Question

Find the value of $k$ so that the function $f$ is continuous at the indicated point.
$f\left(x\right)=\{\begin{array}{cc}kx+1,& \text{if}x\le \pi \\ \cos x,& \text{if}x>\pi \end{array}$ at $x=\pi$.

Answer 1

Given : $f\left(x\right)=\{\begin{array}{cc}kx+1,& \text{if}x\le \pi \\ \cos x,& \text{if}x>\pi \end{array}$
If $f\left(x\right)$ is continuous at $x=\pi$ then
$\underset{x\to {\pi }^{-}}{lim}f\left(x\right)=\underset{x\to {\pi }^{+}}{lim}f\left(x\right)=f\left(\pi \right)$

L.H.L.
$=\underset{x\to {\pi }^{-}}{lim}kx+1$
$=\underset{h\to 0}{lim}k(\pi -h)+1$
Putting $h=0$ then we get, $k(\pi -0)+1=k\pi +1$

R.H.L.
$=\underset{x\to {\pi }^{+}}{lim}\cos x$
$=\underset{h\to 0}{lim}\cos (\pi +h)$
$=\underset{h\to 0}{lim}-\cos h$
Putting $h=0$ then we get, $-\cos 0=-1$

Finding $f\left(x\right)$ at $x=\pi$
$f\left(x\right)=kx+1$ at $x=\pi$
$f\left(\pi \right)=\pi k+1$

Solve for $k$
Since ${lim}_{x\to {\pi }^{-}}f\left(x\right)={lim}_{x\to {\pi }^{+}}f\left(x\right)=f\left(\pi \right)$
$\therefore k\pi +=-1$
$\Rightarrow k\pi =-1-1$
$\Rightarrow k\pi =-2$
$\Rightarrow k=\frac{-2}{\pi }$