If the function f ( x ) = * 20 c 2 + ,x - 1 ( - x^2 ) , & x k , ,… - Vidyadip

MCQ

If the function $f\left( x \right) = \left\{ {\begin{array}{*{20}{c}}
{\frac{{\sqrt {2 + \cos \,x} - 1}}{{\left( {\pi - {x^2}} \right)}},}&{x \ne \pi } \\
{k\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,,}&{x = \pi }
\end{array}} \right.$ is continuous at $x\, =\pi $ , then $k$ equals

A
$0$
B
$\frac{1}{2}$
C
$2$
✓
$0.25$

✓

Answer

Correct option: D.

$0.25$

d
Since $f\left( x \right) = \frac{{\sqrt {2 + \cos x }- 1 }}{{{{\left( {\pi - x} \right)}^2}}}$ is

Continuos at $x = \pi $

$\therefore L.H.L = R.H.L = f\left( \pi \right)$

Let $\left( {\pi - x} \right) = \theta ,\theta \to 0$ when $x \to \pi $

$\therefore \mathop {\lim }\limits_{\theta \to 0} \frac{{\sqrt {2 + \cos \theta - 1} }}{{{\theta ^2}}}$

$ = \mathop {\lim }\limits_{\theta \to 0} \frac{{\left( {2 + \cos \theta } \right) - 1}}{{{\theta ^2}}} \times \frac{1}{{\sqrt {2 + \cos \theta } + 1}}$

$ = \mathop {\lim }\limits_{\theta \to 0} \frac{{1 - \cos \theta }}{{{\theta ^2}}}.\frac{1}{2}\,\,$ ($\because $ $\cos 0 = 1$)

$ = \mathop {\lim }\limits_{\theta \to 0} \frac{{2{{\sin }^2}\theta /2}}{{{\theta ^2}}}$

$ = \frac{2}{2}\mathop {\lim }\limits_{\theta \to 0} \frac{{{{\sin }^2}\theta /2}}{{\frac{{{\theta ^2}}}{4}.4}} = \frac{1}{4}$

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