GATE 2026 IN Q.12: solution and answer

Official answer: DSource: Official answer key, GATE 2026 IN (IIT Guwahati)

1. Write the function piecewise: $$f(x)=x|x|=\{\begin{array}{ll}{x}^2,& x\ge 0\\ -x^2,& x<0\end{array}$$

2. Continuity at $x=0$: ${\lim }_{x\to 0^{+}}{x}^2=0$, ${\lim }_{x\to 0^{-}}(-x^2)=0$ and $f(0)=0$. The left limit, the right limit and the value agree, so $f$ is continuous at $x=0$.

3. Differentiability, from the definition of the derivative. For $h\ne 0$, $\frac{h|h|}{h}=|h|$, so $$f^{\prime }(0)=\underset{h\to 0}{\lim }\frac{f(h)-f(0)}{h}=\underset{h\to 0}{\lim }\frac{h|h|}{h}=\underset{h\to 0}{\lim }|h|=0$$

4. The one-sided derivatives agree: the right-hand derivative is ${\lim }_{h\to 0^{+}}h=0$ and the left-hand derivative is ${\lim }_{h\to 0^{-}}(-h)=0$. So $f^{\prime }(0)$ exists and equals $0$.

5. Cross-check: $f^{\prime }(x)=2x$ for $x>0$, $f^{\prime }(x)=-2x$ for $x<0$ and $f^{\prime }(0)=0$, which is $f^{\prime }(x)=2|x|$ for every $x$. This derivative is continuous but has a corner at $0$, so $f$ is differentiable once at $0$ but not twice.

6. Eliminate the options. A and B both claim $f$ is discontinuous at $0$, which step 2 disproves (B is impossible anyway, since a function differentiable at a point is continuous there). C claims $f$ is not differentiable at $0$, which steps 3 and 4 disprove. Option D is correct.