# Find the values of K. So that the function $f(x) = \left\{ \begin{array}{l l} \large\frac{k \cos x}{\pi -2x} & \quad if \quad x \neq \large\frac{\pi}{2} \\ 3 & \quad if \quad x =\large\frac{\pi}{2} \end{array} \right.$ is continuous at $x =\large\frac{\pi}{2}$ .

$\begin{array}{1 1}(A)\;6 \\(B)\;4 \\(C)\;5 \\(D)\;8 \end{array}$

Since $f(x)$ is continuous at $x= \large\frac{\pi}{2}$
$f(\large\frac{\pi}{2})$$=\lim \limits_{x \to \large\frac{\pi}{2} } f(x) =\lim \limits _{x \to \large\frac{\pi}{2}} \large=\frac{K \cos x }{\pi-2x} Here we need not find left hand and right hand separately because f(x) is not different when x < \large\frac{\pi}{2} and x > \large\frac{\pi}{2} => \lim \limits_{h \to 0} \large\frac{k \cos (\large\frac{\pi}{2} +h)}{\pi -2 (\large\frac{\pi}{2} +h)} Putting x =\large\frac{\pi}{2}$$+h$
So that $x \to \large\frac{\pi}{2} $$h \to 0 \qquad= \lim \limits _{h \to 0} \large\frac{-K \sin h}{-2h} \qquad= \large\frac{K}{2}$$ \lim \limits_{h \to 0} \bigg( \large\frac{\sin h}{h}\bigg)$
$\qquad= \large\frac{k}{2}$$.1 \qquad= \large\frac{k}{2} \large\frac{k}{2}$$=3$
$k=6$
Hence A is the correct answer.
answered Apr 22, 2014 by