JEE Main 2016 · 3 Apr · Shift 1 · Mathematics Q72

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MathematicsSingle correct+4 / -1Question 72 of 90

Consider $f(x)=\tan^{-1}\left(\sqrt{\dfrac{1+\sin x}{1-\sin x}}\right)$ , $x\in\left(0,\dfrac\pi2\right)$ . A normal to y = f(x) at $x=\dfrac\pi6$ also passes through the point :

(0, 0)

\left(0,\dfrac{2\pi}3\right)

\left(\dfrac\pi6,0\right)

\left(\dfrac\pi4,0\right)

Answer and worked solution

Official answer: B

$\dfrac{1+\sin x}{1-\sin x}=\tan^2\left(\frac\pi4+\frac x2\right)$ , so $f(x)=\frac\pi4+\frac x2$ and the slope is $\frac12$ . At $x=\frac\pi6$ , $y=\frac\pi3$ , and the normal $y-\frac\pi3=-2\left(x-\frac\pi6\right)$ passes through $\left(0,\frac{2\pi}3\right)$ .

Topics:Tangents and normals Inverse trigonometric functions

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JEE Main 2016 · 3 Apr · Shift 1, Mathematics question 72