Question

The function is defined by f(x) =$\{\begin{array}{c}Kx=1,ifx\le 5\\ 3x-5,ifx>5\end{array}$ at x = 5.

Answer 1

Here, f(x) =$\{\begin{array}{c}Kx+1,ifx\le 5\\ 3x-5,ifx>5\end{array}$

LHL = $\underset{x\to 5^{-}}{lim}f\left(x\right)=\underset{x\to 5^{-}}{lim}\lambda (kx+1)$

Putting x=5-h as $x\to 5^{-}$ when $h\to 0$
$\therefore \underset{h\to 0}{lim}=\left[k\right(5-h)+1]\underset{h\to 0}{lim}[5k-kh+1]=5k+1$

RHL = $\underset{x\to 5^{+}}{lim}f\left(x\right)=\underset{x\to 5^{+}}{lim}\lambda (kx+1)$

Putting x=5-h as $x\to 5^{+}$ when $h\to 0$

$\therefore \underset{h\to 0}{lim}=\left[3\right(5-h)-5]\underset{h\to 0}{lim}[10+3h]=10$

Also, f(x)=5k+1 [$\therefore$ f(x)=kx+1]

Since, f(x) is continuous at x = 5.

$\therefore LHL=RHL=f\left(5\right)\Rightarrow 5k+1=10\Rightarrow k=\frac{9}{5}$