QUESTION PAPER CODE 65/2 EXPECTED ANSWERS/VALUE POINTS SECTION - A Marks 1-10. 1. 3

2. – 2

3. x sin x

5. 1

6. – I

7. p = −

4. {1, 2, 3} 1 3

8. r = (3ˆi − 4ˆj + 3kˆ) + λ (–5ˆi + 7ˆj + 2kˆ)

9. log 2

10. a = 5ˆi + 5kˆ

1×10 = 10 m

SECTION - B

11.

[

y = [x (x – 2) ] = x 2 – 2x 2

dy = 0 dx

(

)

dy = 2 x 2 – 2x (2x – 2 ) dx

∴

⇒

1m

1m

x = 0, x = 1, x = 2

½m

Intervals are (– ∞ , 0), (0, 1), (1, 2), (2, ∞ )

½m

since ∴

2

dy = 4 x (x – 1) (x – 2 ) dx

⇒

∴

]

dy > 0 in (0, 1) or (2, ∞ ) dx

f(x) is increasing in (0, 1) U (2, ∞ )

1m

OR

x2 y2 2x 2y dy dy b2 x – = 1 ⇒ – = 0 ⇒ = a2 b2 a 2 b 2 dx dx a2y

slope of tangent at

(

2a , b =

)

slope of normal at

(

2a , b = –

)

2b a

1m

½m

a 2b

½m

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(

2b x– 2a a

Equation of tangent is y – b =

(

π

I =

½m

4x sin x dx 2 x π

π

4 (π – x) sin (π – x) 4 (π – x) sin x dx = ∫ dx 2 1 + cos (π – x) 1 + cos 2 x 0

∫

x → (π – x) gives I =

0

π

sin x

∫ 1 + cos

2I = 4 π

2

0

x

dx

–1

I = 2π

1

∫ 1

– dt 1+ t2

[

]

= 2 π tan –1 t

1m

½m Put cos x = t ∴ sin x dx = – dt

∴

½m

∫ 1 + cos 0

∴

)

2 (a2 + b2)

2 by =

i.e. ax +

½m

a x– 2a 2b

and equation of normal is y – b = –

Let

½m

2 bx – ay = ab

i.e.

12.

)

dt

∫ 1+ t

or 2 π

½m

1m

2

–1

1 –1

 π  π  = 2 π  –  –  = π 2  4  4 

1m

OR I=∫

=

=

1 2

x +2 x 2 + 5x + 6

∫

dx =

2x + 5 x 2 + 5x + 6

x 2 + 5x + 6 –

∫

dx –

1 1 (2x + 5) – 2 2 dx 2 x + 5x + 6 1 2

∫

1m

dx

(x + 5 2 ) –  12  2

2

1 5  log  x +  + x 2 + 5x + 6 + c 2 2 

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½+½ m

1+1 m

13.

y = P e ax + Q e bx ⇒

dy = a P e ax + b Q e bx dx

1m

d2y = a 2 P e ax + b 2 Q e bx 2 dx

∴ LHS =

1m

d2y dy – (a + b) + aby 2 dx dx

{

}

{

= a 2 P e ax + b 2 Q e bx – (a + b) a P e ax + b Q e bx + ab P e ax + Q e bx

{

}

{

}

= P e ax a 2 – a 2 – ab + ab + Q e bx b 2 – ab – b 2 + ab

}

1m 1m

= 0 + 0 = 0. = R.H.S.

14.

Putting x = cos θ in LHS, We get  1 + cos θ – 1 – cos θ  LHS = tan–1    1 + cos θ + 1 – cos θ 

1m

 2 cos θ – 2 sin θ  2 2 = tan–1  θ θ   2 cos + 2 sin 2 2 

1m

 1 – tan θ  2  = tan −1  tan  π − θ  = tan–1   2  1 + tan θ  4  2  =

π 1 π 1 − θ = − cos −1 x = R.H.S 4 2 4 2

½+1 m

½m

OR Given equation can be written as  x – 2  = tan–1 1 – tan–1 tan–1  x – 4  

 x + 2    x + 4 16

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½m

= tan–1

∴

⇒

15.

 1–   1 + 

x+2  x+4 –1 x + 2  = tan x + 4 

 2     2x + 6 

x–2 1 = x–4 x+3

1+½ m

½m

x 2 + x – 6 = x – 4 or x 2 = 2 ∴ x = ± 2

½+1 m

Given differential equation can be written as –1 dy 1 1 + ⋅y = ⋅ e tan x 2 2 dx 1 + x 1+ x

1m

1

Integrating factor = e ∴ solution is, y ⋅ e tan

⇒

y ⋅ e tan

or

16.

