101
Hence P(1) is true.
CASE N = k
Assume P(k) is true. That is,
P(k) = 1! + (2 × 2!) + (3 × 3!) + ... + (k × k!) = (k + 1)! − 1.
CASE N = k+1
1! + (2 × 2!) + (3 × 3!) + ... + ((k + 1) × (k + 1)!) = 1! + (2 × 2!) + (3 × 3!) + ... + (k × k!) + ((k + 1) × (k + 1)!)
= P(k) + ((k + 1) × (k + 1)!)
= (k + 1)! − 1 + ((k + 1) × (k + 1)!)
= (k + 1)! × (1 + k + 1) − 1
= (k + 2)! − 1
= P(k + 1)
This implies, P(k + 1) is true whenever P(k) is true. Hence, by the principle of mathematical
induction, P(n) is true for all natural numbers.
10. Using the Mathematical induction, show that for any natural number n, x 2n − y 2n is divisible by
x + y.
Solution:
Let P(n) = x 2n − y 2n = k(x + y). Where k is a factor.
CASE N = 1
2
2
x − y = (x − y)(x + y)
= k(x + y)
LHS = RHS
Hence P(1) is true.
CASE N = k
Assume P(k) is true. That is,
P(k) = x 2k − y 2k = k(x + y).
CASE N = k+1
102
2k 2k
2k 2k
x 2k+2 − y 2k+2 = x 2k+2 − x y + x y − y 2k+2
2k
2
2
= x (x − y ) + y 2 x 2k − y 2k
2k
2
2
2
= x (x − y ) + y (P(k))
2k
2
= x (x + y)(x − y) + y (k)(x + y)
2k
2
= (x + y)(x (x − y) + y (k))
= P(k + 1)
This implies, P(k + 1) is true whenever P(k) is true. Hence, by the principle of mathematical
induction, P(n) is true for all natural numbers.
11. By the principle of mathematical induction, prove that, for n ≥ 1
n 3
2
2
2
2
1 + 2 + 3 + · · · + n > .
3
Solution:
2
2
2
2
Let P(n) = 1 + 2 + 3 + · · · + n .
CASE N = 1
1
2
1 > Hence P(1) is true.
3
CASE N = k
Assume P(k) is true. That is,
k 3
2
2
2
2
P(k) = 1 + 2 + 3 + · · · + k > .
3
CASE N = k+1
2
2
2
2
2
2
2
2
1 + 2 + 3 + · · · + (k + 1) = 1 + 2 + 3 + · · · + k + (k + 1) 2
k 3
> + (k + 1) 2
3
3
k + 3(k + 1) 2
=
3
3
2
k + 3k + 6k + 3
=
3
3
(k + 1) + 3(k + 1)
=
3
(k + 1) 3
= + (k + 1)
3
(k + 1) 3
>
3
= P(k + 1)
This implies, P(k + 1) is true whenever P(k) is true. Hence, by the principle of mathematical
induction, P(n) is true for all natural numbers.
103
3
12. Use induction to prove that n − 7n + 3, is divisible by 3, for all natural numbers n.
3
Let P(n) = n − 7n + 3
CASE N = 1
1 − 7 + 3 = −3
= multiple of 3
Hence P(1) is true.
CASE N = k
Assume P(k) is true. That is,
3
P(k) = k − 7k + 3 = 3p.
CASE N = k+1
2
3
3
(k + 1) − 7(k + 1) + 3 = k + 3k + 3k + 1 − 7k − 7 + 3
3
2
= k − 7k + 3 + 3k + 3k − 6
2
= 3p + 3(k + k − 2)
= multiple of 3
= P(k + 1)
This implies, P(k + 1) is true whenever P(k) is true. Hence, by the principle of mathematical
induction, P(n) is true for all natural numbers.
n
13. Use induction to prove that 5 n+1 + 4 × 6 when divided by 20 leaves a reminder 9, for all natural
numbers n.
Solution:
n
Let P(n) = 5 n+1 + 4 × 6 − 9
CASE N = 1
1
2
5 + 4 × 6 − 9 = 25 + 24 − 9
= 40
= divisible by 20
Hence P(1) is true.
CASE N = k
Assume P(k) is true. That is,
k
P(k) = 5 k+1 + 4 × 6 − 9 = 20p.
CASE N = k+1
104
k
5 k+2 + 4 × 6 k+1 − 9 = (5)5 k+1 + 24 × 6 − 9
k
k
= (4)5 k+1 + 5 k+1 + 4 × 6 + 20 × 6 − 9
k
k
= (20)5 + 5 k+1 + 4 × 6 − 9 + 20 × 6 k
= multiple of 20
This implies, P(k + 1) is true whenever P(k) is true. Hence, by the principle of mathematical
induction, P(n) is true for all natural numbers.
n
14. Use induction to prove that 10 + 3 × 4 n+2 + 5, is divisible by 9, for all natural numbers n.
n
Let P(n) = 10 + 3 × 4 n+2 + 5,
CASE N = 1
2
10 + 3 × 4 + 5 = 10 + 48 + 5
= divisible by 9
Hence P(1) is true.
CASE N = k
Assume P(k) is true. That is,
k
P(k) = 10 + 3 × 4 k+2 + 5 = 9p.
CASE N = k+1
k
10 k+1 + 3 × 4 k+3 + 5 = 10(10 ) + 3 × 4(4 k+2 ) + 5
k
k
= 9(10 ) + 10 + 3 × 3(4 k+2 ) + 3 × (4 k+2 ) + 5
This implies, P(k + 1)
k
k
= 9(10 ) + 9 × (4 k+2 ) + 10 + 3 × (4 k+2 ) + 5
= muliple of 9 + 9p
is true whenever P(k) is true. Hence, by the principle of mathematical induction, P(n) is true for
all natural numbers.
15. Prove that using the Mathematical induction
π 2π
sin(α) + sin α + + sin α + + · · ·
6 6
nπ
sin α + (n−1)π × sin
(n − 1)π 12
+ sin α + = 12 .
6 sin π
12
Solution:
π 2π (n − 1)π
Let P(n) = sin(α) + sin α + + sin α + + · · · + sin α +
6 6 6
CASE N = 1
105
π
sin (α) × sin
sin(α) = π 12
sin
12
LHS = RHS
Hence P(1) is true.
CASE N = k
Assume P(k) is true. That is,
π 2π (k − 1)π
P(k) = sin(α) + sin α + + sin α + + · · · + sin α +
6 6 6
CASE N = k+1
106
π 2π
sin(α) + sin α + + sin α +
6 6
(k)π π 2π
+ · · · + sin α + = sin(α) + sin α + + sin α + + · · ·
6 6 6
(k − 1)π (k)π
+ sin α + + sin α +
6 6
(k)π
= P(k) + sin α +
6
(k − 1)π kπ
sin α + × sin
12 12 (k)π
= π + sin α +
sin 6
12
(k − 1)π kπ π (k)π
sin α + × sin + sin × sin α +
12 12 12 6
= π
sin
12
I+II
= π
sin
12
(k − 1)π kπ
2I = sin α + × sin
12 12
(k − 1)π kπ (k − 1)π kπ
= cos α + − − cos α + +
12 12 12 12
π kπ π
= cos α − − cos α + −
12 6 12
π (k)π
2II = sin × sin α +
12 6
(k)π π (k)π π
= cos α + − − cos α + +
6 12 6 12
π (k)π π
cos α − − cos α + +
I+II 12 6 12
π π
=
sin 2 sin
12 12
(k)π π
sin α + sin (k + 1)
12 12
= π
sin
12
This implies, P(k + 1) is true whenever P(k) is true. Hence, by the principle of mathematical
induction, P(n) is true for all natural numbers.
4.1 Objective type of questions for the Chapter
Choose the best answer
1. The sum of the digits at the ten’s place of all numbers formed with the help of 2, 4, 5, 7 taken all
at a time is
(A) 432 (B) 108 (C) 36 (D) 18
4.1 Objective type of questions for the Chapter 107
Solution:
Given one number in unit’s place, the remaining numbers will appear in ten’s place 3! = 6 times.
Hence sum of digits = (n-1)!sum of numbers = 6(2 + 4 + 5 + 7) = 108.
2. In an examination there are three multiple choice questions and each question has 5 choices .
Number of ways in which a student can fail to get all answer correct is
(A) 125 (B) 124 (C) 64 (D) 63
Solution:
3
Number of ways = 5 = 125. Number of getting all answer correct = 1. Number of ways fail to
get all answers correct is 125 − 1 = 124.
3. The number of ways in which the following prize be given to a class of 30 boys first and second
in mathematics, first and second in physics, first in chemistry and first in English is
2
3
5
4
(A) 30 × 29 2 (B) 30 × 29 3 (C) 30 × 29 4 (D) 30 × 29 .
Number of ways of first and second in Mathematics = 30 × 29
Number of ways of first and second in Physics = 30 × 29
Number of ways of first in Chemistry = 30
Number of ways of first in English = 30
4
Total number of ways = 30 × 29 2
4. The number of 5 digit numbers all digits of which are odd is
(A) 25 (B) 5 5 (C) 5 6 (D) 625. 5
Unit place should be 1, 3, 5, 7, 9. Hence the total number = 5 .
5. In 3 fingers , the number of ways four rings can be worn is · · · · · · · · · ways.
3
(A) 4 − 1 (B) 3 4 (C) 68 (D) 64
Solution:
3
Number of ways = 4 = 64.
6. If (n+5) P (n+1) = ( 11(n−1) (n+3) P n , then the value of n are
)
2
(A) 7 and 11 (B) 6 and 7 (C) 2 and 11 (D) 2 and 6.
Solution:
108
(n + 5)!
(n+5) P (n+1) =
4!
(n + 5)(n + 4)(n + 3)!
=
2(2)(3!)
11(n − 1) (n + 3)! 11(n − 1)
(n+3) =
P n
2 3! 2
(n + 5)(n + 4)(n + 3)! (n + 3)! 11(n − 1)
=
2(2)(3!) 3! 2
(n + 5)(n + 4) = 22(n − 1)
2
n + 9n − 22n + 20 + 22 = 0
2
n − 13n + 42 = 0
n = 6 and 7
7. The product of r consecutive positive integers is divisible by
r
(A) r! (B)(r − 1)! (C) (r + 1)! (D) r .
8. The number of five digit telephone numbers having at least one of their digits repeated is
(A) 90000 (B) 10000 (C) 30240 (D) 69760.
