A) 1
B) \[1/2\]
C) \[-1/2\]
D) \[-1\]
Correct Answer: C
Solution :
Since, \[f(x)=\frac{1-\tan x}{4x-\pi }\] \[\underset{x\to \pi /4}{\mathop{\lim }}\,f(x)=\underset{x\to \pi /4}{\mathop{\lim }}\,\left( \frac{1-\tan x}{4x-\pi } \right)\] By LHospitals rule \[=\underset{x\to \pi /4}{\mathop{\lim }}\,\left( \frac{-{{\sec }^{2}}x}{4} \right)=\frac{-{{\sec }^{2}}(\pi /4)}{4}=-\frac{2}{4}\] \[\Rightarrow \] \[\underset{x\to \pi /4}{\mathop{\lim }}\,f(x)=-\frac{1}{2}\] Also,\[f(x)\] is continuous in\[[0,\pi /2]\], so\[f(x)\]will be continuous at\[\pi /4\]. \[\therefore \]Value of function = Value of limit \[\Rightarrow \] \[f\left( \frac{\pi }{4} \right)=-\frac{1}{2}\]You need to login to perform this action.
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