Question

Let f be a real-valued function of a real variable defined as f(x) = $x^2$ for x $\ge$ 0, and f(x) = -x${}^2$ for x < 0, Which one of the following statements is true?

• A. f(x) is discontinuous at x = 0
• B. f(x) is continuous but not differentiable at x = 0
• C. f(x) is differentiable but its first derivative is not continuous at x = 0
• D. f(x) is differentiable but its first derivative is not differentiable at x = 0

Answer 1

The correct option is D f(x) is differentiable but its first derivative is not differentiable at x = 0

f(x) = $\{\begin{array}{cc}-x^2,& x<0\\ x^2,& x\ge 0\end{array}$

LHL = RHL = 0 = f(0)

So f(x) is continuous at x = 0

f '(x) = $\{\begin{array}{cc}-2x,& x<0\\ 2x,& x>0\end{array}$

LHS = RHD = 0

So, f(x) is differentiable at x = 0

Now, f ''(x) = $\{\begin{array}{cc}-2,& x<0\\ 2,& x>0\end{array}$

$\because LHD=-2andRHD=2$

i.e, LHD $\ne$ RHD
So f '(x) is not differentiable at x = 0