a ± a 2 + b 2 ),0 (d) None of these
b a b
Let the point be (h,0) then a = ± bh + 0 − ab ⇒ bh = ±a a2 + b2 + ab ⇒ h= a (b ± a2 + b2 )
a2 + b2 b
Hence the point is a (b ± a 2 + b 2 ),0
b
Example: 25 The vertex of an equilateral triangle is (2, –1) and the equation of its base is x + 2y = 1 . The length of its sides is [UPSEAT 2003]
(a) 4 (b) 2 A (2, –1)
15 15
(c) 4 (d) None of these
33
Solution: (b) | AD|= 2 − 2 − 1 = 1 5
12 + 22
60o
AD 1/ 5 1 ⇒ BC = 2BD = 2 BD C
BD BD 15 15
tan 60o = ⇒ 3 = ⇒ BD = x+ 2y – 1 = 0
Concurrent Lines.
Three or more lines are said to be concurrent lines if they meet at a point.
First method : Find the point of intersection of any two lines by solving them simultaneously. If the point
satisfies the third equation also, then the given lines are concurrent.
Second method : The three lines a1x + b1y + c1 = 0 , a2x + b2y + c2 = 0 and a3x + b3y + c3 = 0 are
a1 b1 c1
concurrent if, a2 b2 c2 = 0
a3 b3 c3
Third method : The condition for the lines P = 0 , Q = 0 and R = 0 to be concurrent is that three constants
a, b, c (not all zero at the same time) can be obtained such that aP + bQ + cR = 0 .
Example: 26 If the lines ax + by + c = 0 , bx + cy + a = 0 and cx + ay + b = 0 be concurrent, then [IIT 1985; DCE 2002]
Solution: (c) (a) a3 + b3 + c 3 + 3abc = 0 (b) a3 + b3 + c3 − abc = 0 (c) a3 + b3 + c 3 − 3abc = 0 (d) None of these
Here the given lines are, ax + by + c = 0 , bx + cy + a = 0 , cx + ay + b = 0
abc
The lines will be concurrent, iff b c a = 0 ⇒ a3 + b3 + c3 − 3abc = 0
cab
Example: 27 If the lines 4 x + 3y = 1, y = x + 5 and 5y + bx = 3 are concurrent, then b equals
Solution: (c) (a) 1 (b) 3 (c) 6 [Rajasthan PET 1996; MP PET 1997; EAMCET 2003]
Example: 28
(d) 0
If these lines are concurrent then the intersection point of the lines 4 x + 3y = 1 and y = x + 5 , is (–2, 3), which lies on the third line.
Hence, ⇒ 5 × 3 − 2b = 3 ⇒ 15 − 2b = 3 ⇒ 2b = 12 ⇒ b = 6 [Rajasthan PET 2002]
The straight lines 4ax + 3by + c = 0 where a + b + c = 0 , will be concurrent, if point is
(a) (4, 3) (b) 1 , 1 (c) 1 , 1 (d) None of these
4 3 2 3
Solution: (b) The set of lines is 4ax + 3by + c = 0 , where a + b + c = 0
Eliminating c, we get 4ax + 3by − (a + b) = 0 ⇒ a(4 x − 1) + b(3y − 1) = 0
They pass through the intersection of the lines 4x −1= 0 and 3y − 1 = 0 i.e., x = 1 ,y = 1 i.e., 1 , 1
4 3 4 3
Reflection on the Surface.
Here IP = Incident Ray
PN = Normal to the surface I NR
PR = Reflected Ray Normal
Then, ∠IPN = ∠NPR
Reflected
Angle of incidence = Angle of reflection Incident
θθ
α αTangent
P Surfac
Image of a Point in Different cases.
(1) The image of a point with respect to the line mirror : The image of A (x1, y1)
A(x1,y1) with respect to the line mirror ax + by + c = 0 be B (h, k) is given by, ax+by+c = 0
h − x1 = k − y1 = − 2(ax1 + by1 + c)
a b a2 + b2
B (h, k)
(2) The image of a point with respect to x-axis : Let P(x,y) be any point and P′ (x′,y′) its image after
reflection in the x-axis, then Y
x′ = x ( O′ is the mid point of P and P′ )
y′ = – y
P(x, y)
O O′ X
P′(x′, y′)
(3) The image of a point with respect to y-axis : Let P(x,y) be any point and P′(x′,y′) its image after
reflection in the y-axis Y
then x′ = −x ( O′ is the mid point of P and P′ )
y′ = y P′ (x′,y′) P(x, y)
O′
X′ O X
Y′
(4) The image of a point with respect to the origin : Let P(x,y) be Y P(x, y)
any point and P′(x′,y′) be its image after reflection through the origin, then O MX
Y′
x′ = −x ( O′ is the mid point of P and P′ ) N
y′ = −y X′
P′(x′,y′)
(5) The image of a point with respect to the line y = x : Let P(x,y) be any point and P′(x′,y′) be its
image after reflection in the line y = x , then Y
P(x, y)
x′ = y ( O′ is the mid point of P and P′ )
y=x O′
y′ = x X′ 45º P′(x′, y′)
OX
Y′
(6) The image of a point with respect to the line y = x tan θ : Y
Let P(x,y) be any point and P′(x′,y′) be its image after reflection in the P(x, y)
line y = x tanθ then y=x tan θ O′
x′ = x cos 2θ + y sin 2θ ( O′ is the mid point of P and P′ ) X′ θ
P′(x′, y′)
O X
y′ = x sin 2θ − y cos 2θ
Y′
Example: 29 The reflection of the point (4, –13) in the line 5x + y + 6 = 0 is [EAMCET 1994]
Solution: (a)
(a) (–1, –14) (b) (3, 4) (c) (1, 2) (d) (–4, 13)
Example: 30
Solution: (a) Let Q(a,b) be the reflection of P(4,− 13) in the line 5x + y + 6 = 0 . Then the point R a + 4 , b − 13 lies on
2 2
5x + y + 6 = 0 . ∴ 5 a + 4 + b − 13 + 6 = 0 ⇒ 5a + b + 19 = 0 .....(i)
2 2
Also PQ is perpendicular to 5x + y + 6 = 0 . Therefore b + 13 × −5 ⇒ a − 5b − 69 =0 .....(ii)
a−4 1
Solving (i) and (ii), we get a = −1,b = −14 .
