NIMCET 2012 Question Paper with solutions

Let f(z) = $\underset{0}{\overset{\mathit{s}\mathit{i}{\mathit{n}}^2\mathit{x}}{\int }}$ $\mathit{s}\mathit{i}{\mathit{n}}^{-1}\sqrt{\mathit{t}}$ dt + $\underset{0}{\overset{\mathit{c}\mathit{o}{\mathit{s}}^2\mathit{x}}{\int }}$ $\mathit{c}\mathit{o}{\mathit{s}}^{-1}\sqrt{\mathit{t}}$ dt
Differentiating on both sides by Leibnitz rule,
f ' (x) = $\mathit{s}\mathit{i}{\mathit{n}}^{-1}$ (sin x) (2 sin x cos x) + $\mathit{c}\mathit{o}{\mathit{s}}^{-1}$ (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 $\left(\mathit{x}=\frac{\mathit{\pi }}{4}\right)$
f $\left(\frac{\mathit{\pi }}{4}\right)$ = $\underset{0}{\overset{\frac{1}{2}}{\int }}$ $\mathit{s}\mathit{i}{\mathit{n}}^{-1}\sqrt{\mathit{t}}$ dt + $\underset{0}{\overset{\frac{1}{2}}{\int }}$ $\mathit{c}\mathit{o}{\mathit{s}}^{-1}\sqrt{\mathit{t}}$ dt
= $\underset{0}{\overset{\frac{1}{2}}{\int }}$ ($\mathit{s}{\mathit{i}}^{-}\sqrt{\mathit{t}}$ + $\mathit{c}\mathit{o}{\mathit{s}}^{-}\sqrt{\mathit{t}}$) dt = $\underset{0}{\overset{\frac{1}{2}}{\int }}$ $\frac{\mathit{\pi }}{2}$ dt
= $\frac{\mathit{\pi }}{2}\left(\frac{1}{2}-0\right)$ = $\frac{\mathit{\pi }}{4}$
(ii) At (x = 0)
f (0) = 0 + $\underset{0}{\overset{1}{\int }}\mathit{c}\mathit{o}{\mathit{s}}^{-1}\sqrt{\mathit{t}}$ dt
Let t = $\mathit{c}\mathit{o}{\mathit{s}}^2\mathit{\theta }$ , dt = - sin 2θ sθ
= - $\underset{\frac{\mathit{\pi }}{2}}{\overset{0}{\int }}$ θ . sin 2θ dθ
= - $\underset{0}{\overset{\frac{\mathit{\pi }}{2}}{\int }}$ θ . sin 2θ dθ (Since $\underset{\mathit{a}}{\overset{\mathit{b}}{\int }}$ f (x) dx = $\underset{\mathit{b}}{\overset{\mathit{a}}{\int }}$ f (x) dx)
= ${\left[-\mathit{\theta }\frac{\mathit{c}\mathit{o}\mathit{s}2\mathit{\theta }}{2}+\frac{1}{4}\mathit{s}\mathit{i}\mathit{n}2\mathit{\theta }\right]}_0^{\frac{\mathit{\pi }}{2}}$
= $\left[-\frac{\mathit{\pi }}{2.}\frac{1}{2}\left(-1\right)+0\right]$ = $\frac{\mathit{\pi }}{4}$
(iii) At $\left(\mathit{x}=\frac{\mathit{\pi }}{2}\right)$
f $\left(\frac{\mathit{\pi }}{2}\right)$ = $\underset{0}{\overset{1}{\int }}$ $\mathit{s}\mathit{i}{\mathit{n}}^{-1}\sqrt{\mathit{t}}$ dt + 0
Let t = $\mathit{s}\mathit{i}{\mathit{n}}^2\mathit{\theta }$ , dt = sin 2θ dθ
= $\underset{0}{\overset{\frac{\mathit{\pi }}{2}}{\int }}$ θ . sin 2θ . dθ = ${\left[-\mathit{\theta }.\frac{\mathit{c}\mathit{o}\mathit{s}2\mathit{\theta }}{2}+\frac{\mathit{s}\mathit{i}\mathit{n}2\mathit{\theta }}{4}\right]}_0^{\frac{\mathit{\pi }}{2}}$
= $\left[-\frac{\mathit{\pi }}{2}.\frac{1}{2}\left(-1\right)+0\right]$ = $\frac{\mathit{\pi }}{4}$