If $f(x)=\left\{\begin{matrix}ax^2+b, & x<-1\\bx^2+ax+4, & x≥-1\end{matrix}\right.$ is everywhere differentiable, then :
Answer & explanation
Correct answer: option 1
The correct answer is option (1) → $a=2, b=3$
$\lim\limits_{-1^-}(ax^2+b)=\lim\limits_{-1^+}(bx^2+ax+4)$
$a+b=b-a+4$
$a=2$
$f(x)=\left\{\begin{matrix}2ax& x<-1\\2bx+a& x>-1\end{matrix}\right.$
LHD at (-1) → -2a
RHD at (-1) → -2b+a
so $-2b+a=-2a$
$3a=2b$
$3×2=2b⇒b=3$
$a=2$