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Solved Guess Paper - 1

II PUC Mathematics

One Mark Questions

1. Give an example of a relation which is reflective and transitive but net symmetric .

Ans.

A = { 1 , 2, 3}

R = { (1 , 1), (2, 2), (3, 3), (2, 3)}

2. Find the principal value of cos-1

1-

2

Ans. π 3π

π =4 4

3. Constant a 2 2 matrix A = [aij] whose element are given by aij =i

j

Ans.

11

A = 2

2 1

4. If

2 3A =

-1 2 find |2A |

Ans. | A | = 4 + 3 = 7

| 2 A | = 4 |A| = 28

5. If y = e3 log x then show that

2dy= 3x

dx.

Ans.

3logx

3

2

y =

y = x

dy= 3x

dx



6.

2

2

1Evaluate x 1- dx

x

Ans.

2

3

(x -1)dx

x- x + c

3

7. Define the feasible region in L . P . P

Ans. A region which is common to all the constraints

8. Find the unit vector in the direction a = 2i + 3j+ 6K

of the vector .

Ans. 2i + 3j + 6K 2i + 3j + 6Kn = =

74 + 9 + 36

9. Find the direction ratio of the line x -1 2Z + 3

= 3y =2 4

Ans.

3Z +

x -1 y 2= =12 2

3

Direction of ratio 1

(2, , 2)3

.

10. If p(F) = 0.6 and P(E nF) = 0.2 find P(E / F) .

Ans. P(EnF) 0.2 1P(E / F) = = =

P(F) 0.6 3

PART – B

Two Mark Questions

11. Define binary operation on the set verify. whether the operation defined m Z by a

b = ab + 1 is binary or not .

Ans. A function : A A A de fined by (a, b ) = a b is called a binary

operation an a set A .

Given ab = ab + 1 on Z



on Z is a function.

is a binary operation on Z

12. Write

-1

2

1cot

x -1

, x > 1 in the simplest form

Ans .

Put x = sec = sec-1

x

-1

2

-1

2

-1

-1

-1

1cot

x -1

1=cot

sec θ -1

1=cot

tan θ

=cot cotθ

= θ

= sec .x

13. Find the equation of line passing through the points (3 , 2) and (-1 , -3) by using

determinants.

Ans. Equation of the line is 1 1

2 2

x y 1

x y 1 = 0

x y 1

x y 1

3 2 1 = 0

-1 -3 1

x(2+3) –y (3+1)+1(-9+2) = 0

5x – 4y -7 = 0



14. Show that 2 tan-1

x = sin -1

2

2x

1 + x , | x | 1

Ans. Let tan -1

x = = 1 x = tan

2

2tanθsin 2θ =

1 + tan θ

-1

2

-1 -1

2

2xsin(2 + tan x ) =

1 + x

2x2tan x = sin

1 + x

15.

3

-1 3x - x -1 1If y = tan < x <

1- 3x2 3 3

Ans.

3

-1

2

3x - xy = tan

1- 3x

y - = 3 tan x

2

dy 3=

dx 1 + x

16. Find dy

dx if sin

2 x + cos

2 y = K1 where K is constant

Ans sin2

x + cos2 y = K

2sin x + cos2 y = K

2 sin x + cos x + 2cos y - siny

dy -2sin x cos x=

dx -2 sin y cos y

sin2x=

sin 2y

17. If the radius of a sphere is measured as 7cm with an error of 0.02m.then find the

approximate at error in calculating its volume

Ans. given r = 7 cm , r = 0.02



34V = πr

3

dvdv = Δ r

dr

4=

3π × 3

2

2

3

r × Δr

= 4πr × Δr

= 4π × 49× 0.02

= 3.92πm

18. 2

cos 2xEvaluate dx

(cos x +sin x )

Ans.

2

2

2

cos 2xdx

(cos x + sin x )

cos2 x -sin 2xdx

(cos x +sin x )

(cos x + sin x)(cos x -sin x)dx

(cos x +sin x)

cos x -sin xdx

cos x +sin x

= log | cos x + sin x / + C

19. -1

Evauate tan x dx

Ans.



