The value of $\lambda$., for which the function $f\left(x\right)=\left\{\begin{array}{cc}\lambda \left(x^2-2x\right)& \text{ if }x\le 0\\ 4x+1& \text{ if }x>0\end{array}\right.$ is continuous at x = 0, is :

given $f\left(x\right)=\left\{\begin{array}{cc}\lambda \left(x^2-2x\right)& \text{ if }x\le 0\\ 4x+1& \text{ if }x>0\end{array}\right.$ is
continuous at x=0
$\therefore \underset{x\to 0^{+}}{lim} f\left(x\right)=\underset{x\to 0^{-}}{lim} f\left(x\right)=f\left(0\right)$
But R.H.S $=\underset{h\to 0}{lim} 4h+1=1$
$\text{ L.H.L }=\underset{h\to 0}{lim} \lambda \left(h^2-2h\right)=0\text{ and }f\left(0\right)=0$

There is no value of '$\lambda$.' for which the f (x) is continuous at x=0.