Question

For what value of $\lambda$is the function $f\left(x\right)=\{\begin{array}{c}\lambda (x^2-2x)ifx\le 0\\ 4x+1,ifx>0\end{array}$ continuous at x = 0? what about continuity at x = 1?

Answer 1

Here, $f\left(x\right)=\{\begin{array}{c}\lambda (x^2-2x)ifx\le 0\\ 4x+1,ifx>0\end{array}$

At x = 0, $LHL=\underset{x\to 0^{-}}{lim}f\left(x\right)=\underset{x\to 0^{-}}{lim}\lambda (x^2-2x)$

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

$\therefore \underset{h\to 0}{lim}\lambda \left[\right(0-h{)}^2-2(0-h)]=\underset{h\to 0}{lim}[\lambda (h^2+2h)]=0$

$RHL=\underset{x\to 0^{-}}{lim}f\left(x\right)=\underset{x\to 0^{-}}{lim}(4x+1)$

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

$\therefore \underset{h\to 0}{lim}\lambda \left[4\right(0-h)+1]=\underset{h\to 0}{lim}[4h+1]=0+1=1$

$\therefore$ LHL $not=$ RHL.Thus, f(x)is not continuous at x=0 for any value of $\lambda$ .

At x = 1, $LHL=\underset{x\to 0^{-}}{lim}f\left(x\right)=\underset{x\to 0^{-}}{lim}\lambda (4x+1)$

Putting x=0-h as $x\to 1^{-}$ when $h\to 0$

$\therefore \underset{h\to 0}{lim}\left[4\right(1-h)+1]=\underset{h\to 0}{lim}[5-4h]=5+0=0$

$RHL=\underset{x\to 1^{+}}{lim}f\left(x\right)=\underset{x\to 1^{+}}{lim}(4x+1)$

Putting x=1-h as $x\to 1^{+}$ when $h\to 0$

$\therefore \underset{h\to 0}{lim}\left[4\right(1-h)+1]=\underset{h\to 0}{lim}[5-4h]=5+0=5$

Also, f(x)=4 $\times$ 1+1=5 $[\therefore f(x)=4x+1]$

Thus, f(x) is continuous at x=1 for all values of $\lambda$