Question

The function is defined by f(x) = $\{\begin{array}{c}kx+1,ifx\le \pi \\ cosx,ifx>\pi \end{array}$ at x = $\pi$.

Answer 1

Here, f(x) = $\{\begin{array}{c}kx+1,ifx\le \pi \\ cosx,ifx>\pi \end{array}$

LHL = $\underset{x\to {\pi }^{-}}{lim}f\left(x\right)=\underset{x\to {\pi }^{-}}{lim}(kx+1)$

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

$\therefore \underset{h\to 0}{lim}k(\pi -h)+1=\underset{h\to 0}{lim}k\pi -kh+1=k\pi +1$

RHL = $\underset{x\to {\pi }^{+}}{lim}f\left(x\right)=\underset{x\to {\pi }^{+}}{lim}cosx$.

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

$\therefore \underset{h\to 0}{lim}cos(\pi -h)=\underset{h\to 0}{lim}-cosh=-1$

Also, $f\left(\pi \right)$ = $\left(k\pi \right)+1[\therefore (x)=kx+1]$

$\therefore$ Since, f(x) is continuous at x=$\pi$.

$\therefore LHL=RHL=f\left(\pi \right)\Rightarrow k\left(\pi \right)+1=-\Rightarrow k=\frac{-2}{\pi }$