Question

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

• A. $\frac{1}{4}$
• B. $\frac{1}{2}$
• C. $\frac{1}{8}$
• D. None of these

Answer 1

The correct option is C $\frac{1}{8}$

For $f\left(x\right)$ to be continuous at $x=\frac{\pi }{2}$, we must have
$\underset{x\to \frac{\pi }{2}}{lim}f\left(x\right)=f\left(\frac{\pi }{2}\right)=\lambda$
$\Rightarrow \lambda =\underset{x\to \frac{\pi }{2}}{lim}f\left(x\right)=\underset{x\to \frac{\pi }{2}}{lim}\frac{(1-\sin x)}{(\pi -2x{)}^2}$
Using L Hospital's Rule
$=\underset{x\to \frac{\pi }{2}}{lim}\frac{-\cos x}{-4(\pi -2x)}$
$=\frac{1}{4}\cdot \underset{x\to \frac{\pi }{2}}{lim}\frac{-\sin x}{-2}=\frac{1}{8}$