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MATHEMATICS PAPER - 1A

TIME: 3hrs Max. Marks.75

Note: This question paper consists of three sections A, B and C.

m
SECTION A
Very short answer type questions. 10 X 2 =20

co
n.
1. If f : [1, ∞) → [1, ∞) defined by f(x) = 2x(x–1) then find f–1(x).

io
y y
2. If f (y) = and g(y) =
1 − y2 1 + y2
at
then show that (fog)(y) = y.
uc
3. If a ⋅ b = a ⋅ c and a × b = a × c, a ≠ 0 then show that b = c .
ed

4. Prove that 3 csc 20° − sec 20° = 4

5. Find the range of 7 cos x – 24 sin x + 5.
i
sh

6. Find a vector in the direction of vector a = i − 2 j that has magnitude 7 units.

ak

7. If a, b, c are the position vectors of the vertices A, B and C respectively of ∆ABC

.s

then find the vector equation of median through the vertex A.

w

(
8. Prove that cos h −1 x = log e x − x 2 − 1 )
w

1 1 3
w

9. If A =  5 2 6  then find A3.

 −2 −1 −3

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0 2 1
10. If A =  −2 0 −2  is a skew symmetric matrix, find x.
 −1 x 0 

SECTION - B
Short answer type questions.

m
Answer any five of the following. 5 X 4 = 20

co
n.
11. If i , j, k are unit vectors along the positive directions of the coordinate axes, then
show that the four points 4 i + 5 j + k, − j − k , 3i + 9 j + 4k and −4 i + 4 j + 4k are

io
coplanar.

12. Let a = i + j + k and b = 2 i + 3 j + k find

at
uc
i) The projection vector of b on a and its magnitude.
ed

ii) The vector components of b in the direction of a and perpendicular to a .

π 3π 5π 7π 3
13. Prove that sin 4 + sin 4 + sin 4 + sin 4 =
i
sh

8 8 8 8 2.

14. If sin 3x + sin x + 2 cos x = sin 2 x + 2cos2 x find the general solution.
ak

15. It cos−1 p + cos−1 q + cos−1 r = π prove that p 2 + q 2 + r 2 + 2 pqr = 1

If a : b : c = 7 : 8 : 9 then find cos A = cos B = cos C
.s

16.
3 −4  1 + 2n −4n 
w

17. If A =   then show that An =  , n is a positive integer.

1 −1  n 1 − 2n 
w
w

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SECTION - C
Long answer type questions.
Answer any five of the following. 5 X 7 = 35

18. If f : A → B , g : B → C are two bijections then ( gof ) −1 = f −1 og −1 .

m
19. Show that 3.52 n+1 + 23n+1 + 23n+1 is divisible by 17.

co
20. Solve the following equations by Gauss-Jordan method.
3x + 4y + 5z = 18

n.
2x – y + 8z = 13

io
5x – 2y + 7z = 20
1 a2 a3
21.
at
Show that 1 b 2 b3 = (a − b)(b − c)(c − a)(ab + bc + ca)
uc
1 c2 c3
ed

22. If a, b and c are non-zero vectors and a is perpendicular to both b and c. If

2π
| a |= 2,| b |= 3,| c |= 4 and (b, c) = , then find  a b c  .
i
sh

23. Prove that a3cos(B – C) + b3cos (C – A) + c3cos(A – B) = 3 abc.

ak

24. If A, B, C are angles of a triangle then prove that

.s

A B C A B C
sin 2 + sin 2 − sin 2 = 1 − 2 cos cos cos
2 2 2 2 2 2
w
w
w

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Mathematics IA Paper - 2

1. If f: [1, ∞) → [1, ∞) defined by f(x) = 2x(x–1) then find f–1(x).

Sol. f(x): [1 ……∞) → [1 ……∞)

f (x) = 2x(x −1)

m
f (x) = 2x(x −1) = y
x(x − 1) = log 2 y

co
x 2 − x − log 2 y = 0

n.
−b ± b 2 − 4ac
x=
2a

io
1 ± 1 + 4 log 2 y
x=
2

f −1 (x) =
1 ± 1 + 4 log 2 x
at
uc
2
y y
2. If f (y) = and g(y) = then show that (fog)(y) = y.
ed

