If $f(x)=\frac{1-\sin x}{(\pi-2 x)^2}$, when $x \neq \pi / 2$ and $f(\pi / 2)=\lambda$, then f(x) will be continuous function at $x=\pi / 2$, when $\lambda=$
Answer & explanation
Correct answer: option 1
f(x) will be continuous at $x=\pi / 2$, if
$\lim\limits_{x \rightarrow \pi / 2} f(x)=f\left(\frac{\pi}{2}\right)$
$\Rightarrow \lim\limits_{x \rightarrow \pi / 2} f(x) \frac{1-\sin x}{(\pi-2 x)^2}=\lambda$
$\Rightarrow \frac{1}{4} \lim\limits_{x \rightarrow \pi / 2} \frac{1-\cos (\pi / 2-x)}{(\pi / 2-x)^2}=\lambda$
$\Rightarrow \frac{1}{4} \times \frac{1}{2}=\lambda$ $\left[∵ \lim\limits_{x \rightarrow 0} \frac{1-\cos x}{x^2}=\frac{1}{2}\right]$
$\Rightarrow \lambda=\frac{1}{8}$