–1

x

y=

–1

x

=

=

∫ 1+ x 2 dx 1

∫ 1+ x

2

= e tan e 2 tan

–1

–1

x

x

1m dx

1 2 tan –1x e +c 2

1m

1m

–1 1 tan –1x e + c e – tan x 2

A, B, C, D are coplaner, if AB ⋅ AC × AD = 0

AB = – 4ˆi – 6ˆj – 2kˆ, AC = − ˆi + 4ˆj + 3kˆ, AD = − 8ˆi – ˆj + 3kˆ

1m 1½ m

–4 –6 –2 AB ⋅ AC × AD =

–1 –8

4 –1

3 3

½m

= – 4 (15) + 6 ( 21) – 2 (33) = 0

OR

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1m

Given that a ⋅

b+c b+c

=1

½m

a ⋅ b + a ⋅c = b + c

or

½m

(ˆi + ˆj + kˆ )⋅ (2ˆi + 4ˆj – 5kˆ ) + (ˆi + ˆj + kˆ )⋅ (λˆi + 2ˆj + 3kˆ ) = ⇒

(2 + 4 – 5) + (λ + 2 + 3) =

∴

(λ + 6 )2

= (λ + 2 ) + 40

=

b+c

(λ + 2)2 + 36 + 4

⇒ λ =1

2

b+c

Hence

(λ + 2)ˆi + 6ˆj – 2kˆ

3ˆi + 6ˆj – 2kˆ 7

1m ½m

3ˆ 6ˆ 2 ˆ i+ j– k 7 7 7

or

½m

1m

2

17.

x   x  getting fog (x) = f   +2 =  x –1  x –1

1½ m

fog(2) = 6

½m

(

)

getting g of (x) = g x + 2 = 2

x2 + 2 x2 +1

g of (– 3) =

18.

1½ m

11 10

½m

Let probability of success be p and that of failure be q ∴

p = 3 q, and p + q = 1

∴

p=

3 4

1 4

1m

P (atleast 3 successes) = P(r > 3) = P(3) + P(4) + P(5)

½m

2

=

5

and q =

3

1

4

1 3 1 3 3 C3   ⋅   + 5 C 4   ⋅   + 5 C5   4 4 4 4 4

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5

1½ m

=

19.

10.27 5.81 243 918 459 + + = or 1024 1024 1024 1024 512

Operating C1 → C1 − (C 2 + C3 ), we get – 2a

c+a

– 2p r + p – 2x z + x

LHS =

C 2 → C 2 – C1 C3 → C3 – C1

a+b

c+a

a+b

p+q = –2 p r +p x+y x z+x

p+q x+y

a

a c b ⇒ LHS = – 2 p r q

1½+½ m

1m

x z y a

20.

1m

b c

C 2 ↔ C 3 = + 2 p q r = RHS x y z

1m

dx 2 = 2a cos 2t (1 + cos 2t) – 2a sin 2t dt

1m

dy = 2b cos 2t sin 2t – 2b sin 2t (1 – cos 2t) dt

1m

∴

dy b  sin 2t cos 2t – sin 2t (1 – cos 2t)    = dx a  cos 2t (1 + cos 2t) – sin 2 2t 

1m

At t = π 4 , sin 2t = 1 and cos 2t = 0 ∴

21.

(

)

dy b  0 – 1 b = at t = π =  4 dx a  0 – 1  a

1m

Given equation can be written as x y dx – dy = 0 2 1+ x 1 + y2

1m

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Integrating to get ⇒ ∴

(

22.

(

)

(

)

(

1m

)

log 1 + x 2 – log 1 + y 2 = log c12 = log c

½m

(1 + x ) = c (1 + y ) 2

2

x=0 y=1 ⇒ c=

∴

)

1 1 log 1 + x 2 – log 1 + y 2 = log c1 2 2

1 2

1m

1 + y2 = 2 (1 + x2) or y =

2x 2 + 1

½m

Let the D.R’s of the required line be a,b , c ∴

    

a + 2b + 3c = 0

and –3a + 2b + 5c = 0 ⇒

∴

1m

a b c = = ∴ DRs are 2, – 7, 4 4 – 14 8

Equations of line are

1m

x–2 y –1 z –3 = = 2 –7 4

(

) (

which, in vector form is, r = 2ˆi + ˆj + 3kˆ + λ 2ˆi – 7ˆj + 4kˆ

1m

)

1m

SECTION - C 23.

Let number of pieces of type A and type B, manufactured per week be x and y respectivily ∴

L.P.P. is

Maximise P = 80x + 120y

½m

subject to 9x + 12y < 180 or 3x + 4y < 60 x + 3y < 30 x>0 y>0

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    

2m

For correct graph :

2m

Vertices of feasible region are A (0, 10), B (12, 6), C (20, 0) P(A) = 1200, P(B) = 1680, P(C) = 1600 ∴

For Max. P, No. of type A = 12 1m No. of type B = 6

Maximum Profit = Rs. 1680 24.