1− y 2
1 + y2

Sol. Given that

i
sh

y y
f (y) = and g(y) =
1 − y2 1 + y2
ak

 y 
∴ fog(y) = f [g(y)] = f  
.s

 1 − y 2 
w

2
y  y 
= 1−  
w

1 + y2  1 + y2 
 
w

y 1 + y2
= × =y
1 + y2 1 + y2 − y2
∴ fog(y) = y

3. If a ⋅ b = a ⋅ c and a × b = a × c, a ≠ 0 then show that b = c .

Sol. Given that,

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a ⋅ b = a ⋅ c ⇒ a (b − c) = 0 ...(1)
a × b = a × c ⇒ a × (b − c) = 0 ...(2)

From (1) and (2) it is evident that, the vector (b − c) cannot be both perpendicular to
a and parallel to a .

Unless it is zero

m
∴ b − c = 0 (a ≠ 0)
∴b = c

co
4. Prove that 3 csc 20° − sec 20° = 4

n.
Sol. L.H.S. = 3 csc 20° − sec 20°

io
3 1
= −
sin 20° cos 20°

=
3 cos 20° − sin 20°
at
uc
sin 20° cos 20°
3 1
2⋅ sin 20° − sin 20°
ed

= 2 2
1
(2sin 20° cos 20°)
2
i

4(sin 60° cos 20° − cos 60° sin 20°)

sh

=
sin 40°
sin(60° − 20°) sin 40°
=4 = 4⋅
ak

sin 40° sin 40°

= 4 = R.H.S.
.s

5. Find the range of 7 cos x – 24 sin x + 5.

w

Sol. Minimum value of f = C − a 2 + b 2

w

= 5 − (−24)2 + 7 2
w

= 5 − 576 + 49
= 5 − 625 = 5 − 25 = −20

Maximum value of f = C + a 2 + b 2

= 5 + 625 = 5 + 25 = 30

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6. Find a vector in the direction of vector a = i − 2 j that has magnitude 7 units.

Sol. The unit vector in the direction of the given vector ‘a’ is

1 1 1 2
â = a= ( i − 2 j) = i− j
|a | 5 5 5

Therefore, the vector having magnitude equal to 7 and in the direction of a is

m
co
 1 2  7 14
7a = 7  i− j = i− j
 5 5  5 5

n.
7. If a, b, c are the position vectors of the vertices A, B and C respectively of ∆ABC

io
then find the vector equation of median through the vertex A

Sol: OA = a , OB = b , OC = c be the given vertices

at
uc
b+c
Let D be the mid point of BC =
ed

The vector equation of the line passing through the

i
sh

points a , b is r = (1 − t )a + tb
ak

b+c 
∴vector equation is r = (1 = t )a + t   where t ∈ R
 2 
.s

( )
w

8. Prove that cos h −1 x = log e x − x 2 − 1

w

Sol: Let cos h−1 x = y ⇒ x = cos hy

w

e y + e− y 1
x= ⇒ 2x = ey + y
2 e

2 xe y = ( e y ) + 1 ⇒ ( e y ) − 2 xe y + 1
2 2

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2 x ± 4 x2 − 4
e =
y
⇒ e y = x ± x2 −1
2

(
e y = x + x 2 − 1 ⇒ y = log e x + x 2 − 1 )
(
cos h −1 x = log e x + x 2 − 1 )

m
1 1 3
9. If A =  5 2 6  then find A3.

