If f(x) = l , , , , , , , , , 1 - 4x x^2 , ; ; when ,x < 0 , , , … - Vidyadip

MCQ

If $f(x) = \left\{ \begin{array}{l}\,\,\,\,\,\,\,\,\,\frac{{1 - \cos 4x}}{{{x^2}}},\;\;{\rm{when}}\,x < 0\\\,\,\,\,\,\,\,\,\,\,\,a,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{when}}\,\,x = 0\\\frac{{\sqrt x }}{{\sqrt {(16 + \sqrt x )} - 4}},\,\,{\rm{when}}\,\, x > 0\end{array} \right.,$ is continuous at $x = 0$, then the value of $'a\ '$ will be

A
$8$
B
$-8$
C
$4$
D
None of these

✓

Answer

$\mathop {\lim }\limits_{x \to 0 - } f(x) = \mathop {\lim }\limits_{x \to 0 - } \left( {\frac{{2\,{{\sin }^2}2x}}{{{{(2x)}^2}}}} \right)\,4 = 8$
$\mathop {\lim }\limits_{x \to 0 + } f(x) = \mathop {\lim }\limits_{x \to 0 + } \sqrt {16 + \sqrt x + 4} = 8$.
Hence $a = 8$.

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