The image of a point A(3,8) in the line x + 3y − 7 = 0 , is [Rajasthan PET 1991]
(a) (–1, –4) (b) (–3, –8) (c) (1, –4) (d) (3, 8)
Equation of the line passing through (3, 8) and perpendicular to x + 3y − 7 = 0 is 3x − y − 1 = 0 . The intersection point of
both the lines is (1, 2). Now let the image of A(3,8) be A′(x1,y1) .
The point (1, 2) will be the midpoint of AA′ . x1 + 3 =1 ⇒ x1 = −1 and y1 + 8 = 2 ⇒ y1 = 4 . Hence the image is (–1, –4).
2 2
Some Important Results.
(1) Area of the triangle formed by the lines y = m1x + c1 , y = m2 x + c2, y = m3 x + c3 is 1 ∑ (c1 − c2)2 .
2 m1 − m2
(2) Area of the triangle made by the line ax + by + c =0 with the co-ordinate axes is c2 | .
2 | ab
(3) Area of the rhombus formed by the lines ax ± by ± c = 0 is 2c 2
ab
(4) Area of the parallelogram formed by the lines a1x + b1y + c1 = 0 ; a2 x + b2y + c2 = 0 , a1 x + b1y + d1 and
a2x + b2y + d2 = 0 is (d1 − c1)(d2 − c2) .
a1b2 − a2b1
(5) The foot of the perpendicular (h,k) from (x1,y1) to the line ax + by + c = 0 is given by
h − x1 = k − y1 = − (ax1 + by1 + c) . Hence, the coordinates of the foot of perpendicular is
a b a2 + b2
b 2 x1 − aby1 − ac , a 2y1 − abx1 − bc
a2 + b2 a2 + b2
(6) Area of parallelogram A= p1 p2 , where p1 and p2 are the distances between parallel sides and θ is the
sinθ
angle between two adjacent sides.
(7) The equation of a line whose mid-point is (x1, y1) in between the axes is x y =2
x1 + y1
(8) The equation of a straight line which makes a triangle with the axes of centroid (x1, y1) is x + y = 1.
3x1 3y1
Example: 31 The coordinates of the foot of perpendicular drawn from (2, 4) to the line x + y = 1 is [Roorkee 1995]
Solution: (b) (a) 1 , 3 (b) − 1 , 3 (c) 4 , 1 (d) 3 , −1
Example: 32 3 2 2 2 3 2 4 2
Applying the formula, the required co-ordinates is 12 × 2 −1×1× 4 + 1 , 12 × 4 −1×1× 2 + 1 = −1 , 3
12 + 12 12 + 12 2 2
The area enclosed within the curve | x |+|y|= 1 is [Rajasthan PET 1990, 97; IIT 1981; UPSEAT 2003]
(a) 2 (b) 1 (c) 3 (d) 2
Solution: (d) The given lines are ±x ± y = 1 i.e., x + y = 1 , x − y = 1 , x + y = −1 and x − y = −1 . These lines form a quadrilateral whose
vertices are A(−1,0) , B(0, − 1) , C(1,0) and D(0,1) . Obviously ABCD is a square. Length of each side of this square is
12 + 12 = 2 . Hence, area of square is 2 × 2 = 2 sq. units.
Example: 33 If x1, x2, x3 and y1, y2, y3 are both in G.P. with the same common ratio, then the point (x1, y1) , (x2, y2 ) and (x3, y3 )
Solution: (a)
Example: 34 [AIEEE 2003]
(a) Lie on a straight line (b) Lie on an ellipse (c) Lie on a circle (d) Are vertices of a triangle
Taking co-ordinates as x , y , (x, y) and (xr ,yr ) . Above co-ordinates satisfy the relation y = mx , ∴ the three points lie
r r
on a straight line.
A square of side a lies above the x-axis and has one vertex at the origin. The side passing through the origin makes an
angle α 0 <α < π with the positive direction of x-axis. The equation of its diagonal not passing through the origin is
4
[AIEEE 2003]
(a) y(cosα − sinα) − x(sinα − cosα) = a (b) y(cosα + sinα) − x(sinα − cosα) = a
(c) y(cosα + sinα) + x(sinα + cosα)= a (d) y(cosα + sinα) + x(sinα − cosα) = a
Solution: (b) Co-ordinates of A = (a cosα,a sinα) ; Equation of OB y = tan π + α x YB
4
CA⊥ to OB ; ∴ slope of CA = − cot π + α C
4 A
Equation of CA, y − a sinα = − cot π +α (x − a cosα) π/4
4 α
⇒ y(sinα + cosα) + x (cos a − sinα) = a . O X
Example: 35 The number of integral points (integral point means both the coordinates should be integer) exactly in the interior of the
Solution: (b)
triangle with vertices (0, 0), (0, 21) and (21, 0) is [IIT Screening 2003]
(a) 133 (b) 190 (c) 233 (d) 105
x + y = 21 (0, 21)
The number of integral solution to the equation x + y < 21 i.e., x < 21 − y B
Number of integral co-ordinates = 19 + 18 + .......... +1 = 19 × 20 = 190 .