2
2
9. If a −a C 2 = a −a C 4 then the value of ’a’ is
(A) 2 (B) 3 (C) 4 (D) 5
10. There are 10 points in a plane and 4 of them are collinear. The number of straight lines joining
any two points is
(A) 45 (B) 40 (C) 39 (D) 38.
11. The number of ways in which a host lady invite for a party of 8 out of 12 people of whom two do
not want to attend the party together is
11 10 11 10 12 10 10
(A) 2 × C 7 + C 8 (B) C 7 + C 8 (C) C 8 − C 6 (D) C 6 + 2!.
12. The number of parallelograms that can be formed from a set of four parallel lines intersecting
another set of three parallel lines.
(A) 6 (B) 9 (C) 12 (D) 18
13. Everybody in a room shakes hands with everybody else. The total number of shake hands is 66.
The number of persons in the room is · · · · · · · · · · · ·
(A) 11 (B) 12 (C) 10 (D) 6
14. Number of sides of a polygon having 44 diagonals is · · · · · ·
(A) 4 (B) 4! (C) 11 (D) 22
15. If 10 lines are drawn in a plane such that no two of them are parallel and no three are concurrent,
then the total number of point of intersection are
(A) 45 (B) 40 (C)10! (D) 2 10
16. In a plane there are 10 points are there out of which 4 points are collinear, then the number of
triangles formed is
10
(A) 110 (B) C 3 (C) 120 (D) 116
4.1 Objective type of questions for the Chapter 109
n
17. In 2n C 3 : C 3 = 11 : 1 then n is
(A) 5 (B) 6 (C)11 (D)7
18. (n−1) C r + (n−1) C (r−1) is
n
n
(A) (n+1) C r (B) (n−1) C r (C) C r (D) C r−1 .
19. The number of all five digit number which have at least one digit repeated is · · ·
5 10 4 9 5 9 5 10
(A) 10 − P 5 (B)9 × 10 − 9 × P 4 (C)10 − 9 × P 4 (D) 9 × 10 − P 5
110
20. The number of ways of choosing 5 cards out of a deck of 52 cards which include at least one king
is
52 48 52 48 52 48
(A) C 5 (B) C 5 (C) C 5 + C 5 (D) C 5 − C 5 .
21. The number of rectangles that a chessboard has · · ·
(A) 81 (B) 9 9 (C)1296 (D) 6561
22. The number of 10 digit number that can be written by using the digits 2 and 3 is
10 9 10 10
(A) C 2 + C 2 (B)2 (C)2 − 2 (D) 10!
r
23. If P r stands for P r then the sum of the series 1 + P 1 + 2P 2 + 3P 3 + · · · + nP n is
(A) P n+1 (B) P n+1 − 1 (C) P n−1 + 1 (D) (n+1) P (n−1)
24. Array is to be typed
n
n
n
25. If C 4 , C 5 , C 6 are in AP the value of n can be
(A)14 (B)11 (C)9 (D)5
26. 1 + 3 + 5 + 7 + · · · + 17 is equal to (A) 101 B) 81 (C) 71 (D) 61
27. 1 + 2 + 3 + 4 + 5 + · · · + 29 is equal to (A) 220 B) 321 (C) 435 (D) 523
Exercise - 5.1
2
1. Expand (i) 2x − 3 3
x
Solution:
3 2 3
3 3 3 3 27
2 2
2 3
6
2
2
3
2x − = (2x ) − 3 (2x ) + 3 (2x ) − = 8x − 36x + 54 +
x x x x x 3
√ 4 √ 4
2
2
(ii) 2x − 3 1 − x 2 + 2x + 3 1 + x 2 .
Solution:
√ 4 √ 4 √ √ 2
2 3
2 4
2 2
2
2
2x − 3 1 − x 2 + 2x + 3 1 + x 2 = (2x ) − 4 (2x ) 3 1 − x 2 + 6 (2x ) 3 1 − x 2
√ 3 √ 4
2 4
2 1
+4 (2x ) 3 1 − x 2 − 3 1 − x 2 + (2x )
√ √ 2
2 2
2 3
+4 (2x ) 3 1 − x 2 + 6 (2x ) 3 1 − x 2
√ 3 √ 4
2 1
+4 (2x ) 3 1 − x 2 − 3 1 − x 2
√ 4
4
8
2
= 2 (16x ) + (54)(4)x (1 − x ) + 3 1 − x 2
2 2
4
8
2
= 2 (16x ) + 216x (1 − x ) + 81 (1 − x )
4
4
7
2. Compute (i) 102 (ii) 99 (iii) 9 .
Solution:
4
3
4
3
4
4
2
2
(102) = (100 + 2) = (100) + 4(100) (2) + 6(100) (2) + 4(100)(2) + 2 = 108243216
4
4
3
2
(99) 4 = (100 − 1) = (100) − 4(100) + 6(100) − 4(100) + 1 = 96059601
5
6
7
3
4
9 7 = (10 − 1) 7 = (10) − 7(10) + 21(10) − 35(10) + 35(10) − 21(10) 2
+7(10) − 1 = 4782969
4.1 Objective type of questions for the Chapter 111
3. Using binomial theorem, indicate which of the following two number is larger: (1.01) 1000000 ,
10000.
Solution:
(1.01) 1000000 = (1 + 0.01) 1000000
2
= (1 + 1000000 C 1 (0.01) + 1000000 C 2 (0.01) + · · · + (0.01) 1000000 )
= 1 + 100000 + · · · + (0.01) 1000000
> 10000
2
15
4. Find the coefficient of x in x + 1 10 .
x 3
Solution:
10 r
1 1
2 10−r
2
x + General Term = T r+1 = n C r a n−r r 10 C r (x )
b =
x 3 x 3
To find r
x 20−2r−3r = x 15
20 − 5r = 15
5 = 5r
r = 1
15
T 2 = 10 C 1 x . Hence the coefficient of x 15 is 10.
6
1
6
2
2
5. Find the coefficient of x and the coefficient of x in x − .
x 3
Solution:
1 r
2 6−r
r
r 12−5r
r 12−2r−3r
b =
T r+1 = n C r a n−r r 6 C r (x ) (−1) = 6 C r (−1) x = 6 C r (−1) x
x 3
2
6
Since 12 − 5r = 6 has no integer solution, coefficient of x term is zero. Solving for x we get
6
2
12 − 5r = 2 ⇒ r = 2. Hence T 3 = C 2 (−1) = 15.
1 5
3 50
2
4
6. Find the coefficient of x in the expansion of (1 + x ) x + .
x
Solution:
r
3 r
2 5−r
1
T r+1 = n C r a n−r r 50 C r (x ) 5 C r (x )
b =
x
5 1
4
7
3
50
6
T r+1 = (1 + 50(x ) + 50(x ) + . . . ) x + 5x + 10x + 10x + +
x 2 x 5
5
3
4
6
= 1(10)x + 50(x )(10)x + 50(x ) + . . .
x 2
4
= (10 + 500 + 250)x + . . .
4
Hence coefficient of x is 760.
112
3
7. Find the constant term of 2x − 1 5 .
3x 2
Solution:
1 r 2 (5−r)
3 5−r
r
r 5
T r+1 = 5 C r (2x ) (−1) = (−1) C r x (15−5r)
3x 2 3 r
When r = 3, we obtain constant term.
2 2 −(10)4 −40
3 5
T 4 = (−1) C 3 = =
3 3 27 27
8. Find the last two digits of the number 3 600 .
Solution:
2 300
3 600 = (3 ) = (9) 300 = (10 − 1) 300
(10 − 1) 300 = 300 C 0 10 300 − 300 C 1 10 299 + 300 C 2 10 298 + · · · + 1
Except the last term all other terms are multiples of 100.Hence the last two digits are ”01”.
9. If n is a positive integer, show that, 9 n+1 − 8n − 9 is always divisible by 64.
Solution:
2
3
9 n+1 = (1 + 8) n+1 = n+1 C 0 1 + n+1 C 1 8 + n+1 C 2 8 + n+1 C 3 8 + · · · + 8 n+1
2
= 1 + 8n + 8 + 8 ( n+1 C 2 + n+1 C 3 8 + · · · + 8 n−1 )
9 n+1 − 8n − 9 = 64 ( n+1 C 2 + n+1 C 3 8 + · · · + 8 n−1 )
= 64k
We can also prove by Mathematical Induction.
2
Case 1: When n = 1, we have 9 − 8 − 9 = 81 − 17 = 64 and hence it is true for case 1.
Case 2: Let for n = k it is divisible by 64. That is, 9 k+1 − 8k − 9 = 64m.
Case 3: Consider for n = k + 1.
9 k+2 − 8(k + 1) − 9 = 9 k+1+1 − 8k − 9 − 8
= (9)(9 k+1 ) − 72k + 64k − 81 + 64
= (9) 9 k+1 − 8k − 9 + 64k + 64
= (9)(64m) + 64k + 64.
Hence by mathematical induction it is always divisible by 64.
10. If n is an odd positive integer, prove that the coefficients of the middle terms in the expansion of
n
(x + y) are equal.
Solution: Since n is an odd positive integer and the expansion has n + 1 terms, the middle terms
n + 1 n + 1
n
y .
are and + 1. The general term is given by T r+1 = C r x n−r r
2 2
4.1 Objective type of questions for the Chapter 113
n + 1
n
When r + 1 = we have Coefficient of T n+1 is C n−1 and
2 2 2
n + 1
n
when r + 1 = + 1 we have the coefficient of T n+1 is C n+1 .
2 2 +1 2
n
n
Since C r = C n−r the result follows.
r
11. If n is a positive integer and r is a nonnegative integer, prove that the coefficients of x and x n−r
n
in the expansion of (1 + x) are equal.
n
n
n
2
n
Solution: The expansion is (1 + x) = 1 + C 1 x + C 2 x + · · · + x . Hence the coefficients of
n
n
r
x and x n−r are C r and C n−r which are equal.
n
n
12. If a and b are distinct integers, prove that a − b is a factor of a − b , whenever n is a positive
n
n
integer. [Hint: write a = (a − b + b) and expand]
Solution:
n
n
n
a − b = (a − b + b) − b n
n
n
n
n
b + · · · + (a − b) C n−1 b
= (a − b) + C 1 (a − b) n−1 b + C 2 (a − b) n−2 2 n n−1 + b − b n
n
n
= (a − b) ((a − b) n−1 + C 1 (a − b) n−2 b + C 2 (a − b) n−3 2 n n−1 )
b + · · · + C n−1 b
n
13. In the binomial expansion of (a + b) , the coefficients of the fourth and thirteenth terms are equal
to each other, find n.
