MCQ
Function $f(x) = \frac{{1 - \cos 4x}}{{8{x^2}}},$ where $x \ne 0$and $f(x) = k$ where $x = 0$ is a continous function at $x = 0$ then the value of $k$ will be $k = $ ........
- A$0$
- ✓$ 1$
- C$-1$
- DNone of these
If $f(x)$ is continous function at point $x = 0$ then
$\mathop {\lim }\limits_{x \to 0 + } [f(x) = \mathop {\lim }\limits_{x \to 0 - } [f(x)]$$\mathop {\lim }\limits_{x \to 0} [f(x)] = \mathop {\lim }\limits_{h \to 0 - } [f(0 + h)]$
$ = \mathop {\lim }\limits_{h \to 0} [f(h)] = \mathop {\lim }\limits_{h \to 0} \frac{{1 - \cos 4h}}{{8{h^2}}} = \mathop {\lim }\limits_{h \to 0} \frac{{2{{\sin }^2}2h}}{{8{h^2}}} = \mathop {\lim }\limits_{h \to 0} \frac{{{{\sin }^2}2h}}{{4{h^2}}}$
$ = \mathop {\lim }\limits_{h \to 0} {\left( {\frac{{\sin 2h}}{{2h}}} \right)^2} = {(1)^2} = 1$
$\mathop {\lim }\limits_{x \to {0^ - }} f(x)\mathop { = \lim }\limits_{\,\,\,\,\,h \to 0} [f(0 - h)$
$ = \mathop {\lim }\limits_{h \to 0} [f( - h)] = \mathop {\lim }\limits_{h \to 0} \frac{{1 - \cos 4( - h)}}{{8{{( - h)}^2}}}$
$ = \mathop {\lim }\limits_{h \to 0} \frac{{1 - \cos 4h}}{{8{h^2}}}$$ = 1$
$f(0) = 1 \Rightarrow k = 1$.
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