2
(0O, 0) A (21, 0)
Conic sections
Introduction.
The curves obtained by intersection of a plane and a double cone in different orientation are called conic section.
In other words “Graph of a quadratic equation (in two variables) is a “Conic section”.
A conic section or conic is the locus of a point P, which moves in such a way that its distance from a fixed
point S always bears a constant ratio to its distance from a fixed straight line, all
being in the same plane. MP
SP = constant = e (eccentricity) Directrix
PM
S
or SP= e. PM
Definitions of Various important Terms.
(1) Focus : The fixed point is called the focus of the conic-section.
(2) Directrix : The fixed straight line is called the directrix of the conic section.
In general, every central conic has four foci, two of which are real and the other two are imaginary. Due to
two real foci, every conic has two directrices corresponding to each real focus.
(3) Eccentricity : The constant ratio is called the eccentricity of the conic section and is denoted by e.
If e = 1 , the conic is called Parabola.
If e < 1, the conic is called Ellipse.
If e > 1 , the conic is called Hyperbola.
If e = 0 , the conic is called Circle.
If e = ∞ , the conic is called Pair of the straight lines.
(4) Axis: The straight line passing through the focus and perpendicular to the directrix is called the axis of the
conic section. A conic is always symmetric about its axis.
(5) Vertex: The points of intersection of the conic section and the axis are called vertices of conic section.
(6) Centre: The point which bisects every chord of the conic passing through it, is called the centre of conic.
(7) Latus-rectum: The latus-rectum of a conic is the chord passing through the focus and perpendicular to
the axis.
(8) Double ordinate: The double ordinate of a conic is a chord perpendicular to the axis.
(9) Focal chord: A chord passing through the focus of the conic is called a focal chord.
(10) Focal distance: The distance of any point on the conic from the focus is called the focal distance of the point.
General equation of a Conic section when its Focus, Directrix and Eccentricity are given
Let S(α, β ) be the focus, Ax + By + C = 0 be the directrix and e be the eccentricity of a conic. Let P(h,k) be
any point on the conic. Let PM be the perpendicular from P, on the directrix. Then by Z
definition M P(x, y)
SP = ePM ⇒ SP 2 = e 2PM 2 Ax+By+C=0 S(α, β )
⇒ (h − α)2 + (k − β )2 = e 2 Ah + Bk + C 2 Z′
A2 + B 2
Thus the locus of (h,k) is (x − α)2 + (y − β )2 = e2 (Ax + By + C)2 this is the cartesian equation of the conic
(A2 + Β2)
section which, when simplified, can be written in the form ax 2 + 2hxy + by2 + 2gx + 2 fy + c = 0 , which is general
equation of second degree.
Recognisation of Conics.
The equation of conics is represented by the general equation of second degree
ax 2 + 2hxy + by2 + 2gx + 2 fy + c = 0 ......(i)
and discriminant of above equation is represented by ∆ , where
∆ = abc + 2 fgh − af 2 − bg 2 − ch2
Case I: When ∆ = 0
In this case equation (i) represents the degenerate conic whose nature is given in the following table.
S. No. Condition Nature of conic
1. ∆ = 0 and ab − h2 = 0` A pair of coincident straight lines
2. ∆ = 0 and ab − h2 < 0 A pair of intersecting straight lines
3. ∆ = 0 and ab − h2 > 0 A point
Case II: When ∆ ≠ 0
In this case equation (i) represents the non-degenerate conic whose nature is given in the following table.
S. No. Condition Nature of conic
1. ∆ ≠ 0, h = 0, a = b A circle
2. A parabola
3. ∆ ≠ 0, ab − h2 = 0 An ellipse
4. A hyperbola
5. ∆ ≠ 0, ab − h2 > 0 A rectangular hyperbola
∆ ≠ 0, ab − h2 < 0
∆ ≠ 0, ab − h2 < 0 and a + b = 0
Method to find centre of a Conic.
Let S = ax2 + 2hxy + by2 + 2gx + 2 fy + c be the given conic. Find ∂S ; ∂S
∂x ∂y
Solve ∂S = 0, ∂S = 0 for x, y we shall get the required centre (x, y)
∂x ∂y
(x, y) = hf − bg , gh − af
ab − h2 ab − h2
Example: 1 The equation x 2 − 2xy + y 2 + 3x + 2 = 0 represents [UPSEAT 2001]
Solution: (a)
(a) A parabola (b) An ellipse (c) A hyperbola (d) A circle
Example: 2
Solution: (a) Comparing the given equation with ax 2 + 2hxy + by 2 + 2gx + 2 fy + c = 0
Here, a = 1, b = 1, h = – 1, g= 3 , f = 0, c = 2
2
Now ∆ = abc + 2 fgh − af 2 − bg 2 − ch2
⇒ ∆ = (1)(1)(2) + 2 3 (0)(−1) − (1)(0)2 − 1 3 2 − 2(−1)2 ⇒ ∆ = −9 i.e., ∆≠0 and h2 − ab = 1 − 1 = 0 i.e., h2 = ab
2 2 4
So given equation represents a parabola.
