Mathematics · 2014
JEE Main · 11 April 2014, Shift 1 · Q70
If lim_x → 2 (tan (x-2){x^2+( k -2) x-2 k })/(x^2-4 x+4)=5, then k is equal to:
If
$$\lim _{x \rightarrow 2} \frac{\tan (x-2)\left\{x^2+(\mathrm{k}-2) x-2 \mathrm{k}\right\}}{x^2-4 x+4}=5,
$$
then $\displaystyle k$ is equal to :
Official answer
(4)
$\displaystyle 3$
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JEE Main 2014 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.