Mathematics · 2014
JEE Main · 19 April 2014, Shift 1 · Q70
If the function f(x)= {(√(2+ cos x)-1)/((π-x)^2), x ≠ π; k, x=π} is continuous at x=π, then k equals:
If the function
$$f(x)= \begin{cases}\frac{\sqrt{2+\cos x}-1}{(\pi-x)^2} & , x \neq \pi \\ \mathrm{k} & , x=\pi\end{cases}
$$
is continuous at $\displaystyle x=\pi$, then k equals :
Official answer
(4)
$\displaystyle \frac{1}{4}$
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JEE Main 2014 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.