Solution:
Coefficient of T 4 = Coefficient of T 13
n n
C 3 = C 12
n! n!
=
(3)!(n − 3)! 12!(n − 12)!
(n − 3)! 12!
=
(n − 12)! 3!
(n−3)(n−4)(n−5)(n−6)(n−7)(n−8)(n−9)(n−10)(n−11)(n−12)! = 12×11×10×9×8×7×6×5×4×3!
(n−12)! 3!
(n − 3)(n − 4)(n − 5)(n − 6)(n − 7)(n − 8)(n − 9)(n − 10)(n − 11) = 12 × 11 × 10 × 9 × 8 × 7 × 6 × 5 × 4
n − 3 = 12 ⇒ n = 15.
n
14. If the coefficients of three consecutive terms in the expansion of (a+x) are in the ratio 1 : 7 : 42,
then find n.
Solution:
n n n
C r−1 : C r : C r+1 = 1 : 7 : 42
114
n
C r n − r + 1 7
= = ⇒ n − r + 1 = 7r (1)
n r 1
C r−1
n
C r+1 n − r 42
= = ⇒ n − r = 6r + 6 (2)
n r + 1 7
C r
Solving (1) and (2) we get r = 7 and n = 55.
n
15. In the binomial coefficients of (1 + x) , the coefficients of the fifth, sixth and seventh terms are in
AP Find all values of n.
Solution:
n
n
n
The coefficients of the fifth, sixth and seventh terms are C 4 , C 5 , C 6 are in A.P. Hence we have
n n n
2 C 5 = C 4 + C 6
n! n! n!
2 = +
5!(n − 5)! 4!(n − 4)! 6!(n − 6)!
1 30 + (n − 5)(n − 4)
2 =
5!(n − 5)! 6!(n − 4)!
2
12(n − 4) = n − 9n + 50
2
n − 21n + 98 = 0
The values of n are 7, 14
2
2
2
2n!
2
16. Prove that C + C + C + · · · + C = (n!) 2 .
2
n
1
0
Solution:
2 2 2 2
C + C + C + · · · + C = C 0 Cn + C 1 C n−1 + C 2 C n−2 + · · · + C n C 0
0 1 2 n
n o
= 2 C 0 Cn + C 1 C n−1 + C 2 C n−2 + · · · + C n C n
2 2
=
Exercise - 5.2
1. Write the first 6 terms of the sequences whose n-th terms are given below and classify them
as arithmetic progression, geometric progression, arithmetico-geometric progression, harmonic
progression and none of them.
4.1 Objective type of questions for the Chapter 115
1 1 1 1 1 1 1
(i) = , , , , , , . . . G.P
2 n+1 2 2 2 3 2 4 2 5 2 6 2 7
(n + 1)(n + 2) 2 × 3 3 × 4 4 × 5 5 × 6 6 × 7 7 × 8
(ii) = , , , , , , . . . None of the given
n + 3(n + 4) 4 × 5 5 × 6 6 × 7 7 × 8 8 × 9 9 × 10
n 2 3 4 5 6
1 1 1 1 1 1 1
(iii) 4 = 4 , 4 , 4 , 4 , 4 , 4 , . . . G.P
2 2 2 2 2 2 2
(−1) n 1 1 1 1 1
(iv) = −1, , − , , − , None of them
n 2 3 4 5 6
2n + 3 5 7 9 11 13 15
(v) = , , , , ,
3n + 4 7 10 13 16 19 22
(vi) 2018 = 2018, 2018, 2018, 2018, 2018, 2018, . . . Both A.P, G.P hence A.G.P
3n − 2 1 4 7 10 13 16
(vii) = , , , , A.G.P
3 n−1 1 3 3 2 3 3 4 3 5
3
2. Write the first 6 terms of the sequences whose n-th term a n is given below.
Solution:
n + 1 if n is odd
(i) a n = 2, 2, 4, 4, 6, 6, 8, 8, . . .
n if n is even
1 if n = 1
(ii) a n = 2 if n = 2 1, 2, 3, 5, 8, 13, 21, 34, . . .
a n−1 + a n−2 if n > 2
n if n is 1, 2 or 3
(iii) a n = 1, 2, 3, 6, 9, 18, 33, . . .
a n−1 + a n−2 + a n−3 if n > 3
3. Write the n-th term of the following sequences.
Solution:
n + 1 if n is odd
i. 2, 2, 4, 4, 6, 6, . . . a n =
n if n is even.
1 2 3 4 5 n
ii. , , , , , . . . a n =
2 3 4 5 6 n + 1
1 3 5 7 9 2n − 1
iii. , , , , , . . . a n =
2 4 6 8 10 n
6, 4, 2, 0, . . . 7 − n if n is odd.
iv. 6, 10, 4, 12, 2, 14, 0, 16, −2, . . . a n =
10, 12, 14, 16, . . . 8 + n if n is even.
4. The product of three increasing numbers in GP is 5832. If we add 6 to the second number and 9 to
the third number, then resulting numbers form an AP. Find the numbers in GP.
Solution:
a
3
Let , a, ar are in G.P. The product is a = 5832 ⇒ a = 18.
r
18
, 24, 18r + 9 form A.P.
r
116
18
48 = + 18r + 9
r
18
39 = + 18r
r
2
0 = 18r − 39r + 18
2
0 = 6r − 13r + 6
3 2
Solving we get r = or
2 3
2 3
The Numbers in G.P are 27, 18, 12 when r = and 12, 18, 27 when r = .
3 2
3 5 7
5. Find the n-th term of the sequence , , , . . . .
2 2
2 2
2 2
1 2 2 3 3 4
Solution:
2
2
2k + 1 2k + 1 + k − k 2 (k + 1) − k 2 1 1
= = = −
2
2
2
k (k + 1) 2 k (k + 1) 2 k (k + 1) 2 k 2 (k + 1) 2
1 1
Hence T n = −
n 2 (n + 1) 2
6. If t k is the k-th term of a GP, then show that t n−k , t n , t n+k also forms a GP for any positive integer
k.
Solution:
2
n 2
t n−k t n+k = ar n−k−1 ar n+k−1 = (ar ) = (t n ) Hence they form a G.P
1 1 1
7. If a, b, c are in geometric progression, and if a x = b = c z , then prove that x, y, z are in arithmetic
y
progression.
Solution:
1 1 1
x y z x y z
Let a = b = c = k Then we have a = k , b = k , c = k . Given that a, b, c are in
2
x
y
GP. Hence b = ac. That is, k 2y = (k )(k ) = k x+z . So we have 2y = x + z. Thus, x, y, z are in
arithmetic progression.
8. The AM of two numbers exceeds their GM by 10 and HM by 16. Find the numbers.
Solution:
Let the numbers be a, b.
4.1 Objective type of questions for the Chapter 117
a + b √ 2ab
A.M = G.M = ab H.M =
2 a + b
A.M - G.M = 10 A.M - H.M = 16 G.M - H.M = 6
a + b √ a + b 2ab √ 2ab
− ab = 10 − = 16 ab − = 6
2 2 a + b a + b
√ √ 2ab
20 + 2 ab = a + b ab − √ = 6
√ 20 + 2 ab √
20 ab + 2ab − 2ab = 120 + 12 ab
√
8 ab = 120
√
ab = 15
20 + 30 = a + b
50 = a + b
p
2
(a + b) − 4ab = a − b
40 = a − b
Solving we get a = 45 and b = 5. The numbers are 5, 45.
2
9. If the roots of the equation (q − r)x + (r − p)x + p − q = 0 are equal, then show that p, q and r
are in AP.
Solution:
Since the roots are equal, the discriminant is zero.
2
(r − p) − 4(q − r)(p − q) = 0
2
(r − p) = 4(q − r)(p − q)
2
2
2
r + p − 2pr = 4pq − 4q − 4pr + 4qr
2
2
2
r + p + 4q − 4pq + 2pr + 4qr = 0
2
(p + r − 2q) = 0
p + r = 2q
Hence p, q and r are in AP.
th
10. If a, b, c are respectively the p , q th and r th terms of a GP, show that
(q − r) log a + (r − p) log b + (p − q) log c = 0.
Solution: Let the first term be x and the common ratio be y.
118
xy p−1 = a
log xy p−1 = log a
log x + log y p−1 = log a
log x + (p − 1) log y = log a (1)
(
xy q − 1) = b
log x + (q − 1) log y = log b (2)
(
xy r − 1) = c
log x + (r − 1) log y = log c (3)
From (1), (2), (3) the expression becomes
(q − r)log x + (p − 1) log y + (r − p)log x + (q − 1) log y + (p − q)log x + (r − 1) log y
= (q − r + r − p + p − q) log x + (q − r)(p − 1) + (r − p)(q − 1) + (p − q)(r − 1) log y
= 0 + pq − pr − q + r + qr − pq − r + p + pr − qr − p + q log y
= 0
Exercise - 5.3
1. Find the sum of the first 20-terms of the arithmetic progression having the sum of first 10 terms as
52 and the sum of the first 15 terms as 77.
Solution:
S 10 = 52 S 15 = 77
15
5(2a + 9d) = 52 (2a + 14d) = 77
2
10a + 45d = 52 15a + 105d = 77
133 2
Solving we get a = d = −
25 75
266 2 304
Hence S 20 = 10(2a + 19d) = 10 − 19 = .
25 75 3
2. Find the sum up to the 17-th term of the series
3
3
3
1 3 1 + 2 3 1 + 2 + 3 3
+ + + · · · .
1 1 + 3 1 + 3 + 5
Solution:
2
n(n + 1)
P 3 2 2 2
n 2 n (n + 1) (n + 1)
= = =
T n = P
(2n + 1) n 2 4n 2 4
4.1 Objective type of questions for the Chapter 119
17
1 17 1 17 17 17 1 n(n + 1)(2n + 1) n(n + 1)
2
2
Hence Σ (n + 1) = Σ n + Σ (2n) + Σ (1) = + 2 + n
4 n=1 4 n=1 n=1 n=1 4 6 2
1
1 17 × 18 × 35 2 × 17 × 18
= + + 17
4 6 2
17
= [105 + 18 + 1]
4
= 527
3. Compute the sum of first n terms of the following series:
i. 8 + 88 + 888 + 8888 + · · · ii. 6 + 66 + 666 + 6666 + · · ·
S n = 8 + 88 + 888 + 8888 + . . .
