Question
If $\mathop {\lim }\limits_{x \to 2} \frac{{\tan (x - 2)({x^2} + (a - 2)x - 2a)}}{{({x^2} - 4x + 4)}} = 7$,then $'a'$  is -

Answer

b
$\mathop {\lim }\limits_{x \to 2} \frac{{\tan (x - 2)({x^2} + (a - 2)x - 2a)}}{{{{(x - 2)}^2}}} = 7$

$\Rightarrow a+2=7 \Rightarrow a=5$

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