co
 −2 −1 −3

n.
1 1 3 1 1 3
Sol. A = A.A =  5 2 6 
2 5 2 6
 

io
 −2 −1 −3  −2 −1 −3

 1+ 5 − 6 1+ 2 − 3 3+ 6−9 
= 5 + 10 − 12 5 + 4 − 6 15 + 12 − 18

at
uc
 −2 − 5 + 6 −2 − 2 + 3 −6 − 6 + 9 

0 0 0
ed

=  3 3 9 
 −1 −1 −3
i
sh

 0 0 0  1 1 3 
A = A ⋅ A =  3 3 9   5 2 6 
3 2
ak

 −1 −1 −3  −2 −1 −3

 0+0+0 0+0+0 0+0+0 

.s

= 3 + 15 − 18 3 + 6 − 9 9 + 18 − 27 

w

 −1 − 5 + 6 −1 − 2 + 3 −3 − 6 + 9 
w

0 0 0 
= 0 0 0 
w

0 0 0 

0 2 1
10. If A =  −2 0 −2  is a skew symmetric matrix, find x.
 −1 x 0 

Hint : A is a skew symmetric matrix

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⇒ AT= -A
Sol. A is a skew symmetric matrix
⇒ AT = A
 0 −2 −1 0 2 1  0 −2 −1
 2 0 x  = −  −2 0 −2  =  2 0 2 
     
1 −2 0   −1 x 0  1 − x 0 

m
Equating second row third column elements we get x = 2.

co
11. If i , j, k are unit vectors along the positive directions of the coordinate axes, then

n.
show that the four points 4 i + 5 j + k, − j − k , 3i + 9 j + 4k and −4 i + 4 j + 4k are

io
coplanar.

Sol. Let O be a origin, then at

uc
OA = 4 i + 5 j + k, OB − j − k
OC = 3i + 9 j + 4k, and OD = − 4 i + 4 j + 4k
ed

AB = OB − OA = −4 i − 6 j − 2k
i
sh

AC = OC − OA = − i + 4 j + 3k
AD = OD − OA = −8 i − j + 3k
ak

−4 −6 −2
 AB AC AD  = −1 4 3
 
.s

−8 −1 3
w

= −4[12 + 3] + 6[−3 + 24] − 2[1 + 32]

w

= −4 ×15 + 6 × 21 − 2 × 33
w

= −60 + 126 − 66
= −126 + 126 = 0

Hence given vectors are coplanar.

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12. Let a = i + j + k and b = 2 i + 3 j + k find

i) The projection vector of b on a and its magnitude.

ii) The vector components of b in the direction of a and perpendicular to a .
Sol. Given that a = i + j + k , b = 2 i + 3 j + k

a ⋅b
i) Then projection of b on a = ⋅a

m
| a |2

co
( i + j + k) ⋅ (2 i + 3 j + 3k)
= ⋅| i + j + k |
| i + j + k |2
2 + 3 +1

n.
= ⋅i + j+k
( 3)2

io
6( i + j + k)
= = 2( i + j + k)
3

Magnitude = =
at
| a ⋅ b | | ( i + j + k) ⋅ (2 i + 3 j + k) |
uc
|a| | i + j+k|

| 2 + 3 + 1| 6
= = = 2 3 unit
ed

| 3| 3

(a ⋅ b)
i

ii) The component vector of b in the direction of a = ⋅a

sh

| a |2

= 2( i + j + k) (∵ from 10(i))
ak

The vector component of b perpen-dicularto a .

.s

(a ⋅ b)a
=b− = (2 i + 3 j + k) − 2( i + j + k)
| a |2
w

= 2 i + 3 j + k − 2 i − 2 j − 2k = j − k
w

π 3π 5π 7π 3
+ sin 4 + sin 4 + sin 4 =
w

13. Prove that sin 4

8 8 8 8 2
π 3π 5π 7π
Sol: sin 4 + sin 4 + sin 4 + sin 4
8 8 8 8
2 4
 π   2 π π   2  π π    π 
4 4

 sin  + sin −  + sin  +   + sin  π −  

 8  2 8   2 8    8 

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π π π π
sin 4 + cos 4 + cos 4 + sin 4
8 8 8 8