= 8[1 + 11 + 111 + 1111 + . . . ]
8
= [9 + 99 + 999 + . . . ]
9
8
3
2
= [(10 − 1) + (10 − 1) + (10 − 1) + . . . ]
9
8
3
2
= [(10 + 10 + 10 + . . . n terms) − n]
9
n
8 10(10 − 1)
= − n
9 9
8
n
= [10(10 − 1) − 9n]
81
S n = 6 + 66 + 666 + . . .
= 6[1 + 11 + 1111 + . . . ]
6
= [9 + 99 + 999 + . . . ]
9
6
2
3
= [(10 − 1) + (10 − 1) + (10 − 1) + . . . ]
9
6
2
3
= [(10 + 10 + 10 + . . . n terms) − n]
9
n
6 10(10 − 1)
= − n
9 9
2
n
= [10(10 − 1) − 9n]
27
3
2
2
4. Compute the sum of first n terms of 1 + (1 + 4) + (1 + 4 + 4 ) + (1 + 4 + 4 + 4 ) + · · · .
Solution:
Since each term is in GP of n terms, we have,
n
n
n
4 − 1 1 1 4(4 − 1) 4(4 − 1) n
n
T n = and S n = [Σ4 − Σ1] = − n = −
3 3 3 3 9 3
4 7 10
5. Find the general term and sum to n terms of the sequence 1, , , , . . . .
3 9 27
120
Solution:
3n − 2 n 3n − 2 n k n 1
General term T n = and sum is, S n = Σ = Σ − 2 Σ
3 n−1 n=1 3 n−1 k=1 3 k−1 k=1 3 k−1
√ √ √ √
6. Find the value of n, if the sum to n terms of the series 3 + 75 + 243 + · · · is 435 3.
Solution:
√ √ √ n √
3 + 75 + 243 + · · · = Σ (4k − 3) 3
k=1
√ n √
435 3 = Σ (4k − 3) 3
k=1
n
435 = Σ (4k − 3)
k=1
n
= 4 Σ k − 3n
k=1
n(n + 1)
= 4 − 3n
2
= 2n(n + 1) − 3n
2
0 = 2n − n − 435
Solving we get n = 15 or negative value which is not possible. Hence n = 15.
th
th
th
7. Show that the sum of (m + n) and (m − n) term of an AP. is equal to twice the m term.
Solution:
T m+n + T m−n = a + (m + n − 1)d + a + (m − n − 1)d = 2a + (2m − 2)d
= 2 (a + (m − 1) d)
= 2T m
8. A man repays a loan of Rs.3250 by paying Rs.20 in the first month and then increases the payment
by Rs.15 per month. How long will it take him to clear the loan?
Solution:
d = 15, S n = 3250 and a = 20
n
S n = [2a + (n − 1)d]
2
n
3250 = [40 + (n − 1)15]
2
6500 = 40n + 15n(n − 1)
2
0 = 15n + 25n − 6500
2
0 = 3n + 5n − 1300
5 ± 135
Solving, we obtain , n = = 70, or negative value. The time to clear the loan is 70 months
2
or 5 years and 10 months.
4.1 Objective type of questions for the Chapter 121
9. In a race, 20 balls are placed in a line at intervals of 4 meters with the first ball, 24 meters away
from the starting point. A contestant is required to bring the balls back to the starting place one at
a time. How far would the contestant run to bring back all balls?
Solution: Here, d = 4, n = 20, a = 24.
S 20 = 10{48 + 19(4)} = 10{48 + 76} = 1240
Since he has to bring back each time,total distance he run = 2480 m.
10. The number of bacteria in a certain culture doubles every hour. If there were 30 bacteria present
in the culture originally, how many bacteria will be present at the end of 2 nd hour, 4 th hour and
n th hour?
Solution:a = 30, r = 2.
4
2
At the end of 2 nd hour, S 2 = 30(2 − 1) = 90 and at the end of 4 th hour, S 4 = 30(2 − 1) = 450.
2 − 1 2 − 1
11. What will |.500 amounts to in 10 years after its deposit in a bank which pays annual interest rate
of 10% compounded annually?
Solution:
n
Amount = P(1 + r) = 500(1 + 0.1) 10 = 500(1.1) 10 = |.1296.87
12. In a certain town, a viral disease caused severe health hazards upon its people disturbing their
normal life. It was found that on each day, the virus which caused the disease spread in Geometric
Progression. The amount of infectious virus particle gets doubled each day, being 5 particles on
the first day. Find the day when the infectious virus particles just grow over 1,50,000 units?
Solution:a = 5, r = 2, S n > 1, 50, 000
S n > 1, 50, 000
n
5(2 − 1) > 1, 50, 000
n
2 − 1 > 30, 000
15
Since 2 − 1 = 32767, the infectious virus particles just grow over 1, 50, 000 units on 15 th day.
Exercise - 5.4
1. Expand the following in powers of x and find the condition on x for which the binomial expansion
is valid.
122
1 1
(i) =
5 + x 5 1 + x
5
1 x −1
= 1 +
5 5
x
x
1 2
= 1 − + + . . .
5 5 5
< 1 ⇒ |x| < 5
x
5
2 2
(ii) =
(3 + 4x) 2 4x 2
9 1 +
3
−2
2 4x
= 1 +
9 3
" #
2
2 4x 4x
= 1 − 2 + 3 + . . .
9 3 3
2 8x 16 256
2
3
= 1 − + x − x + . . .
9 3 3 27
4x 3
< 1 ⇒ |x| <
3 4
2
2 2 x 2 3
2
(iii) (5 + x ) 3 = 5 3 1 +
5
" #
2 2 2
2 2 x 2 − 1 x 2
= 5 3 1 + + 3 3
3 5 2 5
" #
2 3
2 2 x 2 1 2 1 x 2 1 2 1 4 x 2
= 5 3 1 + + − + − − + . . .
3 5 2 3 3 5 6 3 3 3 5
2 2 1 4
2
2 2
2 3
= 5 3 1 + x − (x ) + (x ) + . . .
15 225 10125
2
x
2
< 1 ⇒ x < 5
5
2 1
(iv) (x + 2) − 3 =
2
(x + 2) 3
1
=
2 x
2
2 3 1 + 3
2 2
2 h x i − 3
−
= 2 3 1 +
2
" #
x
x
2 2 x 2 5 2 2 5 8 3
−
= 2 3 1 − + 3 3 − 3 3 3 + . . .
3 2 2 2 2 2
2 x 5 5
−
3
2
= 2 3 1 − + x − x + . . .
3 36 81
< 1 ⇒ |x| < 2
x
2
4.1 Objective type of questions for the Chapter 123
√
2. Find 3 1001 approximately.
Solution:
1
√ 1 1 1 1 3
3 3
1001 = (1001) 3 = (1000 + 1) 3 = (10 + 1) 3 = 10 1 +
10 3
" 1 2 #
1 −
2
= 10 1 + (0.001) + 3 3 (0.001) + . . .
3 2
= 10 [1 + 0.333(0.001) + . . . ]
= 10[1.00033] ≈ 10.00333 ≈ 10.003 approximately.
√ √ 1
3
3
3. Prove that 3 x + 6 − 3 x + 3 is approximately equal to when x is large.
x 2
Solution:
1 1
√ √ 6 3 3 3
x + 6 − x + 3 = x 1 + − x 1 +
3 3 3 3
x 3 x 3
1 6 1 3
≈ x 1 + − x 1 +
3 x 3 3 x 3
2 1
≈ x + − x −
x 2 x 2
1
≈
x 2
r 2
1 − x x
4. Prove that is approximately equal to 1 − x + when x is very small.
1 + x 2
Solution:
r
1 − x 1 − 1
= (1 − x) 2 (1 + x) 2
1 + x
2 2
1 1 1 x 1 1 1 x
≈ 1 − x + − 1 1 − x + − − + 1
2 2 2 2 2 2 2 2
1 1 1 x 1 1 1 x
2 2
≈ 1 − x + − 1 − x + −
2 2 2 2 2 2 2 2
2 2
1 1 1 x 1 1 1 1 x
≈ 1 − x + − − x 1 − x + −
2 2 2 2 2 2 2 2 2
2 2 2
x x x x x
≈ 1 − − − + −
2 8 2 4 8
1
x
5x
5. Write the first 6 terms of the exponential series (i.) e , (ii.) e −2x , (iii.) e 2 .
Solution:
124
5x (5x) 2 (5x) 3 (5x) 4 (5x) 5 (5x) 6
e 5x = 1 + + + + + + + . . .
1! 2! 3! 4! 5! 6!
25(x) 2 125(x) 3 625(x) 4 625(x) 5 3125(x) 6
= 1 + 5x + + + + + + . . .
2 6 24 24 144
2x (2x) 2 (2x) 3 (2x) 4 (2x) 5 (2x) 6
e −2x = 1 − + − + − + + . . .
1! 2! 3! 4! 5! 6!
4x 3 2(x) 4 4(x) 5 4(x) 6
2
= 1 − 2x + 2x − + − + + . . .
3 3 15 45
1 1 2 1 3 1 4 1 5 1 6
1 x x x x x x x
e 2 = 1 + 2 + 2 + 2 + 2 + 2 + 2 + . . .
1! 2! 3! 4! 5! 6!
x x 2 x 3 x 4 x 5
= 1 + + + + + + . . .
2 8 48 384 3840
6. Write the first 4 terms of the logarithmic series (i.) log(1+4x), (ii.) log(1−2x), (iii.) log 1+3x ,
1−3x
(iv.) log 1−2x . Find the intervals on which the expansions are valid.
1+2x
Solution:
4.1 Objective type of questions for the Chapter 125
(4x) 2 (4x) 3 (4x) 4
(i) log(1 + 4x) = 4x − + − + . . .
2 3 4
64 1024 1
3
5
2
4
= 4x − 8x + x − 64x + x + . . . for |x| <
3 5 4
2 3 4
(2x) (2x) (2x)
(ii) log(1 − 2x) = −2x − − − − . . .
2 3 4
8x 3 1
2
4
= −1{2x + 2x + + 4x + . . . } for |x| <
3 2
1 + 3x
(iii) log = log(1 + 3x) − log(1 − 3x)
1 − 3x
( )
2 3 4 5 6 7
(3x) (3x) (3x) (3x) (3x) (3x)
= 3x − + − + − + + . . .
