If $f(x)\left\{\begin{matrix}ax^2+b:&x≤0\\x^2:&x>0\end{matrix}\right.$ possesses derivative at x = 0, then:
Answer & explanation
Correct answer: option 4
$\underset{x→0^+}{\lim}f(x)=\underset{x→0^-}{\lim}f(x)$
$⇒ b = 0$
$\underset{x→0^+}{\lim}f'(x)=\underset{x→0^-}{\lim}f'(x)$
so $2x =2ax$
$⇒ a = 1$
$a=1,b=0$