Question

The function $f\left(x\right)=\frac{1-\sin x+\cos x}{1+\sin x+\cos x}$ is not defined at $x=\pi$. The value of $f\left(\pi \right)$ so that $f\left(x\right)$ is continuous at $x=\pi$ is

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

Answer 1

Answer: (C)

$-1$

$f$ is continuous at $x=\pi$,

$f\left(\pi \right)=\underset{x\to \pi }{lim}f\left(x\right)=\underset{x\to \pi }{lim}\frac{1-\sin x+\cos x}{1+\sin x+\cos x}$

$=\underset{x\to \pi }{lim}\frac{-\cos x-\sin x}{\cos x-\sin x}$ ..... [L-Hospital's Rule]

$=\frac{-\cos \pi -\sin \pi }{\cos \pi -\sin \pi }$

$=\frac{-(-1)-0}{-1-0}=\frac{1}{-1}=-1$

$\therefore f\left(\pi \right)=-1$

Hence, C is the correct option.