π π  π π π π 
2

2 sin 4 + cos 4  = 2  sin 2 + cos 2  − 2sin 2 cos 2 
 8 8  8 8 8 8 

π π
= 2 − 4 sin 2 cos 2
8 8

m
π π π
2
 1 3
= 2 −  2 sin cos  = 2 − sin 2 = 2 − =
 8 8 4 2 2

co
14. If sin 3x + sin x + 2 cos x = sin 2 x + 2cos2 x find the general solution.

n.
Sol: sin 3x + sin x + 2cos = sin 2 x + 2cos2 x

io
( 2sin 2 x + 2cos x ) = 2sin cos x + 2 cos2 x
2sin 2 x cos x− 2sin x cos x + 2cos x − 2 cos2 x = 0

2 cos x {sin 2 x − sin x + 1 −cos x} = 0

at
uc
3x x x
cos x = 0 2 cos . sin + 2 sin 2 = 0
ed

2 2 2
x  3x x
cos x = 0 2 sin cos + sin  = 0
i

2  2 2
sh

x 3x x
sin =0 cos = − sin
2 2 2
ak

x 3x
n = ( 2n + 1) π / 2 = nπ cos = cos (π / 2 + x / 2 )
2 2
.s

3x
x = 2nπ = 2nπ ± (π / 2 + x / 2 )
w

2
π
w

3x x 3x x
− = 2nπ + π / 2 : + = 2nπ −
2 2 2 2 2
w

x = ( 2n + 1) π / 2 ; x = 2nπ ; x = 2nπ + π / 2 ; x = nπ − π / 4 n ∈ z

15. It cos−1 p + cos−1 q + cos−1 r = π prove that p 2 + q 2 + r 2 + 2 pqr = 1

Sol: Let cos−1 p = α cos−1 q = β cos −1 r = δ

p = cos α q = cos β r = cos δ

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Given α + β + γ = π cos (α + β ) = cos (π − γ )

cos α cos β − sin α sin γ = cos γ

pq = − r + 1 − p 2 1 − q 2 ⇒ pq + r = 1 − p 2 1 − q 2

Squaring on both such

p 2 q 2 + r 2 + 2 pqr = 1 − p 2 − q 2 + p 2 q 2 ⇒ p 2 + q 2 + r 2 + 2 pqr = 1

m
16. If a : b : c = 7 : 8 : 9 then find cos A = cos B = cos C

co
Sol:
12
b 2 + c 2 − a 2 64k 2 + 81k 2 − 49k 2 96k 2
cos A = = =

n.
2bc 2 ( 8k )( 9k ) 2 × 8k × 9 k

io
a 2 + c 2 − b 2 49k 2 + 81k 2 − 64k 2 6633 11
cos B = = = =
2ac 2 × 7 k × 9k 2 × 63 21

cos C =
a 2 + b 2 − c 2 49k 2 + 64k 2 − 81k 2
= =
at
32k 2
=
2
uc
2ab 2 × 7 k × 8k 20k × 8k 7

2 11 2 2 11 2
ed

∴ cos A = cos B = co C = = = = × 21 = × 21 = × 21
3 21 7 3 21 7
= 14 :11: 6
i
sh

3 −4  1 + 2n −4n 
17. If A =   then show that An=  , n is a positive integer.
1 −1  n 1 − 2n 
ak

Sol. We shall prove the result by Mathematical Induction.

1 + 2n −4n 
.s

An = 
 n 1 − 2n 
w

1 + 2 −4  3 −4 
n = 1 ⇒ A′ =  = 
w

 1 1 − 2  1 −1
w

The result is true for n = 1

Suppose the result is true for n = k
1 + 2k −4k 
Ak = 
 k 1 − 2k 

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1 + 2k −4k  3 −4 
A k +1 = A k ⋅ A = 
 k 1 − 2k  1 −1
3 + 6k − 4k −4 − 8k + 4k 
= 
 3k + 1 − 2k −4k − 1 + 2k 
 2k + 3 −4k − 4 
= 
 k + 1 −2k − 1 

m
1 + 2(k + 1) −4(k + 1) 
=
 k +1 1 − 2(k + 1) 

co
∴ The given result is true for n = k + 1

n.
By Mathematical Induction, given result is true for all positive integral values of n.
18. If f : A → B , g : B → C are two bijections then ( gof ) −1 = f −1 og −1 .

