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Find the values of \(k\) so that the function \(f\) is continuous at the indicated point in $ f(x) = \left\{ \begin{array} {1 1} \large\frac{k\cos x}{\pi - 2x}\normalsize,& \quad { if \; x \neq \large\frac{\pi}{2} }\\ \normalsize 3 ,& \quad { if \; x = \large\frac{\pi}{2} }\\ \end{array} \right. \qquad at \; x = \; \large\frac{\pi}{2} $

This question has appeared in model paper 2012

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  • If $f$ is a real function on a subset of the real numbers and $c$ be a point in the domain of $f$, then $f$ is continuous at $c$ if $\lim\limits_{\large x\to c} f(x) = f(c)$.
Step 1:
LHL=$\lim\limits_{\large x\to \Large\frac{\pi}{2}}\large\frac{k\cos x}{\pi-2x}$
Put $x=\large\frac{\pi}{2}$$-h$
LHL=$\lim\limits_{\large h\to 0}\large\frac{k\cos \big(\Large\frac{\pi}{2}-\normalsize h\big)}{\pi-2\big(\Large\frac{\pi}{2}-\normalsize h\big)}$
$\quad\;=\lim\limits_{\large h\to 0}\large\frac{k\sin h}{\pi-\pi+2h}$
$\quad\;=\lim\limits_{\large h\to 0}\large\frac{k\sin h}{2h}$
$\quad\;=\lim\limits_{\large h\to 0}\large\frac{k}{2}\frac{\sin h}{h}$
Step 2:
RHL=$\lim\limits_{\large x\to \Large\frac{\pi}{2}}\large\frac{k\cos 2x}{\pi-2x}$
Put $x=\large\frac{\pi}{2}$$+h$
RHL=$\lim\limits_{\large h\to 0}\large\frac{k\cos \big(\Large\frac{\pi}{2}+\normalsize h\big)}{\pi-2\big(\Large\frac{\pi}{2}+\normalsize h\big)}$
$\quad\;=\lim\limits_{\large h\to 0}\large\frac{-k\sin h}{-2h}$
$\quad\;=\lim\limits_{\large h\to 0}\large\frac{k}{2}\frac{\sin h}{h}$
$f\big(\large\frac{\pi}{2}\big)$$=3\rightarrow $Given
$f$ is continuous if $\large\frac{k}{2}$$=3$ or $k=6$
answered May 29, 2013 by sreemathi.v

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