The centre of 14 x 2 − 4 xy + 11y 2 − 44 x − 58y + 71 = 0 is [BIT Ranchi1986]
(a) (2, 3) (b) (2, –3) (c) (–2, 3) (d) (–2, –3)
Centre of conic is hf − bg , gh − af
ab − h2 ab − h2
Here, a = 14, h = −2 , b = 11 , g = −22 , f = −29 , c = 71
Centre ≡ (−2)(−29) − (11)(−22) , (−22)(−2) − (14)(−29)
(14)(11) − (−2)2 (14)(11) − (−2)2
Centre ≡ (2, 3) .
Conic sections
Definition .
A parabola is the locus of a point which moves in a plane such that its distance from a fixed point (i.e., focus)
in the plane is always equal to its distance from a fixed straight line (i.e., directrix) in the same plane.
General equation of a parabola : Let S be the focus, ZZ' be the directrix and let P be any point on the
parabola. Then by definition, ( e = 1) Z
SP = PM M P(x,y)
Ax+By+C=0
(x − α)2 + (y − β )2 = Ax + By + C Directrix
A2 + B2
{ }Or (A2 + B2 ) (x − α)2 + (y − β )2 = (Ax + By + C)2 S(α,β)
Focus
Z′
Example: 1 The equation of parabola whose focus is (5, 3) and directrix is 3x − 4y + 1 = 0 , is [MP PET 2002]
Solution: (a)
(a) (4 x + 3y)2 − 256x − 142y + 849 = 0 (b) (4 x − 3y)2 − 256x − 142y + 849 = 0
(c) (3x + 4y)2 − 142x − 256y + 849 = 0 (d) (3x − 4y)2 − 256x − 142y + 849 = 0
3 x − 4 y+ 1 2
9 + 16
PM 2 = PS 2 ⇒ (x − 5)2 + (y − 3)2 = Y
M
X' O P(x,y)
⇒ 25(x 2 + 25 − 10x + y 2 + 9 − 6y)
= 9x 2 + 16y 2 + 1 − 12xy + 6x − 8y − 12xy S(5, 3) X
⇒ 16x 2 + 9y 2 − 256x − 142y + 24 xy + 849 = 0 Y'
⇒ (4 x + 3y)2 − 256x − 142y + 849 = 0
Standard equation of the Parabola .
Let S be the focus ZZ' be the directrix of the parabola and (x, y) be any point on parabola.
Let AS = AK = a(> 0) then coordinate of S is (a, 0) and the equation of KZ is x = −a or x + a = 0
Now SP = PM ⇒ (SP)2 = (PM)2 ZY
⇒ (x − a)2 + (y − 0)2 = (a + x)2
∴ y 2 = 4ax L P(x,y X
which is the equation of the parabola in its standard form. X′ K A S (a,0)
Directrix
L′ P′ y2=4ax
Z′ Y′
Some terms related to parabola Y (a, 2a) Q F(h,2 ah)
For the parabola y 2 = 4ax , L
M P x=a Focal chord
Directrix Double ordinate
Focal distance
Vertex Focus Axis X
Z A S(a,0) Latus rectum
x+a=0 F′ L′
Y′ (a,–2a) Q'(h,−2 ah )
(1) Axis : A straight line passes through the focus and perpendicular to the directrix is called the axis of parabola.
For the parabola y 2 = 4ax, x-axis is the axis. Here all powers of y are even in y 2 = 4ax.Hence parabola
y 2 = 4ax is symmetrical about x-axis.
(2) Vertex : The point of intersection of a parabola and its axis is called the vertex of the parabola. The
vertex is the middle point of the focus and the point of intersection of axis and the directrix.
For the parabola y 2 = 4ax, A(0,0) i.e., the origin is the vertex.
(3) Double-ordinate : The chord which is perpendicular to the axis of parabola or parallel to directrix is
called double ordinate of the parabola.
Let QQ' be the double-ordinate. If abscissa of Q is h then ordinate of Q, y 2 = 4ah or y = 2 ah
(for I st Quadrant) and ordinate of Q' is y = −2 ah (for IVth Quadrant). Hence coordinates of Q and Q' are
(h, 2 ah) and (h, − 2 ah) respectively.
(4) Latus-rectum : If the double-ordinate passes through the focus of the parabola, then it is called latus-
rectum of the parabola.
Coordinates of the extremeties of the latus rectum are L(a, 2a) and L'(a,− 2a) respectively.
Since LS = L' S = 2a ∴Length of latus rectum LL' = 2(LS) = 2(L' S) = 4a .
(5) Focal Chord : A chord of a parabola which is passing through the focus is called a focal chord of the
parabola. Here PP' and LL' are the focal chords.
(6) Focal distance (Focal length) : The focal distance of any point P on the parabola is its distance from
the focus S i.e., SP.
Here, Focal distance SP = PM = x + a
Note : If length of any double ordinate of parabola y2 = 4ax is 2l, then coordinates of end points of this
double ordinate are l2 ,l and l2 ,−l .
4a 4a
Important Tips
• The area of the triangle inscribed in the parabola y2 = 4ax is 1 (y1 ~ y2 )(y2 ~ y3 )(y3 ~ y1 ), where y1, y2y3 are the ordinate of the vertices
8a
• The length of the side of an equilateral triangle inscribed in the parabola y 2 = 4ax is 8a 3 (one angular point is at the vertex).
Example: 2 The point on the parabola y 2 = 18x , for which the ordinate is three times the abscissa, is [MP PET 2003]
Solution: (d) [UPSEAT 1998]
Example: 3 (a) (6, 2) (b) (–2, –6) (c) (3, 18) (d) (2, 6)
Solution: (c) Given y = 3x , then (3x)2 = 18x ⇒ 9x 2 = 18x ⇒ x = 2 and y = 6 .