io
at
Proof: f : A → B , g : B → C are bijections ⇒ gof : A → C is bijection ⇒ ( gof ) −1 : C → A is a
uc
bijection.
f : A → B is a bijection ⇒ f −1 : B → A is a bijection
ed

g : B → C is a bijection ⇒ g −1 : C → B is a bijection

g −1 : C → B , g −1 : B → A are bijections ⇒ f −1 og −1 : C → A is a bijection

i
sh

Let z ∈ C
z ∈ C , g : B → C is onto ⇒ ∃ y ∈ B ∋ g ( y ) = z ⇒ g −1 ( z ) = y
ak

y ∈ B, f : A → B is onto ⇒ ∃x ∈ A ∋ f ( x) = y ⇒ f −1 ( y ) = x
.s

( gof ) ( x) = g[ f ( x)] = g ( y ) = z ⇒ ( gof ) −1 ( z ) = x

w

∴ ( gof ) −1 ( z ) = x = f −1 ( y ) = f −1 [ g −1 ( z ) ] = ( f −1og −1 )( z ) ∴ ( gof )−1 = f −1og −1

w

19. Show that 3.52n+1 + 23n+1 + 23n+1 is divisible by 17.

w

Sol: Let S( n ) = 3.52 n +1 + 23n+1 be the given statement.

S(1) = 3.53 + 24 = 375 + 16 = 391 = 17 × 23

This is divisible by 17
Assume Sk is true

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S k = 3.52 k +1 + 23k +1 is divisible by 17

Let 3.52 k +1 + 23k +1 = 17m

3.52 k +1 = 17 m − 23k +1

S k +1 = 3.52 k +3 + 23 k + 4

= 3.52k +1.2 + 23k +1.23

m
= 25 {17 m − 23k +1} + 23k +1.8

co
= 25 × 17 m − 17(23 k +1 ) = 17 {25 m − 23 k +1} is divisible by 17

Hence S k +1 is true

n.
∴ S n is true for all n ∈ N

io
20. at
Solve the following equations by Gauss-Jordan method.
3x + 4y + 5z = 18
uc
2x – y + 8z = 13
ed

5x – 2y + 7z = 20
 3 4 5 18 
The augmented matrix is  2 −1 8 13 
i

Sol:
sh

 5 −2 7 20 

R1 → R1 − R 2
ak

1 5 −3 5 
=  2 −1 8 13 
.s

 5 −2 7 20 
w
w
w

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R 2 → R 2 − 2R 3 , R 3 → R 3 − 5R1
1 5 −3 5 
= 0 −11 14 3 
0 −27 22 −5
R 2 → −5R 2 + 2R 3
1 5 −3 5 
= 0 1 −26 25 


m
0 −27 22 −5

co
R1 → R1 − 5R 2 , R 3 → R 3 + 27R 2
1 0 127 130 

n.
= 0 1 −26 −25 
0 0 −680 −680 

io
R 3 → R 3 + (−680)
1 0 127 135 
at
= 0 1 −26 −25
uc
0 0 1 1 
ed

R1 → R1 − 127R 3 , R 2 → R 2 + 26R 3
 1 0 0 3
= 0 1 0 1
i
sh

0 0 1 1
ak

Hence the solution is x = 3, y = 1, z = 1.

