Board Pattern Maths Paper Ii Answers With Marking Skim

R SAFE HANDS Model marking scheme and answers to Board Pattern Mathematics and statistics Paper II Q1.A (1) 2

x + x + x − 14 x−2 x →2 x + x 2 + x3 − 2 − 4 − 8 = lim x−2 x →2 lim

(1M)

(x − 2) + (x 2 − 4) + (x3 − 8) (x − 2) x →2

= lim

x−2 x 2 − 22 x3 − 23 + lim + lim x →2 x − 2 x →2 x − 2 x →2 x − 2

= lim =1

+ 2.(2)2-1

3(

(2) lim

θ→π 6

+3(2)3-1

(1M)

= 17 (1M)

2 cos θ − 1) 3 Let θ = (π/6) +h (1M) 6θ − π

2 π 3( cos( + h) − 1) ∴ Solving 6 3 lim 6h h →0 3 (cosh− 1) − sinh = lim 6h h →0 − 3 (1 − cosh) sinh = lim = − lim 6h h →0 h →0 6h 0

-

1/6

= -1/6

(1M)

x→π

(1M)

Rationalize then

substitute π-x = h by solving equation becomes –

1 − cosh 1 . = f( π) hence f(π) = 1/4 2 h →0 (h) 2 + ( −1) + 1 lim

Q1 B (1)

5x + 2

 1   2 (2) y = cosec-1  2x 1 − x

5x + 2

∫ x2 − 3x + 2 dx = ∫ (x − 2)(x − 1) dx

use integration by partial fraction

2 d dy dy dt (sin −1 x) = = = 2. dx dx dt dx 1 − x2

∴

5x + 2 A B = + solving A=12,B=-7 (x − 2)(x − 1) (x − 2) (x − 1) 12 -7 5x + 2 + dx solving ∫ x2 − 3x + 2 dx = ∫ x -2 x -1

∫ log(x +

x2 + 1)dx = xlog (x + x 2 + 1) - ( x2 + 1) + c 1 dx (2) To solve ∫ let tan(x/2) = t 3 + 2 sin x 2dt 2.t dx = , problem reduces to – 2 , sinx = 1+ t 1 + t2 2dt 1 2dt 1 + t2 dx =∫ 2 = ∫ 3 + 2 sin x ∫ 2t 3t + 4t + 3 3+2 2 1+ t

= 12log|x-2| - 7log|x-1| + c

∫ sin x.log(cos x)dx let cosx = t, -sinxdx = dt ∴ ∫ sin x. log(cos x)dx = ∫ ( −1). log( t)dt integrating

2  3 tan( x / 2) + 2  tan − 1  5 5  

Q3 (A) (b)(1) π /2

∫

I=

(2)

by parts

0

π sec( − x) 2

0

π π sec( − x) + cos ec( − x) 2 2

∫

=

= -t.logt + t + c = -cosx . log(cosx) + cosx + c Q2 A. (1) f(x) = x3 -3x + 5 ∴f’(x) = 3x2 -3 need to use formula f(a + h) ≅ f(a) + h.f’(a) where a = 2 and h = -0.01 hence f(1.99) ≅ 7 – 0.09 = 6.01

sec x dx sec x + cos ecx

π/ 2

1

∫ ( t ).tdt +c

2  3t + 2  tan −1 = 5  5 

1

∫ 3 + 2 sin xdx =

solving

Let

= -[ logt .t -

    let x = sint then

Q3. (A) (a) (1) Use integration by parts or any valid method to get

x→π

2 + cos x − 1 = f( π) ( π − x) 2

Q2 B (1) Solve to show

y = cosec(cosec(2t)) = 2t

(3) As f(x) is continuous at π ∴ lim f( x) = f( π)

lim

5 sin x + 4 cos x 5 sin x + 4 cos x ) let u = 41 41 5 4 considering cosα= , sinα = ∴ u = sin(x+α) 41 41 dy y = sin-1(sin(x + α)) = x + α hence =1 dx f(x + h) − f(x) (3) Use definition f’(x) = lim h h →0 h a −1 and formula lim = ah . log a h →0 h (2) y = sin-1(

