JEE Main 2014 · 19 Apr · Shift 1 · Mathematics Q70

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If the function
$f(x)=\begin{cases}\dfrac{\sqrt{2+\cos x}-1}{(\pi-x)^2},&x\ne\pi\\k,&x=\pi\end{cases}$
is continuous at $x=\pi$ , then k equals :

\dfrac12

\dfrac14

Answer and worked solution

Official answer: D

Let $t=\pi-x$ : $\cos x=-\cos t\approx-1+\frac{t^2}2$ . So
$\sqrt{2+\cos x}-1\approx\sqrt{1+\tfrac{t^2}2}-1\approx\frac{t^2}4.$
The limit is $\frac14$ , so $k=\frac14$ .

Topics:Continuity Limits

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JEE Main 2014 · 19 Apr · Shift 1, Mathematics question 70