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

• A. $\frac{-1}{2}$
• B. $\frac{1}{2}$
• C. $1$
• D. $-1$

Answer 1

Answer: (D)

$\underset{x\to \pi }{\mathrm{Lim}}\frac{(1+\cos x)-\sin x}{(1+\cos x)+\sin x}$
$=\underset{x\to \pi }{\mathrm{Lim}}\frac{2{\cos }^2\frac{x}{2}-2\sin \frac{x}{2}\cdot \cos \frac{x}{2}}{2{\cos }^2\frac{x}{2}+2\sin \frac{x}{2}\cdot \cos \frac{x}{2}}$
$=\underset{x\to \pi }{\mathrm{Lim}}\tan \left(\frac{\pi }{4}-\frac{x}{2}\right)=-1$
Since, $f(x)$ is continuous at $x=\pi$,
so $f(\pi )=\underset{x\to \pi }{\mathrm{Lim}}f(x)=-1$