Search any question & find its solution
Question:
Answered & Verified by Expert
If $f(x)=\frac{1-\sin x+\cos x}{1+\sin x+\cos x}$, for $x \neq \pi$ is continuous at $x=\pi$, then the value of $f(\pi)$ is
Options:
Solution:
1038 Upvotes
Verified Answer
The correct answer is:
$-1$
$f(x)=\frac{(1+\cos x)-(\sin x)}{(1+\cos x)+(\sin x)}$, is continuous at $x=\pi$
$$
\begin{aligned}
& \lim _{x \rightarrow \pi} f(x)=\lim _{x \rightarrow \pi} \frac{\left(2 \cos ^2 \frac{x}{2}\right)-\left(2 \sin \frac{x}{2} \cos \frac{x}{2}\right)}{\left(2 \cos ^2 \frac{x}{2}\right)+\left(2 \sin \frac{x}{2} \cos \frac{x}{2}\right)} \\
& =\lim _{x \rightarrow \pi} \frac{2 \cos \frac{x}{2}\left(\cos \frac{x}{2}-\sin \frac{x}{2}\right)}{2 \cos \frac{x}{2}\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)} \\
& =\lim _{x \rightarrow \pi} \frac{1-\tan \frac{x}{2}}{1+\tan \frac{x}{2}}=\lim _{x \rightarrow \pi} \tan \left(\frac{\pi}{4}-\frac{x}{2}\right)=\tan \left(\frac{\pi}{4}-\frac{\pi}{2}\right)=-1
\end{aligned}
$$
$$
\begin{aligned}
& \lim _{x \rightarrow \pi} f(x)=\lim _{x \rightarrow \pi} \frac{\left(2 \cos ^2 \frac{x}{2}\right)-\left(2 \sin \frac{x}{2} \cos \frac{x}{2}\right)}{\left(2 \cos ^2 \frac{x}{2}\right)+\left(2 \sin \frac{x}{2} \cos \frac{x}{2}\right)} \\
& =\lim _{x \rightarrow \pi} \frac{2 \cos \frac{x}{2}\left(\cos \frac{x}{2}-\sin \frac{x}{2}\right)}{2 \cos \frac{x}{2}\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)} \\
& =\lim _{x \rightarrow \pi} \frac{1-\tan \frac{x}{2}}{1+\tan \frac{x}{2}}=\lim _{x \rightarrow \pi} \tan \left(\frac{\pi}{4}-\frac{x}{2}\right)=\tan \left(\frac{\pi}{4}-\frac{\pi}{2}\right)=-1
\end{aligned}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.