MCQ
If $f(x) = \left\{ \begin{array}{l}a{x^2} + b;\,\,x \le 0\\\,\,\,\,\,\,\,\,\,{x^2};x > 0\,\end{array} \right.$ possesses derivative at $x = 0$, then
- A$a = 0,b = 0$
- B$a > 0, = 0$
- ✓$a \in R, = 0$
- DNone of these
Now $f(0 + 0) = 0$, $f(0 - 0) = b,f(0) = b$,
$\therefore \,\,b = 0$
Also $Rf'(0) = 0,Lf'(0) = 0,\forall a \in R$
$\therefore \,f'(0) = 0$ if $b = 0$.
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