2 3 4 5 6 7
( 2 3 4 5 6 7 )
(3x) (3x) − (3x) (3x) − (3x) (3x)
− −3x − − − − − − + . . .
2 3 4 5 6 7
2 3 4 5 6 7
9x 27x 81x 243x 729x 2187x
= 3x − + − + − + + . . .
2 3 4 5 6 7
2 3 4 5 6 7
9x 27x 81x 243x 729x 2187x
− −3x − − − − − − + . . .
2 3 4 5 6 7
243 2187x 1
7
3
5
= 2 3x + 9x + x + . . . for |x| <
5 7 3
1 − 2x
(iv) log = log(1 − 2x) − log(1 + 2x)
1 + 2x
( 2 3 4 5 6 7 8 )
(2x) (2x) (2x) (2x) (2x) (2x) (2x)
= −2x − − − − − − − − . . .
2 3 4 5 6 7 8
( )
2 3 4 5 6 7 8
(2x) (2x) (2x) (2x) (2x) (2x) (2x)
− 2x − + − + − + − + . . .
2 3 4 5 6 7 8
8 32 32 128
8
6
4
2
7
2
5
= −2x − x − x − 4x − x − x − x − 128x + . . .
3 5 3 7
8 32 32 128
2
4
5
6
2
7
8
+ −2x + x − x + 4x − x + x − x + 128x + . . .
3 5 3 7
16 64 256 1
5
7
3
= −4x − x − x − x − . . . for |x| <
3 5 7 2
x 2 x 3 x 4 y 2 y 3 y 4
7. If y = x + + + + · · · , then show that x = y − + − + · · · .
2 3 4 2! 3! 4!
Solution:
x 2 x 3 x 4
y = x + + + + · · · = − log(1 − x). Hence we have
2 3 4
y 2 y 3 y 4
y
e = (1 − x) −1 ⇒ x = 1 − e −y = y − + − + · · ·
2! 3! 4!
q
q (n+1)p+(n−1)q
8. If p − q is small compared to either p or q, then show that n p = . Hence find 8 15 .
q (n−1)p+(n+1)q 16
p
Solution: Since p − q is small,1 −
q
126
4
9. Find the coefficient of x in the expansion of 3−4x+x 2 .
e 2x
Solution:
3 − 4x + x 2 4x 3 2(x) 4
2
2
2
= (3 − 4x + x )(e −2x ) = (3 − 4x + x ) 1 − 2x + 2x − + + . . .
e 2x 3 3
3(2) 4 4(2) 3 2 2 16 28
4
Coefficient of x = + + = 2 + + 2 =
1! 3! 2! 3 3
∞
1
1
10. Find the value of Σ 1 n−1 + 2n−1 .
n=1 2n−1 9 9
Solution:
∞ 1 1 1 1 1 1 1 1 1 1 1
Σ + = 1 + + + + + + . . .
n=1 2n − 1 9 n−1 9 2n−1 1 9 3 9 9 3 5 9 2 9 5
( ) ( )
2 4 3 5
1 1 1 1 1 1 1 1 1 1 1
= + + + . . . + + + + . . .
1 3 3 5 3 1 9 3 9 5 9
( ) ( )
3 5 3 5
1 1 1 1 1 1 1 1 1 1
= 3 + + + . . . + + + + . . .
3 3 3 5 3 9 3 9 5 9
3 1 + 1 1 1 + 1
= log 3 + log 9
2 1 − 1 2 1 − 1
3 9
3 1 5
= log 2 + log
2 2 4
1 1 5
= log 8 + log
2 2 4
1
= log 10
e
2
11. After striking the floor a certain ball rebounds 4 th of the height from which it has fallen. Find
5
the total distance that it travels before coming to rest, if it is gently dropped from a height of 120
meters.
Objective Type Questions
Choose the correct answer.
1. The value of 2 + 4 + 6 + · · · + 2n is
a. n(n−1) b. n(n+1) c. 2n(2n+1) d. n(n + 1)
2 2 2
n(n + 1)
2 + 4 + 6 + · · · + 2n = 2 (1 + 2 + 3 + · · · + n) = 2 = n(n + 1)
2
6
2. The coefficient of x in (2 + 2x) 10 is
10
a. 10 C 6 b. 2 6 c. 10 C 6 d. 10 C 6 2 .
10
Given (2 + 2x) 10 = 2 (1 + x) 10
= 10 C r (10) r
General Term is T r+1
Coefficient of x 6 = 10 C 6
= 10 C 6 2 10
4.1 Objective type of questions for the Chapter 127
8 12
20
3. The coefficient of x y in the expansion of (2x + 3y) is
8 12
8 12
12 8
8 12
a. 0 b. 2 3 c. 2 3 + 2 3 d. 20 C 8 2 3 .
General Term is T r+1 = 20 C r (2x) 20−r (3y) r
8 12
To find r x y 20 − r = 18 ⇒ r = 12
12
8
8 12
Coefficient of x y = 20 C 8 (2 )(3 )
n
n
4. If C 10 > C r for all possible r, then a value of n is
a. 10 b. 21 c. 19 d. 20.
n
n
The greatest coefficient is C n . Hence = 10 ⇒ n = 20.
2 2
5. If a is the arithmetic mean and g is the geometric mean of two numbers, then
a. a ≤ g b. a ≥ g c. a = g d. a > g.
a > g
2 2
2
n
6. If (1 + x ) (1 + x) = a 0 + a 1 x + a 2 x + · · · + x n+4 and if a 0 , a 1 , a 2 are in AP, then n is
a. 1 b. 2 c. 3 d. 4.
2 2 n 2 n+4
(1 + x ) (1 + x) = a 0 + a 1 x + a 2 x + · · · + x
n(n − 1)
4
4
n
2
2
2
(1 + 2x + x ) (1 + x) = (1 + 2x + x ) 1 + nx + x + · · ·
2
n(n − 1)
2
= 1 + nx + 2 + x + . . .
2
n(n − 1)
Here a 0 = 1, a 1 = n, a 2 = 2 + .
2
2a 1 = a 0 + a 2
n(n − 1)
2n = 1 + 2 +
2
6 + n(n − 1)
=
2
2
4n = 6 + n − n
2
0 = n − 5n + 6
n = 2 or 3
7. If a, 8, b are in AP, a, 4, b are in GP, and if a, x, b are in HP then x is
a. 2 b. 1 c. 4 d. 16.
128
Using Property of AP a + b = 16 (1)
Using Property of GP ab = 16 (2)
Using above (1) & (2) a + b = ab (3)
2ab
Using Property of HP x =
a + b
Using (3) x = 2
1 1 1
8. The sequence √ , √ √ , √ √ , · · · form an
3 3 + 2 3 + 2 2
a. AP b. GP c. HP d. AGP.
Since Numerator is constant and Denominator is in AP, the series is HP. We can also verify using
the property of HP. Let the terms be a, b, c.
1 1
2 √ √ √
2ac 3 3 + 2 2 2 1
= = √ √ √ = √ √ = b
a + c 1 1 3 + 2 2 + 3 3 + 2
√ + √ √
3 3 + 2 2
9. The HM of two positive numbers whose AM and GM are 16, 8 respectively is
a. 10 b. 6 c. 5 d. 4.
a + b
AM = = 16 ⇒ a + b = 32
√ 2
GM = ab = 8 ⇒ ab = 64
2ab 2(64)
HM = = ⇒ HM = 4
a + b 32
10. If S n denotes the sum of n terms of an AP whose common difference is d, the value of
S n − 2S n−1 + S n−2 is
2
a. d b. 2d c. 4d d. d .
S n − 2S n−1 + S n−2 = [S n − S n−1 ] − [S n−1 − S n−2 ] = t n − t n−1 = d
11. The remainder when 38 15 is divided by 13 is
a. 12 b. 1 c. 11 d. 5.
15
15
38 ÷ 13 = (26 + 12) ÷ 13 = 13k + 12. Hence the remainder is 12.
th
12. The n term of the sequence 1, 2, 4, 7, 11, · · · is
2
3
3
2
2
a. n + 3n + 2n b. n − 3n + 3n c. n(n+1)(n+2) d. n −n+2 .
3 2
Method 1:
2 − 1, 4 − 2, 7 − 4, 11 − 7, . . . gives a new series 1, 2, 3, 4, . . . which is in AP. In other words, First
common difference is 1 then 2 then 3 and so on. We obtain a new series. The second common
4.1 Objective type of questions for the Chapter 129
difference is 1. As there is a constant second difference we know the n th term follows the equation
2
an + bn + c.By putting values 1, 2, 3, in the equation, we obtain,
T 1 = a + b + c
T 2 = 4a + 2b + c
T 3 = 9a + 3b + c
1 1 n(n − 1)
Solving we get a = , b = − , c = 1. Hence T n = + 1.
2 2 2
Method 2:
By cutting a circle into pieces by drawing non-parallel lines also generate this series. Given a
circle that is any size, I have to ”cut” the square n times and see how many pieces I can obtain.
I have to obtain as many pieces as possible. For example, a circle can be cut into two with one
straight cut. That same circle can be cut into four pieces with two straight cuts. However, I can
obtain seven pieces with three straight cuts. So I could obtain the series as shown in below figure.
Figure 5.3
This is also called as Central polygonal numbers (the Lazy Caterer’s sequence) or, maximal
number of pieces formed when slicing a pancake with n cuts.
Method 3:
The formula for the sum of a general quadratic sequence is:
2
n(cn + 2c + 3nd 1 + 6s − 3cn − 3d 1 )
S 2 (n, s, d 1 , c) = where n is the number of terms to be
6
summed, s is the starting term of the series, d 1 is the first difference (subtracting the first term from
the second term) and c is the constant difference between the differences.Here s = 1, d 1 = 1, c = 1.
To find the n th term, T n = S n − S n−1 = S 2 (n, 1, 1, 1) − S 2 (n − 1, 1, 1, 1) = n(n − 1) + 1.
2
130
Method 4:
2 = 1 + (1)
4 = 1 + (1 + 2)
7 = 1 + (1 + 2 + 3)
11 = 1 + (1 + 2 + 3 + 4)
Hence T n = 1 + (1 + 2 + 3 + ...n).
Method 5:
It is the number of Dyck n-paths all of whose ascents and descents have lengths equal to 1(mod10).
