NIMCET 2012 Question Paper with solutions

Let f(z) = $\int\limits_{0}^{\sin^2 x}$ $\sin^{-1}\sqrt{t}$ dt + $\int\limits_{0}^{\cos^2 x}$ $\cos^{-1} \sqrt{t}$ 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 $\left(x=\frac{\pi}{4}\right)$ f $\left(\frac{\pi}{4}\right)$ = $\int\limits_{0}^{\frac{1}{2}}$ $\sin^{-1} \sqrt{t}$ dt + $\int\limits_{0}^{\frac{1}{2}}$ $\cos^{-1} \sqrt{t}$ dt = $\int\limits_{0}^{\frac{1}{2}}$ ($si^{-}\sqrt{t}$ + $\cos^{-}\sqrt{t}$) dt = $\int\limits_{0}^{\frac{1}{2}}$ $\frac{\pi}{2}$ dt = $\frac{\pi}{2}\left(\frac{1}{2}-0\right)$ = $\frac{\pi}{4}$ (ii) At (x = 0) f (0) = 0 + $\int\limits_{0}^{1} \cos^{-1} \sqrt{t}$ dt Let t = $\cos^2 \theta$ , dt = - sin 2θ sθ = - $\int\limits_{\frac{\pi}{2}}^{0}$ θ . sin 2θ dθ = - $\int\limits_{0}^{\frac{\pi}{2}}$ θ . sin 2θ dθ (Since $\int\limits_{a}^{b}$ f (x) dx = $\int\limits_{b}^{a}$ f (x) dx) = $\left[-\theta \frac{\cos 2\theta}{2} + \frac{1}{4} \sin 2\theta\right]_{0}^{\frac{\pi}{2}}$ = $\left[-\frac{\pi}{2} \cdot \frac{1}{2}(-1)+0\right]$ = $\frac{\pi}{4}$ (iii) At $\left(x=\frac{\pi}{2}\right)$ f $\left(\frac{\pi}{2}\right)$ = $\int\limits_{0}^{1}$ $\sin^{-1}\sqrt{t}$ dt + 0 Let t = $\sin^2 \theta$ , dt = sin 2θ dθ = $\int\limits_{0}^{\frac{\pi}{2}}$ θ . sin 2θ . dθ = $\left[-\theta \cdot \frac{\cos 2\theta}{2} + \frac{\sin 2\theta}{4}\right]_{0}^{\frac{\pi}{2}}$ = $\left[-\frac{\pi}{2} \cdot \frac{1}{2}(-1)+0\right]$ = $\frac{\pi}{4}$