question_answer

Let\[f(x)=\frac{1-\tan x}{4x-\pi },x\ne \frac{\pi }{4},x\in \left[ 0,\frac{\pi }{2} \right]\]. If f(x) is continuous in\[\left[ 0,\frac{\pi }{2} \right]\],then\[f\left( \frac{\pi }{4} \right)\]is

A) 1   

B)                        1/2   

C) \[-1/2\]               

D)        \[-1\]

Correct Answer: C

Solution :

A functionis said to be continuous at if . Since,      By L' Hospital rule,             \[\underset{x\to \pi /4}{\mathop{\lim }}\,\,f(x)=-\frac{1}{2}\] Also,is continuous in \[[0,\,\pi /2]\]. So,will be continuous at. Value of function = Value of limit