1 1 1
13. The sum up to n terms of the series √ √ + √ √ + √ √ + · · · is
1 + 3 3 + 5 5 + 7
√
√ 2n + 1 √ √
a. 2n + 1 b. c. 2n + 1 − 1 d. 2n+1−1 .
2 2
√ √ √ √
1 1 2n − 1 − 2n + 1 2n − 1 − 2n + 1
T n = √ √ = √ √ × √ √ =
(2n + 1) − (2n − 1)
2n − 1 + 2n + 1 2n − 1 + 2n + 1 2n − 1 − 2n + 1 √ √
2n − 1 − 2n + 1
=
2
√ √
2T 1 = 3 − 1
√ √
2T 2 = 5 − 3
√ √
2T 3 = 7 − 5
· · · · · · · · · · · · · · · · · · · · ·
√ √
2Σ = 2n + 1 − 1
√ √
2n + 1 − 1
Σ =
2
1 3 7 15
th
14. The n term of the sequence , , , , · · · is
2 4 8 16
n
a. 2 − n − 1 b. 1 − 2 −n c. 2 −n + n − 1 d. 2 n−1 .
n
2 − 1
T n = = 1 − 2 −n
2 n
√ √ √ √
15. The sum up to n terms of the series 2 + 8 + 18 + 32 + · · · is
n(n+1) n(n+1)
a. b. 2n(n + 1) c. d. 1.
2 2
√ √ √
The sequence is 1 2 + 2 2 + 3 2 + . . . Hence,
√ √ n(n + 1) n(n + 1)
S n = 2 (1 + 2 + 3 + · · · + n) = 2 = √
2 2
7
16. The value of the series 1 + + 13 + 19 + · · · is
2 4 8 16
4.1 Objective type of questions for the Chapter 131
a. 14 b. 7 c. 4 d. 6.
The numerator is in A.P and the denominator is in G.P. Hence it is Arithmetico Geometric
1
Progression. Here a = 1, d = 6, r =
2
a dr 1 6 1
S ∞ = + = + 2 = 2 + 3(4) = 14
1 − r (1 − r) 2 1 − 1 1 − 1 2
2 2
17. The sum of an infinite GP is 18. If the first term is 6, the common ratio is
a. 1 b. 2 c. 1 d. 3 .
3 3 6 4
a
S ∞ = = 18
1 − r
6
= 18
1 − r
3(1 − r) = 1
2
r =
3
5
18. The coefficient of x in the series e −2x is
a. 2 b. 3 c. −4 d. 4 .
3 2 15 15
4x 3 2(x) 4 4(x) 5 4(x) 6
2
e −2x = 1 − 2x + 2x − + − + + . . .
3 3 15 45
4
5
Coefficient of x is − .
15
1 1 1
19. The value of + + + · · · is
2! 4! 6!
2
2
a. e +1 b. (e+1) 2 c. (e−1) 2 d. e +1 .
2e 2e 2e 2e
2
1 1 1 e + e −1 e + 1 − 2e (e − 1) 2
+ + + · · · = − 1 = =
2! 4! 6! 2 2e 2e
2 3
2 1 2 1 2
20. The value of 1 − 1 + − + · · · is
2 3 3 3 4 3
5 3 5 5 5 2 2
a. log b. log c. log d. log .
3 2 3 3 3 3 3
2 3
2 1 2 1 2 2 3 5
1 − 1 + − + · · · = log 1 + = log
2 3 3 3 4 3 3 2 3
Exercise - 6.1
1. Find the locus of P, if for all values of α, the co-ordinates of a moving point P is
Solution:
132
2
2
(i) (9 cos α, 9 sin α) x + y = 9 circle
x 2 y 2
(ii) (9 cos α, 6 sin α) + = 1 ellipse
81 36
2. Find the locus of a point P that moves at a constant distant of (i) two units from the x-axis (ii)
three units from the y-axis.
Solution:
(i) two units from the x-axis |y| = 2
(ii) three units from the y-axis |x| = 3
3. If θ is a parameter, find the equation of the locus of a moving point, whose coordinates are
3
3
x = a cos θ, y = a sin θ.
Solution:
2 2
x 3 y 3
2 2 2
2
2
Since = cos θ, and = sin θ, we have x 3 + y 3 = a 3 .
a a
2
4. Find the value of k and b, if the points P(-3,1) and Q(2,b) lie on the locus of x − 5x + ky = 0.
Solution:
2
2
Since (-3,1) lies on x −5x+ky = 0, we obtain k = −24. Similarly,(2, b) lies on x −5x+ky = 0,
1
we obtain 4 − 10 − 24b = 0 ⇒ b = −
4
5. A straight rod of length 8 units slides with its ends A and B always on the x and y axes respectively.
Find the locus of the mid point of the line segmentAB
Solution:
Let AB be the rod of length 8 units.Let the points A and B be (a, 0) and (0, b) respectively. Let P
a b
be the midpoint of AB. Then the coordinates of P(x 1 , y 1 ) is , Then, a = 2x 1 , b = 2y 1 .
2 2
2
2
OA + OB = AB 2
2
2
a + b = 64
2
2
4x + 4y = 64
1
1
2
2
x + y = 16
1 1
2
2
Hence the locus is a circle given by x + y = 16
6. Find the equation of the locus of a point such that the sum of the squares of the distance from the
points (3,5), (1,-1) is equal to 20
Solution:
Let the point be (x, y).
4.1 Objective type of questions for the Chapter 133
2
2
2
2
(x − 3) + (y − 5) + (x − 1) + (y + 1) = 20
2
2
2
2
x + y − 6x − 10y + 34 + x + y − 4x + 2y + 2 = 20
2
2
2x + 2y − 10x − 8y + 16 = 0
2
2
4x + 4y − 20x − 16y + 32 = 0
2
2
{(2x) − 20x + 25 + (2y) − 16y + 16} − 25 − 16 + 32 = 0
2 2
(2x − 5) + (2y − 4) = 9
7. Find the equation of the locus of the point P such that the line segment AB, joining the points
A(1,-6) and B(4,-2), subtends a right angle at P.
Solution:Let the point P be (x, y).
2 2 2
(AP) + (BP) = (AB)
2
2
2
2
(x − 1) + (y + 6) + (x − 4) + (y + 2) = 25
2
2
2
2
x − 2x + 1 + y + 12y + 36 + x − 8x + 16 + y + 4y + 4 = 25
2
2
2x + 2y − 10x + 16y + 16 = 0
2
2
x + y − 5x + 8y + 8 = 0
2
8. If O is origin and R is a variable point on y = 4x, then find the equation of the locus of the
mid-point of the line segment OR.
Solution:Let R be (a, b). The mid-point of OR is P(h, k).
a b
But the mid-point of OR is , .
2 2
a b
Hence h = ⇒ a = 2h.Also k = ⇒ b = 2k.
2 2
2
2
2
2
Since R lies on y = 4x, b = 4a ⇒ 4k = 8h ⇒ k = 2h.
2
Hence the locus is y = 2x.
a b
9. The coordinates of a moving point P are cosecθ + sin θ, (cosecθ − sin θ) , where θ is a
2 2
2 2
2 2
2 2
variable parameter. Show that the equation of the locus P is b x − a y = a b .
Solution:
a b
Let x = (cosecθ + sin θ) and y = (cosecθ − sin θ).
2 2
134
a
x = (cosecθ + sin θ)
2
2x 1
= + sin θ
a sin θ
2
1 + sin θ
=
sin θ
b
y = (cosecθ − sin θ)
2
2y 1
= − sin θ
b sin θ
2
1 − sin θ
=
sin θ
2x 2y 2
+ = (1)
a b sin θ
2
2x 2y 2 sin θ
− = (2)
a b sin θ
(1) × (2) gives
4x 2 4y 2
− = 4
a 2 b 2
2 2
2 2
b x − a y
= 1
2 2
a b
2 2
2 2
2 2
b x − a y = a b
2
2
10. If P(2,-7) is a given point and Q is a point on 2x + 9y = 18, then find the equations of the locus
of the mid-point of PQ.
Solution:
2 + h −7 + k
Let the mid-point of PQ be (a, b) Let Q be (h, k) and hence mid-point of PQ is ( , ).
2 2
2
2
Hence h = 2a − 2, k = 2b + 7. Since Q lies on 2x + 9y = 18,
2
2
2h + 9k = 18
2
2
2(2a − 2) + 9(2b + 7) = 18
(2a − 2) 2 (2b + 7) 2
+ = 1
9 2
(2x − 2) 2 (2y + 7) 2
+ = 1
9 2
11. If R is any point on the x-axis and Q is any point on the y-axis and P is a variable point on RQ
with RP = b, PQ = a. then find the equation of locus of P.
Solution: Let P be (x, y).Let R be (h, 0). Q be (0, k).
4.1 Objective type of questions for the Chapter 135
RP = b
2
2
(x − h) + y = b 2
2
2
2
x + y = 2hx − h + b 2
PQ = a
2
2
x + (y − k) = a 2
2
2
2
x + y = a + 2yk − k 2
2
2
2
2hx − h + b = a + 2yk − k 2
2
2
2
2
h − k − 2hx + 2yk + a − b = 0
12. If the points P(6,2) and Q(-2,1) and R are the vertices of a ∆PQR and R is the point on the locus
2
y = x − 3x + 4, then find the equation of the locus of centroid of ∆PQR
Solution:
Let the coordinates of the centroid be (h, k).R be (a, b).
6 − 2 + a
h = ⇒ 3h = 4 + a ⇒ a = 3h − 4.
3
2 + 1 + b
k = ⇒ 3k = 3 + b ⇒ b = 3k − 3.
3
2
Since R is a point on the locus y = x − 3x + 4,
2
2
we have b = a − 3a + 4 ⇒ (3k − 3) = (3h − 4) − 3(3h − 4) + 4
2
2
(3k − 3) = 9h + 16 − 24h − 9h + 12 + 4 = 9h − 33h + 32.
2
2
Hence 9h − 33h − 3k + 35 = 0 ⇒ 9x − 33x − 3y + 35 = 0.
2
2
13. If Q is a point on the locus of x + y + 4x − 3y + 7 = 0, then find the equation of locus of P
which divides segment OQ externally in the ratio 3:4,where O is origin.
Solution: Let the coordinates of Q be (h, k). Let P which divides OQ externally in the ratio 3:4 be
3h −4k a b
(a, b) = , . Hence h = − , k = .
