Question

The function is defined by f(x)=$\{\begin{array}{c}k{x}^2,ifx\le 2\\ 3,ifx>2\end{array}$

Answer 1

Here, f(x)=$\{\begin{array}{c}k{x}^2,ifx\le 2\\ 3,ifx>2\end{array}$

$LHL=\underset{x\to 2^{-}}{lim}f\left(x\right)=\underset{x\to 2^{-}}{lim}\left(k{x}^2\right)$.

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

$\underset{h\to 0}{lim}k(2-h{)}^2=\underset{h\to 0}{lim}k(4+h^2-4h)=4k$
$RHL=\underset{x\to 2^{+}}{lim}f\left(x\right)=\underset{x\to 2^{+}}{lim}\left(3\right)=\left(3\right)$.

Also, f(2) = k $\times$ (2)^2=4k $[\therefore f(x)=k{x}^2]$

Since, f(x) is continuous at x=2.

$\therefore$ LHL = RHL=f(x) $\Rightarrow 4k=3\Rightarrow k=\frac{3}{4}$