Example: 4
Solution: (d) The equation of the directrix of parabola 5y 2 = 4 x is
(a) 4 x − 1 = 0 (b) 4 x + 1 = 0 (c) 5x + 1 = 0 (d) 5x −1 = 0
The given parabola is y2 = 4 x . Here a= 1 . Directrix is x = −a = −1 ⇒ 5x + 1 = 0
5 5 5
The point on the parabola y 2 = 8x . Whose distance from the focus is 8, has x-coordinate as
(a) 0 (b) 2 (c) 4 (d) 6
If P(x1 , y1 ) is a point on the parabola y 2 = 4ax and S is its focus, then SP = x1 + a
Here 4a = 8 ⇒ a = 2; SP = 8
∴ 8 = x1 + 2 ⇒ x1 = 6
Example: 5 If the parabola y 2 = 4ax passes through (–3, 2), then length of its latus rectum is [Rajasthan PET 1986, 95]
Solution: (c)
(a) 2/3 (b) 1/3 (c) 4/3 (d) 4
The point (−3, 2) will satisfy the equation y2 = 4ax ⇒ 4 = −12a ⇒ Latus rectum = 4 |a |= 4×| − 1 |= 4
3 3
Some other standard forms of Parabola a
(1) Parabola opening to left (2) Parabola opening upwards (3) Parabola opening down wards
(i.e. x 2 = −4ay ); (a > 0)
(i.e. y 2 = −4ax ); (a > 0) (i.e. x 2 = 4ay) ; (a>0)
Y M x-a=0 Y Y
LP Z M y–a=0
X′ ZX S(0, a) L
S(–a,0) A L′ A
X′ P
L′ Y′ P X′ X
A L′
X
L
S(0, -a)
y+a=0 Z M Y′
Y′
Important terms y2 = 4ax y2 = −4ax x 2 = 4ay x 2 = −4ay
(0, 0) (0, 0) (0, 0) (0, 0)
Coordinates of vertex (a, 0) (–a, 0) (0, a) (0, –a)
Coordinates of focus x = −a x=a y = −a y=a
Equation of the directrix y=0 y=0
Equation of the axis x=0 x=0
4a 4a
Length of the latusrectum x+a a−x 4a 4a
Focal distance of a point y+a a−y
P(x, y)
Example: 6 Focus and directrix of the parabola x 2 = −8ay are [Rajasthan PET 2001]
Solution: (a) (a) (0,−2a) and y = 2a (b) (0, 2a) and y = −2a
(c) (2a, 0) and x = −2a (d) (−2a, 0) and x = 2a
Example: 7
Solution: (c) Given equation is x 2 = −8ay
Comparing the given equation with x2 = −4 AY , A = 2a
Focus of parabola (0,–A) i.e. (0, –2a)
Directrix y = A , i.e. y = 2a
The equation of the parabola with its vertex at the origin, axis on the y-axis and passing through the point (6, –3) is
[MP PET 2001]
(a) y 2 = 12x + 6 (b) x 2 = 12y (c) x 2 = −12y (d) y 2 = −12x + 6
Since the axis of parabola is y-axis with its vertex at origin.
∴Equation of parabola x 2 = 4ay . Since it passes through (6, –3) ; ∴36= –12a ⇒ a = −3
∴ Equation of parabola is x 2 = −12y .
Special form of Parabola (y – k)2 = 4a(x – h)=a
The equation of a parabola with its vertex at (h, k) and axis as parallel to Y x=h–a X′′
x-axis is (y − k)2 = 4a(x − h) X
a
If the vertex of the parabola is (p,q) and its axis is parallel to y-axis, then Z′ A′ (h,k) S
the equation of the parabola is (x − p)2 = 4b(y − q)
Directrix
When origin is shifted at A'(h,k)without changing the direction of axes, X′ A
its equation becomes (y − k)2 = 4a(x − h) or (x − p)2 = 4b(y − q)
Y′
Equation of Parabola Vertex Axis Focus Directrix Equation Length of
y=k (h + a,k) of L.R. L.R.
(y − K)2 = 4a(x − h) (h, k) x=p x+a−h= 0
(x − p)2 = 4b(y − q) (p, q) x =a+h 4a
(p, b + q) y + b − q = 0 y = b + q 4b
Important Tips
• y 2 = 4a(x + a) is the equation of the parabola whose focus is the origin and the axis is x-axis.
• y 2 = 4a(x − a) is the equation of parabola whose axis is x-axis and y-axis is directrix.
• x 2 = 4a(y + a) is the equation of parabola whose focus is the origin and the axis is y-axis.
• x 2 = 4a(y − a) is the equation of parabola whose axis is y-axis and the directrix is x-axis.
• The equation to the parabola whose vertex and focus are on x-axis at a distance a and a' respectively from the origin is
y 2 = 4(a'−a)(x − a) .
• The equation of parabola whose axis is parallel to x-axis is x = Ay2 + By + C and y = Ax 2 + Bx + C is a parabola with its axis parallel to
y-axis.
Example: 8 Vertex of the parabola x 2 + 4 x + 2y − 7 = 0 is [MP PET 1990]
Solution: (a)
Example: 9 (a) (−2,11 / 2) (b) (−2,2) (c) (−2,11) (d) (2,11)
Solution: (b)
Equation of the parabola is (x + 2)2 = −2y + 7 + 4 ⇒ (x + 2)2 = −2 y − 11 . Hence vertex is − 2, 11 .