−1 −1 3 4
2
2
Since (h, k) lies on the locus, we have h + k + 4h − 3k + 7 = 0.
2
a b a b
2
− + + 4 − − 3 + 7 = 0
3 4 3 4
a 2 b 2 4a 3b
+ − − + 7 = 0
9 16 3 4
x 2 y 2 4x 3y
+ − − + 7 = 0
9 16 3 4
14. Find the points on the locus of points that are 3 units from x-axis and 5 units from the point (5, 1).
Solution:
136
The locus of points that are 3 units from x-axis are in the line y = 3. Any point on this line is
2
2
(h, 3). The locus of points 5 units from the point (5, 1) are on the circle (x − 5) + (y − 1) = 25.
√
Solving, we get h = 21 + 5. The points are (0.4, 3) and (9.6, 3) which are 3 units from x-axis
and 5 units from the point (5, 1).
15. The sum of the distance of a moving point from the points (4,0) and (-4, 0) is always 10 units.
Find the equation of the locus of the moving point.
Solution: Let S(4, 0) and T(−4, 0) be the two given points. Given the sum of distances is 10 units.
q q
2 2
(x − 4) + y 2 + (x + 4) + y 2 = 10
q q
2 2
(x − 4) + y 2 = 10 − (x + 4) + y 2
Squaring both sides
q
2
2
2
2
2
(x − 4) + y = 10 + (x + 4) + y + 20 (x + 4) + y 2
q
2
−8x + 16 = 10 + 8x + 16 + 20 (x + 4) + y 2
q
2
−16x − 10 = 20 (x + 4) + y 2
2
2
(−16x − 10) = 400 (x + 4) + y 2
2
2
2
256x + 100 + 320x = 400(x + 8x + 16 + y )
2
2
144x + 400y + 2880x + 6300 = 0
2
2
36x + 100y + 720x + 1575 = 0
Exercise - 6.2
1. Find the equation of the lines passing through the point (1,1)
(i) with y-intercept (−4)
Let the equation be y = mx + b, where m is the slope and b is the y- intercept. It becomes
y = mx − 4. Since it passes through (1, 1), we have m − 4 = 1 ⇒ m = 5.Hence the
equation is y = 5x − 4.
(ii) with slope 3
Let the equation be y = mx + b, where m is the slope and b is the y- intercept. It becomes
y = 3x+b. Since it passes through (1, 1), we have b = −2. Hence the equation is y = 3x−2.
(iii) and (-2, 3)
y − 1
y − y 1 x − x 1
Equation of a line passing through two points is = . Here, =
y 2 − y 1 x 2 − x 1 2
x − 1
⇒ 2x + 3y − 5 = 0.
−3
◦
(iv) and the perpendicular from the origin makes an angle 60 with x- axis.
4.1 Objective type of questions for the Chapter 137
◦
Let the equation be x cos α + y sin α = p. Here α = 60 . Hence the equation becomes
√
√ 3 + 1
x + y 3 = 2p. since it passes through (1, 1), we have p = . Hence the equation is
√ √ 2
2x + 2y 3 = 3 + 1.
x y
2. If P(r, c) is mid point of a line segment between the axes, then show that + = 2.
r c
Solution: Clearly the end points of the line segment be (2r, 0) and (0, 2c). Hence the intercepts
x y x y
are 2r and 2c. The equation is + = 1 ⇒ + = 2.
2r 2c r c
3. Find the equation of the line passing through the point (1, 5)and also divides the line segment
between the co-ordinate axes in the ratio 2:3.
x y
Solution: Let the equation of the line be + = 1. Since it passes through (1, 5), we have
2a 3a
1 5 3 + 10 13
+ = 1 ⇒ = 1 ⇒ a = .
2a 3a 6a 6
x y 13
+ = ⇒ 3x + 2y = 13.
2 3 6
4. If p is length of perpendicular from origin to the line whose intercepts on the axes are a and b, then
1 1 1
show that = + .
p 2 a 2 b 2
Solution: The equation of the line whose intercepts on the axes are a and b, is,
x y
+ = 1
a b
bx + ay − ab = 0
The perpendicular distance (d) of a line Ax + By + C = 0 from a point (x 1 , y 1 ) is
|Ax 1 + By 1 + C|
d = √
2
A + B 2
|b(0) + a(0) − ab|
p = √
2
b + a 2
(ab) 2
2
p =
2
2
(b + a )
2 2 2 2
(ab) = p (b + a )
2
2
1 (b + a )
=
p 2 (ab) 2
1 1 1
= +
p 2 a 2 b 2
◦
◦
◦
5. The normal boiling point of water is 100 Cor 212 F and the freezing point of water is 0 C or
◦
◦
32 F. Find the linear relationship between C and F. Find (i) the value of C for 98.6 F (ii) the
◦
value ofF for 38 C
Solution:
The equation for a line that converts Celsius to Farenheit is:F = mC + b
138
where C is the Celsius temperature
m is the ”slope” of the line
F is the Fahrenheit temperature
and b is the ”intercept” of the line
So if we use our two ”points” to fill in this equation we get:
Farenheit Celsius equation
32 0 32 = 0m + b
212 100 212 = 100m + b
Now we have two equations and two unknown values (m and b) so we can solve for these values
and make an equation that converts Celsius to Farenheit:The two equations:
32 = 0m + b
212 = 100m + b
9
Solving we get m = , b = 32
5
9
Hence F = C + 32
5
◦
◦
(i) the value of C for 98.6 F is 37 C
◦
◦
(ii) the value ofF for 38 C is 100.4 F
6. An object was launched from a place P in constant speed to hit a target. At the 15th second it was
1400m away from the target and at the 18th second 800m away. Find (i) the distance between the
place and the target (ii) the distance covered by it in 15 seconds.(iii) time taken to hit the target.
Solution: Let the points be (15, 1400) and (18, 800). The equation of the line be
y − y 1 x − x 1
=
y 2 − y 1 x 2 − x 1
y − 1400 x − 15
=
−600 3
3y − 4200 = −600x + 9000
3y + 600x − 5100 = 0
y + 200x − 1700 = 0
(i) the distance between the place and the target is 1700m.
(ii) the distance covered by it in 15 seconds is1300 m.
(iii) time taken to hit the target is 85 seconds.
7. Population of a city in the years 2005 and 2010 are 1,35,000 and 1,45,000 respectively.Find the
approximate population in the year 2015.(the growth of population is constant)
4.1 Objective type of questions for the Chapter 139
Solution:Let the points be (2005, 135000) and (2010, 145000). The equation of the line be
y − y 1 x − x 1
=
y 2 − y 1 x 2 − x 1
y − 135000 x − 2005
=
10000 5
5y − 675000 = 10000x − 20050000
y − 135000 = 2000x − 4010000
y = 2000x − 3875000
The approximate population in the year 2015 is 155000
◦
8. Find the equation of the line, if the perpendicular drawn from the origin makes an angle 30 with
x-axis and its length is 12.
Solution:
◦
The equation of the line is x cos α + y sin α = p where α = 30 and p = 12. Hence the equation
√
3 1 √
becomes x + y = 12 ⇒ x 3 + y = 24.
2 2
9. Find the equation of the straight lines passing through (8, 3) and having intercepts whose sum is 1
Solution:
x y
Let the intercepts be a and 1 − a. The equation of the line is + = 1. Since it pass through
a 1 − a
8 3
2
2
(8, 3), we have, + = 1. 8(1 − a) + 3a = a(1 − a) ⇒ 8 − 5a = a − a ⇒ a − 6a + 8 = 0.
a 1 − a
x y x y
Solving, a = 2, 4. The equation of the straight lines are − = 1 and − = 1.
2 1 4 3
1
10. Show that the points (1, 3), (2, 1) and , 4 are collinear, by using (i) concept of slope (ii) using
2
a straight line and (iii) any other method
Solution:
(i) concept of slope
−2
Slope of (1, 3) and (2, 1) = −2
1
y 2 − y 1 Since slopes are same,
1 1
x 2 − x 1 Slope of (1, 3) and , 4 = −2 points are collinear
2 1
−
2
(ii) using straight line
−2
Equation of (1, 3) and (2, 1) (y − 3) = (x − 1)
1
y 2 − y 1 Equations are same
(y − y 1 ) = 1 1
x 2 − x 1 Equation of (1, 3) and , 4 (y − 3) = (x − 1) hence collinear
2 1
−
2
140
11. A straight line is passing through the point A(1,2) with slope 5 . Find points on the line which are
12
13 units away from A.
Solution:
12. A 150m long train is moving with constant velocity of 45km / h. (i) time takento cross a
pole.(ii)The time to cross the bridge of length 850m is ?
Solution:
13. A spring was hung from a hook in the ceiling. A number of different weights were attached to the
spring to make it stretch, and the total length of the spring was measured each time shown in the
followng table.
Weight, (kg) 2 4 5 8
Length, (cm) 3 4 4.5 6
(a) Draw a graph showing the results.
(b) Find the equation relating the length of the spring to the weight on it.
(c) What is the actual length of the spring.
(d) If the spring to stretch to 9 cm long, how much weight should be added?
(e) How long will the spring be when 6 kilograms of weight on it?
14. A family is using Liquefied petroleum gas (LPG) of weight 14.2 kg for consumption. (Full weight
29.5kg includes the empty cylinders tare weight of 15.3kg.). If it is use with constant rate then it
lasts for 24 days. Then the new cylinder is replaced (i) Find the equation relating the quantity of
gas in the cylinder to the days. (ii) Draw the graph for first 96days.
15. In a shopping mall there is a hall of cuboid shape with
dimension800 × 800 × 720 units, which needs to be added the
facility of an escalator in the path as shown by the dotted line in
the figure. Find (i)the minimum total length of the escalator. (ii)
the heights at which the escalator changes its direction. (iii) the
slopes of the escalator at the turning points.
Exercise - 6.3
1. Show that the lines are 3x + 2y + 9 = 0 and 12x + 8y − 15 = 0 are parallel lines.
2. Find the equation of the straight line parallel to 5x − 4y + 3 = 0 and having x-intercept −3.
3. Find the equations of two straight lines which are parallel to the line 12x + 5y + 2 = 0 and at a
unit distance from the point (1, − 1).
4. Find the equations of straight lines which are perpendicular to the line 3x + 4y − 6 = 0 and are at
a distance of 4 units from (2, 1).
5. Find the equation of a straight line parallel to 2x + 3y = 10 and which is such that the sum of its
intercepts on the axes is 15.