3

π/2

∫

I=

0 π /2

2I=

∫

0

cos ec(x) cos ec(x) + sec(x) sec x dx + sec x + cos ecx

π /2

∫

0

dx

dx hence

cos ecx dx cos ecx + sec x

I = π/4 (2) Use partial fraction = 10

dx ∫ (x − 1).(x − 2) = 5 10

=

∫(

5

10

A

B

∫ ( x − 1 + x − 2 )dx

5

10

−1 1 x−2 32 + )dx = log ) = log( ( x −1 x − 2 x −15 27

Q3(B) (1) f(x) = x2 + (100-x)2 to determine minimum find f’(x) = 2x - 2(100-x) to determine critical points let f’(x) = 0 hence 4x – 200 = 0 ∴ x = 50 & f’’(50) = 4 <0 hence at x = 50 there is minima. Thus numbers are 50and 50. (2) y=A.ex+Be-2x∴

dy d2 y =A.ex–2Be-2x& = Aex+4Be-2x 2 dx dx

substitute to verify

d y dy + -2y = 0 dx 2 dx 2

du dy =9 + dx dx

du − 9 = u2 by variable separable dx du 1 −1 9x + y + 1 ∫ u2 + 9 = ∫ dx ⇒ 3 tan ( 3 ) = x + c

equation becomes

is general solution of differential equation. (2)

∫ dy = ∫ x.(100 − x )

2 1/2

dx hence let 100-x2 = t

− 1 ( 3 / 2) −1 t +c= (100 − x2 )(3 / 2) + c ∴-2xdx = dt ⇒ y = 3 3 −1 (100)(3 / 2) + c given that if x= 0 then y=0 hence 0 = 3 ∴ c = (103)/3 particular solution of given D.E. is

−1 1000 (100 − x2 )(3 / 2) + y= 3 3

(b) (1) Use definition to prove required result. (2) To determine polynomial use formula f(x) =(I+∆)xf(0) = (I+x∆+ x f(x) ∆f(x) ∆2f(x) 0 8 4 1 12 2 6 2 18

dy δy = lim as δx→0 , δy→0 dx δx →0 δx δx 1 = consider δy δy δx

taking limit of both sides as δx→0 we get

δx 1 = lim as δx →0 δy δx →0 δy δx 1 δx = 1 lim δy  =  lim   dy dx δy →0 δy δx → 0 δx  dx 1 = hence as limit on RHS exist and finite so dy dy dx lim

Q4(A)(a) dy = (9x + y + 1)2 Let u = 9x + y + 1 hence (1) dx differentiating with respect to x we get

(2) Let δy be the increments in y corresponding to the increment δx in x as y is differentiable function of x

x(x − 1) 2 ∆ ) f(0) 2

x(x − 1) .2 2 = 8 +4x +x2 –x = x2 +3x +8

F(x) = f(0) + x.4 +

Q4 (B) (1) (a+a) = (a+a).1 = (a+a) . (a + a’) = a + a.a’ = a + 0 = a and a.a = a.a + 0 = a.a + a.a’ = a.(a + a’) = a.1 = a (2) Dual of (1 + a).(b + 0) =b is (0 . a) + (b . 1) = b and dual of a . ( a’ + b) = a + b is a + (a’.b) = a.b Q5.(A) (a) (1) Let δu ,δy be the increments in u and y respectively corresponding to the increment δx in x.

dy δy = lim du δu→0 δu du δu = lim as u is differentiable function of x , dx δx →0 δx δy δy δu = . As , consider limit of both sides as δx→0 δx δu δx δy δy δu δy δu lim = lim . = lim lim . δx →0 δx δx →0 δu δx δx →0 δu δx →0 δx δy δy δu = lim lim . as δx→0 δu→0 Hence lim δx →0 δx δu→0 δu δx →0 δx dy dy du = . as limit of RHS are finite and exist hence dx du dx

of LHS also exist. (b) (1) Assume that b

∫ f(x)dx = F(x) and ∫ g(x)dx = G(x) So b

∫ f(x)dx = F(b) − F(a) , ∫ g(x)dx = G(b) − G(a) as a

a

∫ [f(x) + g(x)]dx = F(x) + G(x) Hence

b

∫ [f(x) + g(x)]dx = (F(b) + G(b)) − (F(a) + G(a)) a

=

(F(b) − F( a) ) + ( G(b) − G( a))

=

b

b

a

a

∫ [f(x)dx + ∫ g(x)dx

similarly we can show b

b

b

a

a

a

∫ [f(x) − g(x)]dx =

(2) I =

∫

∫ [f(x)dx − ∫ g(x)dx

x 2 + a 2 dx = ∫ 1. x2 + a 2 dx integration by

parts solve to get I=

x a2 x2 + a2 + log | x + x 2 + a 2 |+ c 2 2

Q5 (B)

b

(1)

a

c L

As y is differentiable function of u,

limit of LHS also is finite and exists.

(2)

A

A’ A’ • B’

B

B’