If \(\mathrm{f}(x)=\frac{1+\cos \pi x}{\pi(1-x)^2}\), for \(x \neq 1\) is continuous at \(x=1\), then \(\mathrm{f}(1)\) is equal to

\(\begin{aligned} \quad \mathrm{f}(x) & =\frac{1+\cos \pi x}{\pi(1-x)^2} \\ \therefore \quad \mathrm{f}(1) & =\lim _{x \rightarrow 1} \mathrm{f}(x) \\ & =\lim _{x \rightarrow 1} \frac{1+\cos \pi x}{\pi(1-x)^2}\end{aligned}\)
\(\begin{aligned} & \text { Put } 1-x=\mathrm{h} \\ & \Rightarrow 1-\mathrm{h}=x \\ & \text { As } x \rightarrow 1, \mathrm{~h} \rightarrow 0 \\ & \begin{aligned} \mathrm{f}(1) & =\lim _{\mathrm{h} \rightarrow 0} \frac{1+\cos \pi(1-\mathrm{h})}{\pi \mathrm{h}^2} \\ = & \lim _{\mathrm{h} \rightarrow 0} \frac{1+\cos (\pi-\pi h)}{\pi \mathrm{h}^2} \\ = & \lim _{\mathrm{h} \rightarrow 0} \frac{1-\cos \pi \mathrm{h}}{\pi \mathrm{h}^2} \\ = & \lim _{\mathrm{h} \rightarrow 0} \frac{2 \sin ^2 \frac{\pi \mathrm{~h}}{2}}{\pi \mathrm{~h}^2} \\ = & 2 \lim _{\mathrm{h} \rightarrow 0} \frac{\sin ^2 \frac{\pi h}{2}}{\frac{\pi^2 h^2}{4} \times 4} \cdot \pi \\ = & \frac{2 \pi}{4} \lim _{h \rightarrow 0}\left(\frac{\sin \frac{\pi h}{2}}{\frac{\pi h}{2}}\right)^2 \\ \therefore \quad \mathrm{f}(1) & =\frac{\pi}{2}\end{aligned}\end{aligned}\)