6. Find the length of the perpendicular and the co-ordinates of the foot of the perpendicular form
(−10, −2) to the line x + y − 2 = 0
7. If P 1 and P 2 are the lengths of the perpendiculars from the origin to the straight lines x sec θ + y
2
2
2
cosec θ = 2a and x cos θ − y sin θ = a cos 2θ, then prove that P + P = a .
2
1
8. Find the distance between the parallel lines
4.1 Objective type of questions for the Chapter 141
(i) 12x + 5y = 7 and 12x + 5y + 7 = 0
(ii) 3x − 4y + 5 = 0 and 6x − 8y − 15 = 0.
9. Find the family of straight lines (i) Perpendicular (ii) Parallel to 3x + 4y − 12 = 0.
10. If the line joining two points A(2,0) and B(3,1) is rotated about A in anticlockwise direction
o
through an angle of 15 , then find the equation of the line in new position.
11. A ray of light coming from the point (1,2) is reflected at a point A on the x-axis and it passes
through the point (5,3). Find the co-ordinates of the point A.
12. A line is drawn perpendicular to 5x = y+ 7. Find the equation of the line if the area of the triangle
formed by this line with co-ordinate axes is 10 sq. units.
13. Find the of the image point (−2, 3)about the line x + 2y − 9 = 0.
14. A photocopy store charges | 1.50 per copy for the first 10 copies and | 1.00 per copy after the 10th
copy. Let x be the number of copies, and let y be the total cost of photocopying. Draw graph the
cost as x goes from 0 to 50 copies. (b) Find the cost of making40 copies
15. Find the the distance between the line 4x + 3y + 4 = 0,and a point (i) (−2, 4) (ii) (7, −3)
16. Write the equation of the lines through the point (1,-1)
(iii) parallel to x + 3y − 4 = 0
(iv) perpendicular to 3x + 4y = 6
17. If (−4, 7) is one vertex of a rhombus and if the equation of one diagonal is 5x − y + 7 = 0,then
find the equation of another diagonal.
18. Find the equation of the lines passing through the point of intersection lines 4x − y + 3 =
0 and 5x + 2y + 7 = 0, and (i) through the point(−1, 2) (ii) Parallel to x − y + 5 = 0 (iii)
Perpendicular to x − 2y + 1 = 0
19. Find atleast two equations of the straight lines in the family of the linesy = 5x + b, for which band
the x- coordinate of the point of intersection of the lines with 3x − 4y = 6 are integers.
20. Find all the equations of the straight lines in the family of the linesy = mx − 3, for which m and
the x- coordinate of the point of intersection of the lines with x − y = 6 are integers.
Exercise - 6.4
√
1. Find the combined equation of the straight lines whose separate equations are x − 2y − 3 = 2
and x + y + 5 = 0.
2
2
2. Show that 4x + 4xy + y − 6x − 3y − 4 = 0 represents a pair of parallel lines.
2
2
3. Show that 2x + 3xy − 2y + 3x + y + 1 = 0 represents a pair of perpendicular lines.
2
2
4. Show that the equation 2x − xy − 3y − 6x + 19y − 20 = 0 represents a pair of intersecting lines.
−1
Show further that the angle between them is tan (5).
5. Prove that the equation to the straight lines through the origin, each of which makes an angle α
2
2
with the straight line y = x is x − 2xy sec 2α + y = 0
6. Find the equation of the pair of straight lines passing through the point (1, 3) and perpendicular to
the lines2x − 3y + 1 = 0and 5x + y − 3 = 0
7. Find the separate equation of the following pair of straight lines
2
2
(i) 3x + 2xy − y = 0
2
2
(ii) 6(x − 1) + 5(x − 1)(y − 2) − 4(y − 3) = 0
2
2
(iii) 2x − xy − 3y − 6x + 19y − 20 = 0
2
2
8. The slope of one of the straight lines ax + 2hxy + by = 0 is twice that of the other, show that
2
8h = 9ab.
2
2
9. The slope of one of the straight lines ax + 2hxy + by = 0 is three times the other, show that
2
3h = 4ab.
142
2
2
10. A ∆OPQ is formed by the pair of straight lines x − 4xy + y = 0 and the PQ. Find the equation
of PQ is x + y − 2 = 0. Find the equation of the median of the triangle ∆OPQdrawn from the
origin.
11. Findpand q, if the following equation represents a pair of perpendicular lines
2
2
6x + 5xy − py + 7x + qy − 5 = 0
12. Find the value of k, if the following equation represents a pair of straight lines. Further, find
2
2
whether these lines are parallel or intersecting, 12x + 7xy − 12y − x + 7y + k = 0
2
2
13. For what value of k does the equation 12x +2kxy+2y +11x−5y+2 = 0 represent two straight
lines.
2
2
14. Show that the equation 9x −24xy +16y −12x+16y −12 = 0 represents a pair of parallel lines.
Find the distance between them.
2
2
15. Show that the equation 4x + 4xy + y − 6x − 3y − 4 = 0 represents a pair of parallel lines. Find
the distance between them.
2
2
16. Prove that one of the straight lines given by ax + 2hxy + by = 0 will bisect the angle between
2
the co-ordinate axes if (a + b) = 4h 2
2
2
17. If the pair of straight lines x − 2kxy − y = 0 bisect the angle between the pair of straight lines
2
2
x − 2lxy − y = 0, Show that the later pair also bisects the angle between the former.
2
2
18. Prove that the straight lines joining the origin to the points of intersection of 3x + 5xy − 3y +
2x + 3y = 0 and 3x − 2y − 1 = 0 are at right angles.
Exercise - 6.5
Choose the correct or more suitable answer
1. The equation of the locus of the point whose distance from y-axis is half the distance from origin
is
2
2
2
2
2
2
2
2
(a) x + 3y = 0 (b) x − 3y = 0 (c) 3x + y = 0 (d) 3x − y = 0
2
2. Which of the following equation is the locus of (at , 2at)
x 2 y 2 x 2 y 2
2
2
2
(a) − = 1 (b) + = 1 (c) x + y = a 2 (d) y = 4ax
a 2 b 2 a 2 b 2
2
2
3. Which of the following point lie on the locus of 3x + 3y − 8x − 12y + 17 = 0
(a) (0,0) (b) (-2,3) (c) (1,2) (d) (0,-1)
4. If the point (8,-5) lies on the locus x 2 − y 2 = k, then the value of k is
16 25
(a) 0 (b) 1 (c) 2 (d) 3
5. straight line joining the points (2, 3)and (−1, 4)passes through the point (α, β) if
(a) α + 2β = 7 (b) 3α + β = 9 (c)α + 3β = 11 (d)3α + β = 11
◦
6. The slope of the line which makes an angle 45 with the line 3x − y = −5 are
(a) 1, −1 (b) 1 π , −1 (c) 1, 1 (d) 2, − 1
2 3 2 2
7. Equation of the straight line that forms an isosceles triangle with coordinate axes in the I-quadrant
√
with perimeter 4 + 2 2 is
√ √
(a) x + y + 2 = 0 (b) x + y − 2 = 0 (c) x + y − 2 = 0 (d) x + y + 2 = 0
8. The coordinates of the four vertices of a quadrilateral are (−2,4), (−1,2), (1,2) and (2,4) taken in
order.The equation of the line passing through the vertex (−1,2) and dividing the quadrilateral in
4.1 Objective type of questions for the Chapter 143
the equal areas is
(a) x+ 1 = 0 (b) x + y = 1 (c) x + y+ 3 = 0 (d) x – y = 1
9. The intercepts of the perpendicular bisector of the line segment joining (1, 2) and (3,4) with
coordinate axes are
π
(a) 5, −5 (b) 5 , 5 (c) 5, 3 (d) 5, −4
3
10. The equation of the line with slope 2 and the length of the perpendicular from the origin equal to
√
5 is
√ √
(a) x + 2y = 5 (b) 2x + y = 5 (c) 2x + y = 5 (d) 2x + y = −5
11. If a line is perpendicular to the line 5x – y = 0 and forms a triangle, with the coordinate axes of
area 5 sq. units, then its equation is
√ √ √
(a) x+ 5y ± 5 2 = 0 (b) x− 5y ± 5 2 = 0 (c) 5x + y ± 5 2 = 0
√
(d) 5x − y ± 5 2 = 0
12. Equation of the straight line perpendicular to the linex−y+5 = 0, through the point of intersection
the y-axis and the given line
(a) x − y − 5 = 0 (b) x + y − 5 = 0 (c) x + y + 5 = 0
(d) x + y + 10 = 0 (e) x + y = 0
13. If the equation of the base opposite to the vertex (2, 3) of an equilateral triangle is x + y = 2, then
the length of a side is
√ √
q
(a) 3 (b) 6 (c) 6 (d) 3 2
2
14. The line (p + 2q)x + (p − 3q)y = p − q for different values of p and q passes through the point
,
,
(a) 3 5 (b) 2 2 (c) 3 3 (d) 2 3
,
,
2 2 5 5 5 5 5 5
15. The point on the line 2x − 3y = 5 is equidistance from (1,2) and (3, 4) is
(a) (7, 3) (b) (4, 1) (c) (1, −1) (d) (−2, 3)
16. The image of the point (2, 3) in the line y = −x is
(a) (−3, −2) (b) ( −3, 2 ) (c) ( −2, −3) (d) ( 3, 2 )
17. The length of ⊥ from the origin to the line x − y = 1, is
3 4
(a) 11 (b) 5 (c) 12 (d) − 5
5 12 5 12
18. The straight lines represented by 8x − 7y + 18 = 0 and 24x − 21y + 31 = 0 are
(a) parallel to each other (b) perpendicular to each other
◦
(c) inclined at 45 to each other (d) coincident to each straight lines
19. The y-intercept of the straight line passing through (1,3)and perpendicular to 2x − 3y + 1 = 0 is
3 9 2 2
(a) (b) (c) (d)
2 2 3 9
20. If the two straight lines x + (2k − 7)y + 3 = 0 and 3kx + 9y − 5 = 0 are perpendicular then the
value of k is
1 2 3
(a) k = 3 (b) k = (c)k = (d) k =
3 3 2
21. If a vertex of a square is at the origin and its one side lies along the line 4x + 3y − 20 = 0, then
the area of the square is
(a) 20 sq. units (b) 16 sq. units (c) 25 sq. units (d) 4 sq.units
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