Question

Let $f\left(x\right)=\left\{\begin{array}{ll}\frac{k\cos x}{\mathrm{\pi }-2x},& \mathrm{where}x\ne \frac{\mathrm{\pi }}{2}\\ 3,& \mathrm{where}x=\frac{\mathrm{\pi }}{2}\end{array}\right.$ and if $\underset{x\to \frac{\mathrm{\pi }}{2}}{\lim }f\left(x\right)=f\left(\frac{\mathrm{\pi }}{2}\right)$, find the value of k.

Answer 1

We have,

$f\left(x\right)=\left\{\begin{array}{ll}\frac{k\cos x}{\mathrm{\pi }-2x},& \mathrm{where}x\ne \frac{\mathrm{\pi }}{2}\\ 3,& \mathrm{where}x=\frac{\mathrm{\pi }}{2}\end{array}\right.$

It is given that,

$$\begin{gathered} \underset{x\to \frac{\mathrm{\pi }}{2}}{\lim }f\left(x\right)=f\left(\frac{\mathrm{\pi }}{2}\right) \\ \Rightarrow \underset{x\to \frac{\mathrm{\pi }}{2}}{\lim }\frac{k\cos x}{\mathrm{\pi }-2\mathrm{x}}=3 \\ \Rightarrow \frac{k}{2}\times \underset{x-\frac{\mathrm{\pi }}{2}\to 0}{\lim }\frac{\sin \left({\displaystyle \frac{\mathrm{\pi }}{2}}-x\right)}{{\displaystyle \frac{\mathrm{\pi }}{2}}-x}=3 \\ \Rightarrow \frac{k}{2}\times \underset{h\to 0}{\lim }\frac{\sin \left({\displaystyle -h}\right)}{{\displaystyle -h}}=3\left(\mathrm{Put}x-\frac{\mathrm{\pi }}{2}=h\right) \end{gathered}$$
$$\begin{gathered} \Rightarrow \frac{k}{2}\times \underset{h\to 0}{\lim }\frac{\sin h}{h}=3\left[\sin \left(-\theta \right)=-\sin \theta \right] \\ \Rightarrow \frac{k}{2}=3\left(\underset{x\to 0}{\lim }\frac{\sin x}{x}=1\right) \\ \Rightarrow k=6 \end{gathered}$$

Hence, the value of k is 6.