NIMCET 2012 Question Paper with solutions

Let f(z) =

sin2x

∫

0

sin−1√t dt +

cos2x

∫

0

cos−1√t dt
Differentiating on both sides by Leibnitz rule,
f ' (x) = sin−1 (sin x) (2 sin x cos x) + cos−1 (cos x) (- 2 sin x . cos x)
= x . sin 2x - x . sin 2x
= 0
⇒ f (x) = Constant
Now, we check the constant value of this integration ondifferent value of x.
(i) At (x=

π

4

)
f (

π

4

) =

1 2

∫

0

sin−1√t dt +

1 2

∫

0

cos−1√t dt
=

1 2

∫

0

(si−√t + cos−√t) dt =

1 2

∫

0

π

2

dt
=

π

2

(

1

2

−0) =

π

4

(ii) At (x = 0)
f (0) = 0 +

1

∫

0

cos−1√t dt
Let t = cos2θ , dt = - sin 2θ sθ
= -

0

∫

π 2

θ . sin 2θ dθ
= -

π 2

∫

0

θ . sin 2θ dθ (Since

b

∫

a

f (x) dx =

a

∫

b

f (x) dx)
= [−θ

cos2θ

2

+

1

4

sin2θ]0

π

2

= [−

π

2.

1

2

(−1)+0] =

π

4

(iii) At (x=

π

2

)
f (

π

2

) =

1

∫

0

sin−1√t dt + 0
Let t = sin2θ , dt = sin 2θ dθ
=

π 2

∫

0

θ . sin 2θ . dθ = [−θ.

cos2θ

2

+

sin2θ

4

]0

π

2

= [−

π

2

.

1

2

(−1)+0] =

π

4