If $$ f(x) = \begin{cases} \dfrac{1}{|x|}, & |x| > 2 \\ A + Bx^2, & |x| \leq 2 \end{cases} $$ Then $f(x)$ is differentiable at $x = -2$ for
(a) A = $\tfrac{3}{4}$ ,B = $-\tfrac{1}{16}$
(b) A = $-\tfrac{3}{4}$ , B = $\tfrac{1}{16}$
(c) A = $-\tfrac{3}{4}$ ,B = $-\tfrac{1}{16}$
(d) A = $\tfrac{3}{4}$, B = $\tfrac{1}{16}$
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