Search any question & find its solution
Question:
Answered & Verified by Expert
If the function $f(x)=\frac{1-\sin 2 x+\cos 2 x}{1+\sin 2 x+\cos 2 x}$ if $\quad x \neq \frac{\pi}{2}$
$=\mathrm{k} \quad$ if $\quad x=\frac{\pi}{2}$
is continuous at $x=\frac{\pi}{2}$, then $\mathrm{k}=$
Options:
$=\mathrm{k} \quad$ if $\quad x=\frac{\pi}{2}$
is continuous at $x=\frac{\pi}{2}$, then $\mathrm{k}=$
Solution:
1515 Upvotes
Verified Answer
The correct answer is:
-1
$\begin{aligned}
\lim _{x \rightarrow \frac{\pi}{2}} f(x)=& \lim _{x \rightarrow \frac{\pi}{2}} \frac{1-\sin 2 x+\cos 2 x}{1+\sin 2 x+\cos 2 x} \\
=& \lim _{x \rightarrow \frac{\pi}{2}} \frac{(1+\cos 2 x)-\sin 2 x}{(1+\cos 2 x)+\sin 2 x}=\lim _{x \rightarrow \frac{\pi}{2}} \frac{2 \cos ^{2} x-2 \sin x \cos x}{x+2 \sin x \cos x} \\
=& \lim _{x \rightarrow \frac{\pi}{2}} \frac{2 \cos x(\cos x-\sin x)}{2 \cos x(\cos x+\sin x)}=\lim _{x \rightarrow \frac{\pi}{2}} \frac{\cos x-\sin x}{\cos x+\sin x}=\frac{0-1}{0+1}=-1
\end{aligned}$
Since $f(x)$ is continuous at $x=\frac{\pi}{2}$, we get $k=-1$
\lim _{x \rightarrow \frac{\pi}{2}} f(x)=& \lim _{x \rightarrow \frac{\pi}{2}} \frac{1-\sin 2 x+\cos 2 x}{1+\sin 2 x+\cos 2 x} \\
=& \lim _{x \rightarrow \frac{\pi}{2}} \frac{(1+\cos 2 x)-\sin 2 x}{(1+\cos 2 x)+\sin 2 x}=\lim _{x \rightarrow \frac{\pi}{2}} \frac{2 \cos ^{2} x-2 \sin x \cos x}{x+2 \sin x \cos x} \\
=& \lim _{x \rightarrow \frac{\pi}{2}} \frac{2 \cos x(\cos x-\sin x)}{2 \cos x(\cos x+\sin x)}=\lim _{x \rightarrow \frac{\pi}{2}} \frac{\cos x-\sin x}{\cos x+\sin x}=\frac{0-1}{0+1}=-1
\end{aligned}$
Since $f(x)$ is continuous at $x=\frac{\pi}{2}$, we get $k=-1$
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.