Example: 10 2 2
The focus of the parabola 4y 2 − 6x − 4y = 5 is [Rajasthan PET 1997]
(a) (−8 / 5,2) (b) (−5 / 8, 1 / 2) (c) (1 / 2, 5 / 8) (d) (6 / 8, − 1 / 2)
Given equation of parabola when written in standard form, we get
y 1 2 y 1 2 3 Y2 3 1
2 2 2 2 2
4 − = 6(x + 1) ⇒ − = (x + 1) ⇒ = X where, Y = y − , X = x +1
∴ y = Y + 1 , x = X −1 ....(i)
2
Focus ⇒ X = a, Y = 0 ; ∴ 4a = 3 ⇒ a = 3 ⇒ x = 3 − 1 = − 5 ; y = 0 + 1 = 1 ⇒ Focus = − 5 , 1
2 8 8 8 2 2 8 2
The equation of the directrix of the parabola y2 + 4y + 4 x + 2 = 0 is [IIT Screening 2001]
(a) x = −1 (b) x = 1 (c) x = −3 (d) x = 3
2 2
Solution: (d) Here, y2 + 4y + 4 + 4x − 2 = 0 or (y + 2)2 = −4 x − 1
2
Example: 11
Solution: (c) Let y + 2 = Y , 1 − x = X . Then the parabola is Y2 = 4X . ∴ The directrix is X + 1 = 0 or 1 − x +1 = 0 , ∴ x = 3
Example: 12 2 2 2
Solution: (c)
The line x − 1 = 0 is the directrix of the parabola y 2 − kx + 8 = 0 . Then one of the values of k is [IIT Screening 2000]
(a) 1 (b) 8 (c) 4 (d) 1
8 4
The parabola is y2 = 4 k x − 8 . Putting y = Y, x − 8 = X . The equation is Y2 = 4. k .X
4 k k 4
∴ The directrix is X + k = 0 i.e., x − 8 + k = 0 . But x − 1 = 0 is the directrix. So 8 − k =1 ⇒ k = −8, 4 .
4 k 4 k 4
Equation of the parabola with its vertex at (1, 1)and focus (3, 1)is [Karnataka CET 2001, 2002]
(a) (x − 1)2 = 8(y − 1) (b) (y − 1)2 = 8(x − 3) (c) (y − 1)2 = 8(x − 1) (d) (x − 3)2 = 8(y − 1)
Given vertex of parabola (h, k) ≡ (1,1) and its focus (a + h, k) ≡ (3,1) or a + h = 3 or a = 2 . We know that as the
y-coordinates of vertex and focus are same, therefore axis of parabola is parallel to x-axis. Thus equation of the parabola
is (y − k)2 = 4a(x − h) or (y − 1)2 = 4 × 2 (x − 1) or (y − 1)2 = 8(x − 1) .
Parametric equations of a Parabola a
The simplest and the best form of representing the coordinates of a point on the parabola y 2 = 4ax is
(at2 2at) because these coordinates satisfy the equation y2 = 4ax for all values of t. The equations x = at2, y = 2at
taken together are called the parametric equations of the parabola y2 = 4ax , t being the parameter.
The following table gives the parametric coordinates of a point on four standard forms of the parabola and
their parametric equation.
Parabola y 2 = 4ax y 2 = −4ax x 2 = 4ay x 2 = −4ay
Parametric (at 2 ,2at) (−at 2 ,2at) (2at, at 2 ) ( 2at, − at 2 )
Coordinates
Parametric x = at 2 x = −at 2 x = 2at x = 2at ,
Equations y = 2at y = 2at y = at 2
y = −at 2
Note : The parametric equation of parabola (y − k)2 = 4a(x − h) are x = h + at 2 and y = k + 2at
Example: 13 x − 2 = t 2 , y = 2t are the parametric equations of the parabola
Solution: (d)
(a) y 2 = 4 x (b) y 2 = −4 x (c) x 2 = −4y (d) y 2 = 4(x − 2)
y y 2 2
2 2
Here = t and x − 2= t2⇒ (x − 2) = ⇒ y = 4(x − 2)
Position of a point and a Line with respect to a Parabola a Y P(on)
(1) Position of a point with respect to a parabola: The point P P(inside)
(Outside)
P(x1, y1)lies outside on or inside the parabola y 2 = 4ax according as X
y12 − 4ax1 >, =, or < 0
(2) Intersection of a line and a parabola: Let the parabola be y 2 = 4ax .....(i)
And the given line be y = mx + c .....(ii)
Eliminating y from (i) and (ii) then (mx + c)2 = 4ax or m2 x 2 + 2x(mc − 2a) + c 2 = 0 .....(iii)
This equation being quadratic in x, gives two values of x. It shows that every straight line will cut the parabola
in two points, may be real, coincident or imaginary, according as discriminate of (iii) >, = or < 0
∴ The line y = mx +c does not intersect, touches or intersect a parabola y2 = 4ax , according as c >, =, < a
m
Condition of tangency : The line y = mx + c touches the parabola, if c = a
m
Example: 14 The equation of a parabola is y 2 = 4 x . P(1,3) and Q(1,1) are two points in the xy-plane. Then, for the parabola
(a) P and Q are exterior points (b) P is an interior point while Q is an exterior point
(c) P and Q are interior points (d) P is an exterior point while Q is an interior point
Solution: (d) Here, S ≡ y 2 − 4 x = 0 ; S(1, 3) = 3 2 − 4.1 > 0 ⇒ P(1, 3) is an exterior point.
S(1,1) = 12 − 4.1 < 0 ⇒ Q(1,1) is an interior point.
