If $f(x)=\left\{\begin{matrix}ax^2+b,&x≤0\\0,&x>0\end{matrix}\right.$ is differentiable at x = 0, then

a = 1, b =1 a = 1, b = 2 a = 2, b = 0 a = 2, b = 1

a = 2, b = 0

For continuity at x = 0, b = 0 For differentiability 2ax = 0 at x = 0 So a can be any real number , b = 0