The function f defined by \(f(x)=\left\{\begin{matrix}\frac{\sin x^2}{x} &for \, x≠0 \\ 0&for \,  x = 0\end{matrix}\right.\) is

continuous and derivable at x = 0

The function is continuous at x = 0, because

$\underset{x→0}{\lim}f(x)=\underset{x→0}{\lim}\frac{\sin x^2}{x}=\underset{x→0}{\lim}(\frac{\sin x^2}{x^2})$. x = 0 = f(0).

Also, $Rf'(0)=\underset{h→0}{\lim}\frac{f(0+h)-f(0)}{h}=\underset{h→0}{\lim}\frac{\frac{\sin h^2}{h}-0}{h}=\underset{h→0}{\lim}\frac{\sin h^2}{h^2}=1$

and $Lf'(0)=\underset{h→0}{\lim}\frac{f(0-h)-f(0)}{-h}=\underset{h→0}{\lim}\frac{\frac{\sin h^2}{-h}-0}{-h}=\underset{h→0}{\lim}\frac{\sin h^2}{h^2}=1$

so that, f(x) is derivable at x = 0.

Hence (A) is the correct answer.