Example: 15 The ends of a line segment are P(1,3) and Q(1,1) . R is a point on the line segment PQ such that PQ : QR = 1 : λ . If R is
an interior point of the parabola y2 = 4 x , then
(a) λ ∈ (0, 1) (b) λ ∈ − 3 , 1 (c) λ ∈ 1 , 3 (d) None of these
5 2 5
1, 1 + 3λ y2 1 + 3λ 2
1+λ 1+λ
Solution: (a) R = It is an interior point of − 4 x = 0 iff −4<0
⇒ 1 + 3λ − 2 1 + 3λ + 2 < 0 ⇒ λ −1 5λ + 3 < 0 ⇒ (λ − 1) λ + 3 < 0
1+λ 1+λ 1 +λ 1+λ 5
Therefore, − 3 <λ < 1 . But λ > 0 ∴0 < λ < 1 ⇒ λ ∈ (0, 1) .
5
Equation of Tangent in Different formsa
(1) Point Form: The equation of the tangent to the parabola y2 = 4ax at a point (x1, y1 ) is yy1 = 2a(x + x1)
P(x1,y1)
T
Q(x2,y2)
Equation of tangent of all other standard parabolas at (x1, y1)
Equation of parabolas Tangent at (x1, y1)
y2 = −4ax yy1 = −2a(x + x1)
x 2 = 4ay xx1 = 2a(y + y1)
x 2 = −4ay xx1 = −2a(y + y1)
Note : The equation of tangent at (x1, y1) to a curve can also be obtained by replacing x 2 by xx1 , y 2 by
yy1 , x by x + x1 , y by y + y1 and xy by xy1 + x1y provided the equation of curve is a polynomial
2 2 2
of second degree in x and y.
( )(2) Parametric form : The equation of the tangent to the parabola y 2 = 4ax at at 2,2at is ty = x + at 2
Equations of tangent of all other standard parabolas at 't'
Equations of parabolas Parametric co-ordinates 't' Tangent at 't'
y2 = −4ax (−at 2,2at) ty = −x + at 2
x 2 = 4ay (2at,at 2) tx = y + at 2
x 2 = −4ay (2at, − at 2) tx = −y + at 2
(3) Slope Form: The equation of a tangent of slope m to the parabola y2 = 4ax at a , 2a is y = mx + a
m2 m m
Equation of Point of contact in Equation of tangent Condition of
parabolas terms of slope (m) in terms of slope (m) Tangency
y2 = 4ax
a , 2a y = mx + a c = a
y2 = −4ax m2 m m m
x 2 = 4ay − a ,− 2a y = mx − a c = a
x 2 = −4ay m2 m m −m
(2am, am2 ) y = mx − am2 c = −am2
(−2am,−am2) y = mx + am2 c = am2
Important Tips
• If the straight line lx + my + n = 0 touches the parabola y 2 = 4ax then l n = am 2 .
• If the line x cos α + y sin α = p touches the parabola y 2 = 4ax, then P cos α + a sin2 α = 0 and point of contact is (a tan2 α,−2a tan α)
• If the line x + y = 1 touches the parabola y 2 = 4a(x + b) , then m2 (l + b) + al 2 = 0
l m
Point of intersection of Tangents at any two points on the Parabola a
The point of intersection of tangents at two points P(at12 ,2at1 ) and Q(at 2 ,2at 2 ) on the parabola y2 = 4ax is
2
(at1t 2 , a(t1 + t 2 )) .
The locus of the point of intersection of tangents to the parabola y 2 = 4ax which meet at an angle α is
(x + a)2 tan2 α = y 2 − 4ax .
Director circle: The locus of the point of intersection of perpendicular Y (at12 ,2at1 )
tangents to a conic is known as its director circle. The director circle of a parabola P
is its directrix. (at1t2,a(t1+t2))
X′ R O X
Q(at 2 ,2at 2 )
2
Y′
Note : Clearly, x-coordinates of the point of intersection of tangents at P and Q on the parabola is
the G.M of the x-coordinate of P and Q and y-coordinate is the P(at12,2at1)
A.M. of y-coordinate of P and Q.
The equation of the common tangents to the parabola y 2 = 4ax R [at1t2,a(t1+t2)]
1 1 22
and x 2 = 4by is a 3 x + b 3 y + a 3 b 3 = 0
Q(at 2 ,2at 2 )
2
The tangents to the parabola y 2 = 4ax at P(at12,2at1)and
Q(at 2 ,2at 2 ) intersect at R. Then the area of triangle PQR is 1 a 2(t1 − t 2 )3
2 2
Equation of Pair of Tangents from a point to a Parabolaa
If y12 − 4ax1 > 0 , then any point P(x1, y1) lies out side the parabola and a
pair of tangents PQ, PR can be drawn to it from P Y
The combined equation of the pair of the tangents drawn from a point to X′ (x1,y1)P O Q X
a parabola is SS' = T 2 where S = y 2 − 4ax;S' = y12 − 4ax1 and R
T = yy1 − 2a(x + x1)
Y′
Note : The two tangents can be drawn from a point to a parabola. The two tangent are real and distinct
or coincident or imaginary according as the given point lies outside, on or inside the parabola.
Important Tips
• Tangents at the extremities of any focal chord of a parabola meet at right angles on the directrix.
• Area of the triangle formed by three points on a parabola is twice the area of the triangle formed by the tangents at these points.
• If the tangents at the points P and Q on a parabola meet in T, then ST is the geometric mean between SP and SQ, i.e. ST 2 = SP.SQ
• Tangent at one extremity of the focal chord of a parabola is parallel to the normal at the other extremity.