Show that
.s

1 a2 a3
w

21. 1 b 2 b3 = (a − b)(b − c)(c − a)(ab + bc + ca)

1 c2 c3
w
w

1 a2 a3
Sol: L.H.S = 1 b 2 b3
1 c2 c3

By applying R2→R2 – R1 and R3→R3 – R1

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1 a2 a3
= 0 b2 − a 2 b2 − a 3
0 c2 − a 2 c3 − a 3

1 a2 a3
= (b − a)(c − a) 0 b + a b 2 + ba + a 2
0 c+a c 2 + ca + a 2

m
Applying R2→R2 – R3

co
1 a2 a3

n.
= −(a − b)(c − a) 0 b − c b 2 − c 2 + a(b − c)
0 c+a c2 + ca + a 2

io
1 a2 a3
= −(a − b)(c − a)(b − c) 0 1 b+c+a
0 c + a c2 + ca + a 2
at
uc
= −(a − b)(b − c)(c − a)
ed

[(c2 + ca + a 2 ) − (b + c + a)(c + a)]

= −(a − b)(b − c)(c − a)
i

[c2 + ca + a 2 − b(c + a) − (c + a) 2 =
sh

= −(a − b)(b − c)(c − a)

ak

[c2 + ca + a 2 − bc − ab − c2 − 2ca − a 2 ]
= −(a − b)(b − c)(c − a)[−ab − bc − ca]
.s

= (a − b)(b − c)(c − a)(ab + bc + ca)

w

1 a2 a3
∴ 1 b 2 b3 = (a − b)(b − c)(c − a)(ab + bc + ca)
w

1 c2 c3
w

22. If a, b and c are non-zero vectors and a is perpendicular to both b and c. If

2π
| a |= 2,| b |= 3,| c |= 4 and (b, c) = , then find  a b c  .
3

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Sol: If a is perpendicular to b and c .

⇒ a is parallel to b × c

⇒  a, b × c  = 0
⇒ b × c = | b || c | sin(b, c)aˆ
2π
⇒| b × c |= 3 × 4sin aˆ
3

m
3
⇒| b × c |= 12sin120.1 = 12 × =6 3

co
2
∴ | a b c | = | a ⋅ (b × c) |= | a || b × c | cos(a b c)

n.
= (2 ⋅ 6 3) cos 0 = 12 3
∴| a ⋅ b × c | = (2 ⋅ 6 3) = 12 3

io
23. Prove that a3cos(B – C) + b3cos (C – A) + c3cos(A – B) = 3 abc.
Sol. L.H.S. = Σ a3cos(B – C)
at
uc
= Σ a 2 (2 R sin A ) cos(B − C )
ed

= R Σ a 2 ⋅ [2 sin(B + C ) cos(B − C )]
= R Σ a 2 (sin 2 B + sin 2C )
i

= R Σ a 2 (2 sin B cos B + 2 sin C co s C )

sh

= Σ [a 2 (2 R sin B ) cos B + a 2 (2 R sin C ) cos C ]

ak

= Σ (a 2 b cos B + a 2 c cos C )
= (a 2 b cos B + a 2 c cos C )
.s

+ (b 2 c cos C + b 2 a cos A )
w

+ (c 2 a cos A + c 2 b cos B )
= ab (a cos B + b co s A ) + bc(b cos C + c cos B )
w

+ ca (c co s A + a cos C )
w

= ab (c ) + bc (a ) + ca (b )
= 3abc = R .H .S.

24. If A,B,C are angles of a triangle then prove that

A B C A B C
sin 2 + sin 2 − sin 2 = 1 − 2 cos cos cos
2 2 2 2 2 2

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Sol: A+B+C = 1800

A B C
LHS = sin 2 + sin 2 − sin 2
2 2 2
A B C B C
= sin 2 + sin  +  .sin  − 
2 2 2 2 2
A  A B C
= sin 2 + sin  90 −  .sin  − 
2  2 2 2

m
A A B C
=1-cos 2 + cos .sin  − 

co
2 2 2 2

A A  B C 
= 1 − cos  cos − sin  −  

n.
2 2  2 2 
A   B C   B C 

io
=1 − cos  cos  90 −  +   − sin  −  
2   2 2   2 2 

= 1 − cos
A  B C 

 B C 
sin  +  − sin  −  
2 2 2  2 2 
at
uc
A B C
= 1 − cos  2 cos sin 
2 2 2
ed

A B C
= 1 − 2 cos cos sin = RHS
2 2 2
i
sh
ak
.s
w
w
w

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