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