-1

-1

-1

2

-1

2

-1 2

2

2

I = tan x dx

= tan x ×1dx

1= tan x× x - x× dx

x

1 2x= x tan x - dx

2 1 + x

1= xtan x - log | 1 + x | + c

2

1= x tan x - log | 1 + x | +c

2

= x tanx - log | 1 + x | +c

20. Find the projection of the vector i + 3j + 7k on the 7i – j+8k

Ans.

let a = i + 3 j + 7k b = 7 i - j + 8K

a .bpropcehion fa on b =

| b |

7 - 3 + 56=

49 + 1 + 64

60=

114

21. Find the area of parallelogram whose adjacent sides are given by the vectors

a = 3i + j+ 4K and b - i - j+ k

Ans. Area g a parallelogram = a b

Ans. i j k

a × b = 3 1 4

1 -1 1

= I (1+4) -j(3 - 4) + k (-3 -1)

= 5i +j – 4k

a×b = 25 +1+16 = 42

Area of a parallelogram = 42 sq. units .

22. Find the distance of a point ( 21 51 -7) from the plane r . (6i - 3j+ 2k) = 4

Ans. Given plane in Cartesian form is 6x – 3y =+2z = 4



6(2) - 3(5) + 2(-7) - 4Distance =

36 + 9 + 4

12 -15 -14 - 4=

7

-21=

7

= | - 3|

= 3 units

23. Find and the order and the degree of the differential equation 3 2

3 2

d y d y dy+ + = 0

dx dx dx

Ans. Order = 3, degree = 1

24. Given that the event A and B are such that P(A) = 1

2, P(AB)= 3

5 and P(B) = K,

find K if A and B are independent .

Ans.

P(AB) = P (A)P(B)

3 1= K

5 2

6K =

5

PART – C

Three Mark Questions

25. Show that to relation R in the set of all integers Z defined by R = { (a. b) : 2 divides

a – b} is an equivalence relation .

Ans.

(a) Reflective relation :

2 divides a – a ( a , a ) G R

R is reflective

(b) Symmetric relation



( a, b) R 2 divides a – b

2 divides b – a

( b , a)R

R is symmetric

(c) Transitive relation

(a , b ) and (b, c) R 2 divides a –b and 2 divides b – c

2 divides a – b+ b – c

2 divides a – c

( a , c) R

R is transitive

R is an equivalence relation.

26.

-1 2 cos x - 3sin x 2Simplify : tan , tan x > - 1.

3cos x + 2 sin x 3

Ans.

-1

-1 -1

-1

÷ Nr and Dr by 3cos x

2- tan x

3= tan2

1 + tan x3

2= tan - tan tan x

3

2= tan - x

3

27. Express matrix

1 2A =

2 -1as the sum of a symmetric and skew - symmetric matrix .

Ans



.

1

1

1

1 1

1 2A =

2 -1

1 2 1 2 2 4A + A = + =

2 -1 2 -1 4 -2

1 2 1 2A - A = -

2 -1 2 -1

0 0=

0 0

1 1A = (A + A )+ (A - A )

2 2

2 4 0 01 1= +

4 -2 0 02 2

1 2 0 0= +

2 -1 0 0

28. Prove that if the function is differentiable at a point C then it is also continuous at

that point.

Ans. since f is differentiable at c , we have

x c

x c x c

x c x c x c x c

1

x c

f(x) - f(c)lim = f ;(c)

x - c

But for x c we have

f(x)- f(c)f(x)- f(c) = (x - c)

x - c

f(x)f(c)lim f(x) - f(c) = lim (x - c )

x - c

f(x) - f(c)limf(x) - limf(c) = lim lim(x - c)

x - c

= f (c)(o)

= o

lim f(x) = f (c)

Hencef is continous at x = c

29. Verify mean value theorem for the function fx) = x2 – 4x - 3 in the interval [ 1, 4].

Ans. f(x) is continuous in [ 1 , 4 ] and differentiable in (1 , 4) .

f(a) = f(1) = 1-4-3 = 0

f(b) = f(4) = 16-16-3 = -3

f1 (x) = 2x-4

f1( c)= 2c – 4

By mean value tacorem



1f(b) - f(a)= f (c)

b - a

-3 - 0= 2c - 4

4 -1

-3= 2c - 4

3

-1 = 2c - 4

2c = 3

3c = c (1, 4)

2

30. Find the equation of tangent to the curve given by x = a sin3 t, y = b cos

3 t at a point

where t = π

2 .

