Consider the function f(x)=\ array ccc (α x)/(π-2 x) & If & x ≠...

f(x)=\left\{$\begin{array}{ccc} \frac{\alpha \cos x}{\pi-2 x} & \text{ if } & x \neq \frac{\pi}{2} \\ 3 & \text{ if } & x=\frac{\pi}{2} \end{array}$\right. For continuity at $\mathrm{x}=\frac{\pi}{2}\text{ L.H.L. }=\lim _{x \rightarrow \frac{\pi^{-}}{2}} f(x)=\lim _{x \rightarrow \frac{\pi^{-}}{2}} \frac{\alpha \cos x}{\pi-2 x}$ Put $\mathrm{x}=\frac{\pi}{2}-\mathrm{h}$ where $\mathrm{x} \rightarrow \frac{\pi}{2}$, then $\mathrm{h} \rightarrow 0$$\begin{aligned} & \therefore \quad \text{ L.H.L. }=\lim _{h \rightarrow 0} \frac{\alpha \cos \left(\frac{\pi}{2}-h\right)}{\pi-2\left(\frac{\pi}{2}-h\right)}=\lim _{h \rightarrow 0} \frac{\alpha \sin h}{\pi-\pi+2 h} \\ & {\left[\because \cos \left(\frac{\pi}{2}-\theta\right)=\sin \theta\right]} \\ & =\lim _{h \rightarrow 0} \frac{2 \alpha \sin h}{2 h}=\frac{\alpha}{2} \lim _{h \rightarrow 0} \frac{\sin h}{h} \\ & =\frac{\alpha}{2} \cdot 1=\frac{\alpha}{2} \quad\left[\because \lim _{\theta \rightarrow 0} \frac{\sin \theta}{\theta}=1\right] \\ & \text{ and R.H.L. }=\lim _{x \rightarrow \frac{\pi^{+}}{2}} f(x) \lim _{x \rightarrow \frac{\pi^{+}}{2}} \frac{\alpha \cos x}{\pi-2 \pi} \end{aligned}$Put $\mathrm{x}=\frac{\pi}{2}+\mathrm{h}$ when $\mathrm{x} \rightarrow \frac{\pi}{2}$, then $\mathrm{h} \rightarrow 0\therefore \quad \text{ R.H.L. }=\lim _{h \rightarrow 0} \frac{\alpha \cos \left(\frac{\pi}{2}+h\right)}{\pi-2\left(\frac{\pi}{2}+h\right)}=\lim _{h \rightarrow 0} \frac{\alpha(-\sin h)}{\pi-\pi-2 h}\begin{array}{ll} =\lim _{h \rightarrow 0}-\frac{\alpha \sin h}{-2 h} & {\left[\because \cos \left(\frac{\pi}{2}+\theta\right)=-\sin \theta\right]} \\ =\frac{\alpha}{2} \lim _{h \rightarrow 0} \frac{\sin h}{h}=\frac{\alpha}{2} & \left(\because \lim _{\theta \rightarrow 0} \frac{\sin \theta}{\theta}=1\right) \end{array}Also, \mathrm{f}$\left(\frac{\pi}{2}\right)=3$ Since, $f(x)$ is continuous at $x=\frac{\pi}{2}\begin{aligned}& \therefore \quad \text{ L.H.L. }=\text{ R.H.L. }=\mathrm{f} \left(\frac{\pi}{2}\right)\\ \text{ and } \quad \frac{\alpha}{2}=\frac{\alpha}{2}=3 \\ \text{ and } \quad \alpha=6 \end{aligned}$$$Hence, for$\alpha=6$, the given function ( f ) is continuous$\text{ at }}x=\frac{\pi}{2}$