Question
Let $$f\left( x \right) = \frac{{1 - \tan \,x}}{{4x - \pi }},\,\,x \ne \frac{\pi }{4},\,x \in \left[ {0,\,\,\frac{\pi }{2}} \right].$$ If $$f\left( x \right)$$ is continuous in $$\left[ {0,\,\,\frac{\pi }{2}} \right]$$ then $$f\left( {\frac{\pi}{4}} \right)$$ is-
A.
$$-1$$
B.
$$\frac{1}{2}$$
C.
$$ - \frac{1}{2}$$
D.
$$1$$
Answer :
$$ - \frac{1}{2}$$
Solution :
$$f\left( x \right) = \frac{{1 - \tan \,x}}{{4x - \pi }}$$ is continuous in $$\left[ {0,\,\,\frac{\pi }{2}} \right]$$
$$\eqalign{
& \therefore f\left( {\frac{\pi }{4}} \right) = \mathop {\lim }\limits_{x \to \frac{\pi }{4}} f\left( x \right) = \mathop {\lim }\limits_{x \to \frac{{{\pi ^ + }}}{4}} f\left( x \right) = \mathop {\lim }\limits_{h \to 0} f\left( {\frac{\pi }{4} + h} \right) \cr
& = \mathop {\lim }\limits_{h \to 0} \frac{{1 - \tan \left( {\frac{\pi }{4} + h} \right)}}{{4\left( {\frac{\pi }{4} + h} \right) - \pi }},\,h > 0 = \mathop {\lim }\limits_{h \to 0} \frac{{1 - \frac{{1 + \tan \,h}}{{1 - \tan \,h}}}}{{4h}} \cr
& = \mathop {\lim }\limits_{h \to 0} \frac{{ - 2}}{{1 - \tan \,h}}.\frac{{\tan \,h}}{{4h}} \cr
& = \frac{{ - 2}}{4} = - \frac{1}{2} \cr} $$