If the derivative of the function $f(x)=\left\{\begin{matrix}bx^2+ax+4;&x≥-1\\ax^2+b;&x<-1\end{matrix}\right.$ is everywhere continuous, then:

a = 2, b = 3 a = 1, b = 1 a = – 2, b = – 3 a = – 3, b = – 2

a = 1, b = 1

$LHL(-1)=RHL(-1)⇒b-a+4=a+b$ so $2a=4⇒a=2$ $LHD(-1)=RHD(-1)⇒2b(-1)+a=2a(-1)$ $⇒-2b+a=-2a$ so $3a=2b$ so $b=3$