• The angle of intersection of two parabolas y2 = 4ax and x2 = 4by is given by tan −1 3a1 / 3b1 / 3
2(a 2 / 3 + b 2 / 3 )
Example: 16 The straight line y = 2x + λ does not meet the parabola y 2 = 2x, if [MP PET 1993; MNR 1977]
Solution: (b) (a) λ < 1 (b) λ > 1 (c) λ = 4 (d) λ = 1
Example: 17 4 4
Solution: (c)
Example: 18 y = 2x + λ does not meet the parabola y2 = 2x , If λ > a = 1 = 1 ⇒λ > 1
Solution: (a) m 2.2 4 4
If the parabola y 2 = 4ax passes through the point (1, − 2) , then the tangent at this point is [MP PET 1998]
(a) x + y − 1 = 0 (b) x − y − 1 = 0 (c) x + y + 1 = 0 (d) x − y + 1 = 0
Parabola passes through the point (1, –2), then 4 = 4a ⇒ a = 1 . From yy1 = 2a(x + x1) ⇒ −2y = 2(x + 1)
∴ Required tangent is x + y + 1 = 0
The equation of the tangent to the parabola y 2 = 16x , which is perpendicular to the line y = 3x + 7 is [MP PET 1998]
(a) y − 3x + 4 = 0 (b) 3y − x + 36 = 0 (c) 3y + x − 36 = 0 (d) 3y + x + 36 = 0
A line perpendicular to the given line is 3y + x =λ⇒y=− 1 x + λ
3 3
Here m = − 1 , c= λ . If we compare y2 = 16x with y2 = 4ax, then a=4
3 3
Condition for tangency is c = a ⇒ λ = 4 3) ⇒ λ = −36 . ∴ Required equation is x + 3y + 36 = 0 .
m 3 (−1 /
Example: 19 If the tangent to the parabola y 2 = ax makes an angle of 45o with x-axis, then the point of contact is
Solution: (d)
[Rajasthan PET 1985, 90, 2003]
Example: 20
Solution: (d) (a) a , a (b) a , a (c) a , a (d) a , a
Example: 21 2 2 4 4 2 4 4 2
Solution: (a)
Example: 22 Parabola is y2 = ax i.e. y2 = 4 a x .....(i)
Solution: (a) 4
Example: 23 Let point of contact is (x1, y1). ∴ Equation of tangent is y − y1 = 2(a / 4) (x − x1) ⇒ y = a (x) − ax1 + y1
Solution: (b) y1 2y1 2y1
Here, m = a = tan 45o ⇒ a =1⇒ y1 = a . From (i), x1 = a , So point is a , a .
2y1 2y1 2 4 4 2
The line x − y + 2 = 0 touches the parabola y 2 = 8x at the point [Roorkee 1998]
(a) (2, –4) (b) (1, 2 2) (c) (4,−4 2) (d) (2, 4)
The line x − y + 2 = 0 i.e. x = y − 2 meets parabola y 2 = 8x, if
⇒ y2 = 8(y − 2) = 8y − 16 ⇒ y 2 − 8y + 16 = 0 ⇒ (y − 4)2 = 0 ⇒ y = 4,4 [Rajasthan PET 1996]
Roots are equal, ∴ Given line touches the given parabola.
∴ x = 4 − 2 = 2 , Thus the required point is (2, 4).
The equation of the tangent to the parabola at point (a / t 2 ,2a / t) is
(a) ty = xt 2 + a (b) ty = x + at 2 (c) y = tx + at 2 (d) y = tx + (a / t 2 )
Equation of the tangent to the parabola, y 2 = 4ax is yy1 = 2a(x + x1 )
⇒ y. 2a = 2a x + a ⇒ y = x + a ⇒ y = t2x + a ⇒ ty = t 2 x + a
t t2 t t2 t t2
Two tangents are drawn from the point (–2, –1) to the parabola y 2 = 4 x. If α is the angle between these tangents, then
tanα = [EAMCET 1992; Karnataka CET 1992]
(a) 3 (b) 1/3 (c) 2 (d) 1/2
Equation of pair of tangent from (−2,−1) to the parabola is given by SS1 = T 2 i.e. (y 2 − 4 x)(1 + 8) = [y(−1) − 2(x − 2)]2
⇒ 9y 2 − 36x = [−y − 2x + 4]2 ⇒ 9y 2 − 36x = y 2 + 4 x 2 + 16 + 4 xy − 16x − 8y ⇒ 4 x 2 − 8y 2 + 4 xy + 20x − 8y + 16 = 0
∴ tan α = 2 h2 − ab = 2 4 − 4(−8) = 12 =3
a+b 4−8 −4
a 1 / 3 b 1 / 3 3 y2 x2
b a 2
If + = , then the angle of intersection of the parabola = 4ax and = 4by at a point other than the
origin is
(a) π / 4 (b) π / 3 (c) π / 2 (d) None of these
Given parabolas are y 2 = 4ax ....(i) and x 2 = 4by .....(ii)
These meet at the points (0, 0), (4a1 / 3 b2 / 3,4a2 / 3b1 / 3)
Tangent to (i) at (4a1 / 3b 2 / 3 ,4a 2 / 3b1 / 3 ) is y.4a2 / 3b1 / 3 = 2a(x + 4a2 / 3b1 / 3)
Slope of the tangent (m1 ) = 2a a1/ 3
4a 2/ 3b1/ 3 = 2b1 / 3
Tangent to (ii) at (4a1 / 3b 2 / 3 ,4a 2 / 3b1 / 3 ) is x.4a1 / 3b 2 / 3 = 2b(y + 4a 2 / 3b1 / 3 )
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