Question

f(x) $\{\begin{array}{c}\frac{kcosx}{\pi -2x},ifx\ne \frac{\pi }{2}\\ 3,ifx=\frac{\pi }{2}.\end{array}$

Answer 1

Here, f(x) $\{\begin{array}{c}\frac{kcosx}{\pi -2x},ifx\ne \frac{\pi }{2}\\ 3,ifx=\frac{\pi }{2}.\end{array}$

$\therefore LHL=\underset{x\to \frac{{\pi }^1}{2}}{lim}f\left(x\right)=LHL=\underset{x\to \frac{{\pi }^1}{2}}{lim}\frac{kcosx}{\pi -2x}$

Putting x=$\frac{\pi }{2}$-h as $x\to \frac{\pi }{2}$ when $h\to 0$

$\underset{x\to 0}{lim}\frac{kcos(\frac{\pi }{2}-h)}{\pi -2(\frac{\pi }{2}-h)}\underset{x\to 0}{lim}\frac{ksinh}{2h}$

=$\underset{x\to 0}{lim}\frac{k}{2}\times \frac{sinh}{h}=\frac{k}{2}\times 1=\frac{k}{2}$ $\because \underset{x\to 0}{lim}\frac{sinx}{x}=1$

$\therefore RHL=\underset{x\to {\pi }^{+}/2}{lim}f\left(x\right)=LHL=\underset{x\to {\pi }^{+}/2}{lim}\frac{kcosx}{\pi -2x}$

Putting x=$\frac{\pi }{2}$-h as $x\to \frac{{\pi }^{+}}{2}$ when $h\to 0$

$\underset{x\to 0}{lim}\frac{kcos(\frac{\pi }{2}+h)}{\pi -2(\frac{\pi }{2}+h)}\underset{x\to 0}{lim}\frac{ksinh}{-2h}$=$\underset{x\to 0}{lim}\frac{k}{2}\times \frac{sinh}{h}=\frac{k}{2}\times 1=\frac{k}{2}$ $\because \underset{x\to 0}{lim}\frac{sinx}{x}=1$

Also, $\frac{\pi }{2}=3Since,f\left(x\right)iscontinuousatx=\frac{\pi }{2}$

$\therefore$ LHL=RHL= $f\frac{\pi }{2}=\frac{k}{2}=3\Rightarrow k=6$