For what value of λ is the function

Solution:

The given function $f$ is $f(x)= \begin{cases}\lambda\left(x^{2}-2 x\right), & \text { if } x \leq 0 \\ 4 x+1, & \text { if } x>0\end{cases}$

If $f$ is continuous at $x=0$, then

$\lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x)=f(0)$

$\Rightarrow \lim _{x \rightarrow 0^{-}} \lambda\left(x^{2}-2 x\right)=\lim _{x \rightarrow 0^{+}}(4 x+1)=\lambda\left(0^{2}-2 \times 0\right)$

$\Rightarrow \lambda\left(0^{2}-2 \times 0\right)=4 \times 0+1=0$

$\Rightarrow 0=1=0$, which is not possible

Therefore, there is no value of $\lambda$ for which $f(x)$ is continuous at $x=0$.

At $x=1$

$f(1)=4 x+1=4 \times 1+1=5$

$\lim _{x \rightarrow 1}(4 x+1)=4 \times 1+1=5$

$\therefore \lim _{x \rightarrow 1} f(x)=f(1)$

Therefore, for any values of $\lambda, f$ is continuous at $x=1$

At $x=-1$, we have

$f(-1)=\lambda(1+2)=3 \lambda$

$\lim _{x \rightarrow-1} \lambda(1+2)=3 \lambda$

$\therefore \lim _{x \rightarrow-1} f(x)=f(-1)$

Therefore, for any values of $\lambda, f$ is continuous at $x=-1$