Let event E1 : choosing first (two headed) coin E2 : choosing 2nd (biased) coin E3 : choosing 3rd (biased) coin ∴

P(E1) = P(E2) = P(E3) =

    

1 3

½m

½m

1m

A : The coin showing heads. ∴

P(A/E1) = 1, P(A/E2) =

75 3 60 3 = , P(A/E3) = = 100 4 100 5

1 ⋅1 3 P(E1/A) = 1 1 3 1 3 ⋅1 + ⋅ + ⋅ 3 3 4 3 5

=

1½ m

1+1m

20 47

1m OR

Total number of ways of selecting two numbers = 6 C 2 = 15

½m

Values of x (larger of the two) can be 2, 3, 4, 5, 6

1m

P(x = 2) =

1 2 3 , P(x = 3) = , P(x = 4) = 15 15 15

4 P(x = 5) = 15

5 and P(x = 6) = 15

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2½

∴

Distribution can be written as x:

2 3 4 1 2 3 P(x) : 15 15 15 2 6 12 x P(x) : 15 15 15 Mean = ∑ x P(x) =

25.

5 4 15 20 15

6 5 15 30 15

1m

70 14 = 15 3

1m

Equation of plane through the intersection of given two planes is : x + y + z – 1 + λ (2x + 3y + 4z – 5) = 0

1m

or (1 + 2λ ) x + (1 + 3λ ) y + (1 + 4λ ) z – 1 – 5λ = 0 .......................... (i)

½m

Plane (i) is perpendicular to the plane x – y + z = 0,

so, 1 (1 + 2λ ) – 1 (1 + 3λ ) + 1 (1 + 4λ ) = 0 ⇒

3λ = − 1 ∴ λ = –

1½ m

1 3

½m

5  2  4 ∴ Equation of plane is 1 –  x + (1 – 1) y + 1 –  z – 1 + = 0 3  3  3

i.e x – z + 2 = 0 Distance of above plane from origin =

½m 1m

2 = 2 units 2

1m

OR

(

)

Any point on the line r = 2ˆi – 4ˆj + 2kˆ + λ 3ˆi + 4ˆj + 2kˆ is

(2 + 3λ )ˆi + (– 4 + 4λ )ˆj + (2 + 2λ )kˆ

1½ m

For the line to intersect the plane, the above point must satisfy the equation of plane, for some value of λ ∴

{(2 + 3λ )ˆi + (– 4 + 4λ )ˆj + (2 + 2λ )kˆ}⋅ (ˆi – 2ˆj + kˆ ) = 0 22

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1m

⇒

2 + 3λ + 8 – 8λ + 2 + 2λ = 0 ⇒ λ = 4

∴

The point of intersection is 14ˆi + 12ˆj + 10kˆ

Required distance = 12 2 + 0 2 + 52 = 13 units

26.

Here

3x + 2y + z = 1600 4x + y + 3z = 2300 x

∴

+

3 2 1   4 1 3 1 1 1  

y

+

z

=

1½ m 1m 1m

1½

900

 x   1600       y  =  2300  or AX = B  z   900     

| A | = 3(–2) –2 (1) + 1 (3) = – 5 ≠ 0 ∴ X = A –1 B

½m

Cofactors are : A11 = – 2, A12 = – 1, A13 = 3 A 21 = – 1, A 22 = 2, A 23 = –1 A 31 = 5, A 32 = −5, A 33 = –5

1½ m

 – 2 − 1 5   1600   x       1  ∴  y  = –  – 1 2 − 5   2300  5      z   3 – 1 – 5   900    ∴ x = 200, y = 300, z = 400

1½ m

i.e. Rs 200 for sincerity, Rs 300 for truthfulness and Rs 400 for helpfulness One more value like, honesty, kindness etc.

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1m

27.

Correct fiqure AB is : y =

1 (3x + 7) 2

½m

is : y =

1 (11 – x) 2

½m

is : y =

1 (x + 5) 2

½m

  Equation of BC   AC  Required area =

1

1 2

∫ (3x + 7) dx +

–1

1m

1 2

1

3

∫ (11 – x) dx – 1

[

1 1 2 (11 – x) 2 =  (3x + 7)  – 12  –1 4

]

3

1

–

1 2

I =

∫(

)

cot x + tan x dx =

∫

cos x + sin x dx sin x cos x

Putting sin x – cos x = t, so that (cos x + sin x) dx = dt and sin x cos x =

∴ I =

=

29.

2

∫

1 (1 – t2 ) 2

dt 1– t

2

=

∫ (x + 5) dx

1m

[

1½

–1

1 (x + 5) 2 4

= 7 + 9 – 12 = 4 sq. units

28.

3

]

3 –1

1m

1m 1m 1m

2 sin –1t + c

1+1 m

2 sin –1 (sin x – cos x) + c

1m

Correct fiqure let the radius and height of cylinder be r and h respectively

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½m

∴

V = π r2 h ....................... (i)

½m

2 2 But r2 = R – h 4

  h2  h3  π h  R 2 –  = π  R 2 h –  4  4  

∴

 2 3h 2  dv  = π  R – dh 4  

∴

dv = 0 dh

⇒ h2 =

1m

½m

4R 2 2R or h = 3 3

½+1 m

 6h  d2v  < 0 ∴ Volume is maximum and 2 = π  – dh  4 

1m

 2 2R 1  2R 3  4π R 3 –   = Maximum volume = π ⋅ R ⋅ cubic units 3 4  3   3 3 

1m

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