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…
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
- $\frac{-1}{2}$
- $-1$
- $1$
- $\frac{1}{2}$
Solution
$f(x)=\frac{(1+\cos x)-(\sin x)}{(1+\cos x)+(\sin x)}$, is continuous at $x=\pi$
$\begin{aligned}
& \lim _{x \rightarrow \pi} f(x)=\lim _{x \rightarrow \pi} \frac{\left(2 \cos ^2 \frac{x}{2}\right)-\left(2 \sin \frac{x}{2} \cos \frac{x}{2}\right)}{\left(2 \cos ^2 \frac{x}{2}\right)+\left(2 \sin \frac{x}{2} \cos \frac{x}{2}\right)} \\
& =\lim _{x \rightarrow \pi} \frac{2 \cos \frac{x}{2}\left(\cos \frac{x}{2}-\sin \frac{x}{2}\right)}{2 \cos \frac{x}{2}\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)} \\
& =\lim _{x \rightarrow \pi} \frac{1-\tan \frac{x}{2}}{1+\tan \frac{x}{2}}=\lim _{x \rightarrow \pi} \tan \left(\frac{\pi}{4}-\frac{x}{2}\right)=\tan \left(\frac{\pi}{4}-\frac{\pi}{2}\right)=-1
\end{aligned}$
Asked in: MHT CET 2021 (23 Sep Shift 2)
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