Functions Question 458
Question: The function $ f(x)=\frac{1-\sin x+\cos x}{1+\sin x+\cos x} $ is not defined at $ x=\pi . $ The value of $ f(\pi ) $ so that $ f(x) $ is continuous at $ x=\pi $ is
[Orissa JEE 2003]
Options:
A) $ -\frac{1}{2} $
B) $ \frac{1}{2} $
C) - 1
D) 1
Show Answer
Answer:
Correct Answer: C
Solution:
$ f(x)=\frac{2{{\cos }^{2}}\frac{x}{2}-2\sin \frac{x}{2}\cos \frac{x}{2}}{2{{\cos }^{2}}\frac{x}{2}+2\sin \frac{x}{2}\cos \frac{x}{2}}=\frac{\cos \frac{x}{2}-\sin \frac{x}{2}}{\cos \frac{x}{2}+\sin \frac{x}{2}} $ $ =\tan ( \frac{\pi }{4}-\frac{x}{2} ) $ at $ x=\pi $ , $ f(\pi )=-\tan \frac{\pi }{4}=-1 $ .
BETA
