Mathematics · 2018
JEE Main · 15 April 2018, Shift 2 · Q71
Let f(x)= {(x-1)^(1/(2-x)), x>1, x ≠ 2; k, x=2} The value of k for which f is continuous at x=2 is:
Let $\displaystyle f(x)= \begin{cases}(x-1)^{\frac{1}{2-x}} & , x>1, x \neq 2 \\ \mathrm{k} & , x=2\end{cases}$
The value of k for which $\displaystyle f$ is continuous at $\displaystyle x=2$ is :
Official answer
(3)
$\displaystyle \mathrm{e}^{-1}$
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JEE Main 2018 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.