If function f(x) given by $f(x)=\left\{\begin{array}{cc}(\sin x)^{1 /(\pi-2 x)}, & x \neq \pi / 2 \\ \lambda ~~\quad, & x=\pi / 2 \end{array}\right.$ is continuous at $x=\frac{\pi}{2}$, then $\lambda=$

1

For f(x) to be continuous at x = $\frac{\pi}{2}$, we must have

$\lim\limits_{x \rightarrow \pi / 2} f(x)=f\left(\frac{\pi}{2}\right)$

⇒ $\lim\limits_{x \rightarrow \pi / 2}(\sin x)^{1 /(\pi-2 x)}=\lambda$

$\Rightarrow \lim\limits_{x \rightarrow \pi / 2}\{1+(\sin x-1)\}^{1 /(\pi-2 x)}=\lambda$

$\Rightarrow e^{\lim\limits_{x \rightarrow \pi / 2} \frac{\sin x-1}{\pi-2 x}}=\lambda$

$\Rightarrow e^{-\frac{1}{2} \lim\limits_{x \rightarrow \pi / 2} \frac{1-\cos (\pi / 2-x)}{(\pi / 2-x)}}=\lambda \Rightarrow e^0=\lambda \Rightarrow \lambda=1$