Question

If $f\left(x\right)=\{\begin{array}{c}\begin{array}{cc}\frac{1-\sin x}{\pi -2x},& x\ne \frac{\pi }{2}\end{array}\\ \begin{array}{cc}\lambda ,& x=\frac{\pi }{2}\end{array}\end{array}$ is continuous at $x=\frac{\pi }{2}$, then the value of $\lambda$ is

• A. $-1$
• B. $1$
• C. $0$
• D. $2$

Answer 1

The correct option is C $0$

Since $f\left(x\right)$ is continuous at $x=\frac{\pi }{2},$

$\therefore f\left(\frac{\pi }{2}\right)=\underset{x\to \frac{\pi }{2}}{lim}f\left(x\right)$

$\Rightarrow \lambda =\underset{x\to \frac{\pi }{2}}{lim}\frac{1-\sin x}{\pi -2x}$ $(\frac{0}{0}$ from$)$

Applying L Hospital 's rule

$\Rightarrow \lambda =\underset{x\to \frac{\pi }{2}}{lim}\frac{-\cos x}{-2}$

$\Rightarrow \lambda =0$