Ans. Equation of tangent is y - y1 = m (x – x1) x1 = a sin3π

2 = a (1) = a .

y1= b cos 3

π

2 = b(0) = 0

2

2

dy dy / dt 3bcos t × -sint= =

dx dx / dt 3asin t ×cost

bcos t= -

a sin t

b= - cot t

a

dy π -b π -b/ t = = cot = = m

dx 2 a 2 a

-btangent s y = (x - a)

a

Contunetion q.

31. 2

x + 2Evaluate dx

2x + 6x + 5

Ans.



2

2 2

2

2

2

dPut x + 2 = m (2x + 6x + 5)+ n

dx

x + 2 = m(4x + 6) + n

x + 2 = 4mx + 6m + n

4m = 1 6m + n = 2

1 1m = 6× + n = 2

4 4

3+ n = 2

2

3n = 2 -

2

1n =

2

52x + 6x + 5 = 2 x + 3x +

2

3 9 5= 2 x + - +

2 4 2

3 -9 + 10= 2 x + +

2 4

3 1= 2 x + +

2 4

3= 2 x +

2

2 2

2

2 2 2

1+

2

d2x + 6x + 5

x + 2 1dxdx = m dx + n dx2x + 6x + 5 2x + 6x + 5 2x + 6x + 5

2

2 2 2

d(2x + 6x + 5)

1 1 1dx= dx + dx4 2x + 6x + 5 2 3 1

2 x + +2 2

2 -1

3x +

1 1 2= log 2x + 6x + 5 + + tan + c14 4

2



2

2 2 2

2 -1

d(2x + 6x + 5)

1 1 1dx= dx + dx4 2x + 6x + 5 2 3 1

2 x + +2 2

1 1= log 2x + 6x + 5 + + tan (2x + 3) + c

4 4

32.

x 1 + sin x

Evaluate e dx1 + cos x

Ans.

x

x

x

2 2

x 2

x 2

x x 1

x

1 + sin xe dx

1 + cos x

1 sin x= e + dx

1 + cosx 1 + cos x

x x2sin cos

1 2 2= e + dxx x

cos 2cos2 2

1 x x= e sec + tan dx

2 2 2

x 1 x= e tan + sec dx

2 2 2

n= e tan + c e (f(x) + f (x) dx

2

= e ×f(x) + c

33. Find the area boundary by parabola y2 = 4x and the line y = 2x .

Ans. y2 = 4x

(2x)2

= 4x

4x2-4x= 0

4x (x-1) = 0

x = 0 or y = x = 1

y = 0 or y = 2



The f intersection are (0 , 0 ) and ?(1 , 2)

1

0

1

2

0

1

0

13/2

0

1 1

2

0 0

12

0

1 2

y dx

2 x dx y = 4x y = 2 x

3x

2= 23

2

4= x

3

4= 1- 0

3

4=

3

Δ = y dx = 2x dx

x= 2

2

4=1 - 0 = 1 Required Area = -1 A - A

3

1= s.q units .

3

34. Three vectors a, b ,c

are such that a + b + c = 0

and

a =1, | b|= 4, c = 2 find a ,b + b .c + c .a

Ans.

2 22 2

a + b +c = a + | b | - | c | (a . b +b .c +c .a )

0 = 1+ 16+ 4+ 2(a .b + b . c + c . a)

0 = 21+ 2(a .b + b.c + c . a )

-21a .b + b .c + c .a =

2

35. a.= 2i - 3j+ 4k,b = i + 2j- 3k and c= 3i + 4j- k then find a a×b .c

Ans.

