Explain solution RD Sharma class 12 Chapter 8 Continuity exercise 8.1 question 36 subquestion (ii)

$k=\frac{-2}{\pi}$

Hint:

For a function to be continuous at a point, its LHL RHL and value at that point should be equal.

Given:

$f(x)=\left\{\begin{array}{c} (x-1) \tan \frac{\pi x}{2}, \text { if } x \neq 1 \\\\ k, \text { if } x=1 \end{array}\right.$

Solution:

$f(x)=\left\{\begin{array}{c} (x-1) \tan \frac{\pi x}{2}, \text { if } x \neq 1 \\\\ k, \text { if } x=1 \end{array}\right.$

If  $f\left ( x \right )$ is continuous at $x=1$ , then

\begin{aligned} &\lim _{x \rightarrow 1} f(x)=f(1) \\\\ &\lim _{x \rightarrow 1}(x-1) \tan \frac{\pi x}{2}=k \end{aligned}

Putting $x-1=y$ , we get

\begin{aligned} &\lim _{y \rightarrow 0} y \tan \frac{\pi(y+1)}{2}=k \\\\ &\lim _{y \rightarrow 0} y \tan \left(\frac{\pi y}{2}+\frac{\pi}{2}\right)=k \end{aligned}

$\lim _{y \rightarrow 0} y \tan \left(\frac{\pi}{2}+\frac{\pi y}{2}\right)=k$

$-\lim _{y \rightarrow 0} y \cot \left(\frac{\pi y}{2}\right)=k$                                                                                   $[\because \tan (90+\theta)=-\cot \theta]$

$-\frac{2}{\pi} \lim _{y \rightarrow 0} \frac{\frac{\pi y}{2} \cos \left(\frac{\pi y}{2}\right)}{\sin \left(\frac{\pi y}{2}\right)}=k$                                                                   $\left[\because \cot x=\frac{\cos x}{\sin x}\right]$

$-\frac{2}{\pi} \frac{\lim _{y \rightarrow 0} \cos \left(\frac{\pi y}{2}\right)}{\lim _{y \rightarrow 0}\left(\frac{\sin \left(\frac{\pi y}{2}\right)}{\frac{\pi y}{2}}\right)}=k$

\begin{aligned} & \\\\ &\frac{-2}{\pi} \times \frac{1}{1}=k \\\\ &k=\frac{-2}{\pi} \end{aligned}                                                                                         $\left[\because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right]$