JEE Main 2016 (Offline)Question 1
Consider $f(x)=\tan^{-1}\left(\sqrt{\frac{1+\sin x}{1-\sin x}}\right)$, $x\in\left(0,\frac\pi2\right)$. A normal to $y=f(x)$ at $x=\frac\pi6$ also passes through the point
Differentiation and its Application
Solution
Concept. The fraction is a perfect $\tan^2$, so $f$ is linear. Formulas used. (1) $\dfrac{1+\sin x}{1-\sin x}=\tan^2\left(\dfrac\pi4+\dfrac x2\right)$Step 1 - Simplify. For $x\in\left(0,\frac\pi2\right)$, $\frac\pi4+\frac x2\in\left(\frac\pi4,\frac\pi2\right)$, where $\tan>0$. By (1), $$f(x)=\tan^{-1}\left(\tan\left(\tfrac\pi4+\tfrac x2\right)\right)=\frac\pi4+\frac x2.$$ Step 2 - Point and slope. At $x=\frac\pi6$: $y=\frac\pi3$, $f'=\frac12$. The normal slope is $-2$. Step 3 - Normal. $y-\frac\pi3=-2\left(x-\frac\pi6\right)$, i.e. $y=-2x+\frac{2\pi}{3}$. It passes through $\left(0,\frac{2\pi}{3}\right)$. Answer: (D)