2 -3 4

a . b ×c = 1 2 -3

3 4 -1



=2(-2 +12) +3 (-1+9)+4(4 – 6)

= 2 (10)+3(8) +4 (-2)

= 20+24-8

= 36

Since

a .(10) + 3(8) + 4(-2)

= 20 + 24 - 8

= 36

Since a .(b×c) = a×b .c

(a×b) . c = 36

36. Find the shortest distance between the lines.

r = i + j + (2i - j + k)and

r = 2i + j - K + M(3I - 5j + 2K)

Ans. Shortest distance = 2 1 1 2

1 2

(a - a ). a ×b

| b ×b |

2 1a . a = i - K

1 2

1 j k= i(-2 + 5) - j(4 - 3) + k(-10 + 3)

b ×b = 2 -1 1= 3i - j - 7k

3 -5 2

37. From the differential equation of the family of circles touching the x – axis at origin .

Ans. 2 1 1 2(a - a ). b ×b = (1)(3) + (0)(-1) + (-1)(-7)

= 3 + 7

= 10

10

Shortest distance =59

38. An insurance company insured 2000 scooter drivers, 4000 car drivers and 60000

truck drivers, the probability of an accident are 0 .01 , 0.03 and 0.15 respectively. One

of the insured person meets with an accident. What is the probability that he is a

scooter driver?

Ans. E1 : An event is insured scooter drivers .

E2 : An event of insured car drivers .

E3 : An event of insured truck drivers .

A : An event of insured getting accident .



1

2

3

1

2

3

1

1 1

1

1 1 2 2 3 3

2000P(E ) =

12000

4000P(E )=

12000

6000P(E ) =

12000

P(A / E ) = 0.01

P(A / E ) = 0.03

P(A / E ) = 0.15

P(E / A) = ?

P(E )P(A / E )P(E / A) =

P(E )P(A / E ) + P(E )P(A / E ) + P(E )P(A / E )

1

1

2000× 0.01

12000P(E A ) =2000 4000 6000

× 0.01 + × 0.03 + × 0.1512000 12000 12000

2× 0.01=

2× 0.01 + 4× 0.03 + 6× 0.15

1=

1 + 2× 3 + 3×15

1=

1 + 6 + 45

1=

52

PART - D

Five Mark Questions

39. Let f = N R defined by f(x) = 4x2

+ 12x +15, show that f = N S1 where S is the

range of the function is invertible . Also find f -1

.

Ans.

(i)one – one function

1 2

2 2

1 1 2 2

2 2

1 2 1 2

F(x ) = f(x )

4x + 12x + 15 = 4x + 12x + 15

4x - 4x + 12x -12x = 0



2 2

1 2 1 2

1 2 1 2 1 2

1 2 1 2

1 2 1 2

1 2

÷ 4

x - x + x - x = 0

(x + x )(x - x ) + (x - x ) = 0

(x - x )(x + x + 1) = 0

x - x = 0 or x + x + 1 = 0

x = x N

f is one - one function

(ii) onto function for every y Co – domain then there exists x co – domain such

that

2

2

2

2

2

f(x) = y

4x2 + 12x + 15 = y

154 x2 + 3x + = y

4

3 9 154 x + - + = y

2 4 4

34 x + - 9 + 15 = y

2

34 x + + 6 = y

2

34 x + = y - 6

2

3 y - 6x + =

2 4

y - 63 3x + = -

2 2 2

y - 6 - 3x = domain

2

f is onto

F is objective

F is invertible

T o find f – 1

Consider x = y - 6 - 3

2



-1

-1

y - 6 - 3f (y) =

2

x - 6 - 3f (x) =

2

40. Verify the equations (B+C) A = BA + CA

1 2 3 1 1 2 1 0 -1

If A = 2 3 1 , B = 1 2 1 , C = -1 1 0

3 1 2 2 1 1 1 -1 0

Ans.

1 1 2 1 0 -1

B + C = 1 2 1 + -1 1 0

2 1 1 -1 -1 0

2 1 1

= 0 3 1

3 0 1

2 1 1 1 2 3

(B + C) A = 0 3 1 2 3 1

3 0 1 3 1 2

2 + 2 + 3 4 + 3 + 1 6 + 1 + 2

= 0 + 6 + 3 0 + 9 + 1 0 + 3 + 2

3 + 0 + 3 6 + 0 + 1 9 + 0 + 2

7 8 9

= 9 10 5 (1)

6 7 11

1 1 2

BA =

1 2 3

1 2 1 2 3 1

2 1 1 3 1 2

1 + 2 + 6 2 + 3 + 2 3 + 1 + 4

= 1 + 4 + 3 2 + 6 + 1 3 + 2 + 2

2 + 2 + 3 4 + 3 + 1 6 + 1 + 2

9 7 8

= 8 9 7

7 8 9

1 0 -1 1 2 3

CA = -1 1 0 2 3 1

1 -1 0 3 1 2



1 + 0 - 3 2 + 0 -1 3 + 0 - 2

= -1 + 2 + 0 -2 + 3 + 0 -3 + 1 + 0

1- 2 + 0 2 - 3 + 0 3 -1 + 0

-2 1 1

= 1 1 -2

-1 -1 2

9 7 8 -2 1 1

BA +CA = 8 8 7 + 1 1 -2

7 9 9 -1 -1 2

7 8 9

= 8 10 5 (2)

9 7 11

from (1) and (2)

(B + C) A = BA + CA

41. Solve the equations x – y + 3z = 10, x – y = - 22 x + 3y + 4z = 4 by Matrix Method .

Ans.

1 -1 3 x 10

A = 1 -1 -1 X = y B = -2

2 3 4 z 4

| A| =1(-4+ 3) + 1)(4 + 2) + 3(3 + 2)

= -1+ 6+ 15 = 20

-(-4 + 3) -(4 + 2 +(3 + 2)

Co -factor matrix = -(-4 - 9) +(4 - 6) -(3 + 2)

+(1 + 3) -(-1 - 3) +(-1 + 1)

-1

-1 -6 5

= 13 -2 -5

4 4 0

-1 -6 5

adj A = 13 -2 -5

4 4 0

1A = adjA

| A |



Ans.

-1

-1

-1 13 41

= -6 -2 420

5 -5 0

X = A B

-1 13 4 101

A = -6 -2 4 -220

5 -5 0 4

-10 -26 +161

= -60 +4 +1620

50 +10 +0

-201

= -4020

60

-1

= -2 x = 1, y = - 2 , z = 3

3

42. If g = Ae mx

+ Be nx

+ then prove that 2

2

d y dy- (m + n) + mn y = 0

dx dx

Ans.

mx nx

mx nx

2

2 2

2

2

2

2 mx 2 nx mn nx mx nx

2 mx 2 nx 2 mx nx mx

2 nx mx nx

y = Ae + Be

dymAe + n Be

dx

d y= m Aemx + n Benx

dx

d y dyLHS = - (m + n) + mny

dx dx

= m Ae + n Be - (m + n)(mAe + nBe ) + mn(Ae + Be )

= m Ae + n Be - m Ae - mn Be - nm Ae

-n Be + mn Ae + mn Be

= 0 = RHS



43. A ladder 5 m long is leaning against a wall. The bottom of the ladder is pulled along

the ground away from the wall at the rate of 2 cm / sec . how fast is its height on the

wall decreasing when the foot of the ladder is 4 m away from the wall?

Ans.

2 2 2

2

dy= 2cm / sec

dt

dy= ? x = 4 m

dt

x + y = 5

16 + y = 25

2

2

2 2 2

y = 25 -16

y = 9

y = 3

x + y = 5

dx dy2x + 2y = 0

dt dt

÷2

dx dyx + y = 0

dt dt

dy4(2) + (3) = 0

dt

dy -8= cm / sec

dx 3

44. Find the integeral of2 2

1

x - a

with respect to x and hence evaluate 2

1dx

x - 2x

Ans.



2 2 2

2 2

xLet I = Put x = a sec θ = sec θ

a

dx a secθ tanθ dθ

a secθ tanθdθI =

a sec θ - a

a sec θtanθ dθ=

a sec θ-1)

asecθ tan θ= dθ

atanθ

= sec θ dθ

2

2 2

2 2

2 2

2 2 1

2 2 2

2 2 2

2

= log sec θ + tanθ +c

= log secθ + sec θ -1 + c

x x - a= log + + c

a a

x + x - a= log + c

a

= log x + x - a - log a +c

I = log x + x - a + c

Consider x - 2x = (x -1) -1

1 1dx = dx

x - 2x (x -1) -1

= log x -1 + x - 2x + c

45. Using integration find the area bounded by the curve x2 – y

2 = 16 and the parabola y

2

= 6x.

Ans.

X2 +y

2 = 16

X2 +6x = 16

X2

+6x - 16 = 0



X2 +8x – 2x – 16 = 0

X(x+8) -2(x+8) = 0

(x+8)(x-2)-= 0

X = - 8 or x = 2

2

2

0

2

3/2

0

b n dx y = 6x

y = 6 x

x=

3

2

4

2 2 2

2

42

2 2 -1

2

-1 -1

2 6= 8 -0

3

8 3=

3

A = 4 - x dx

x 4 x= 4 - x + sin

2 2 4

16 1= 0 + sin 1 - 12 + 8sin

2 2

π π= 8× - 12 - 8×

2 6

4π= 4π - 12 -

3

1 2

8π- 2 3

3

Required area

= 2 A + A

8 3 8π= 2 + - 2 3

3 3

2 3 8π= 2 +

3 3

4= 3 + 4π Sq. units

3



46. Derive the equation of the line in space passing through a point and parallel to a

vector both in the vector and Cartesian form .

Ans.

(a) Vector form

Let a plane be passing through a point A (a )

And parallel to a given vector b

p(r )

Be any point on a line .

AP | | b

AP = λ b

op - OA = λb

r - a = λ b

r = a = λ (1)

(b)Cartesian form

1 1 1

Let r = xi + yj + zk

a = x i + y j + z k

b = ai + bj + ck

Using equation (1)

1 1 1

1 1 1

1 1 1

xi + yj + zk = (x i + y j + z k) + λ(ai + bj + ck)

xi + yj + zk = (x aλ)i + (y + bλ)j + (z + cλ)k

comparing efficients f , i, j,k

x = x + aλ,y = y + bλ,zzz + cλ

Elimineting λfrom above equations

1 1 1x - x b - b z - z= =

a b c



47. Solve the differential equation x log x dy 2

+ y = log xdx x

Ans.

2

1dx

log(log x)x log x

2

-2

-1 -1

2

dy 2x log x + y = log x

dx x

÷ x log x

dy y 2+ =

dx log x x

I.F =e = e = log x

Solution

2y = log x = log x dx

x

= 2 log x × x dx

1= 2 log x ×(x ) - -x × dx

x

-log x 1= 2 + dx

x x

-log x 1= 2 + - + c

x x

-2 log x 1y log x = - + c

x x

-y =

2 1 c- +

x x log x log x

48. A pair of dice is thrown 4 times . if getting a doublet is considered a success find

the probability of two successes .

Ans.



2 2

2

2

6 1 1 5n = 4 , p = = ,q = 1- =

36 6 6 6

5 1p (x = 2 )4c

6 6

4=

× 3

22 2

25 1× ×

6 6×1

25=

216

PART - E

Ten Mark Question

49.

a

0a

a

2 f(x)dx if f is even

(a)Prove that f(x)dx = 0

0 if f is odd

1 + a 1 1

(b) prove that 1 1 + b 1 = ab + bc + ca + abc

1 1 1 + c

Ans

.

a

-a

a a

-a 0

0

-a

LHS = f(x)dx

= f(x)dx + f(x)dx (1)

consider f(x)dx

put x = - t

dx = - dt



0 0

-a a

a

0

a

0

x = - a r = a

x = 0 t = 0

f(x)dx = f(-t)(-dt)

= f(t)dt

= f(-x)dx (2)

f(-x) = f(x) f is even

(3)

f(-x)= -f(x) if f is odd

a a a

0 0 0

using (1) , (2) and(3)

LHS = f(x)dx + f(x)dx = 2 f(x)dx if f is even

a a

0 0

a

a

0

0

π

9 92

LHS = -f(x)dx + f(x)dx = 0 if f isodd

2 f(x)dx if f (x) isevenf(x)dx =

0 if f(x)x) is odd

tan x dx = 0 tan x is a odd

-π

2

1 1 2 2 2 3

1 + a 1 1

(b) LHS = 1 1 + b 1

1 1 1 + c

R R R and R R R

a -b 0

= o b -c

1 1 1 + c

= a(b + bc + c)+ b(o + c) + o

= ab + abc + ac + bc

= ab + bc + ca + abc

= RHS

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