GATE 2024 EE Q.44: solution and answer

Official answer: BSource: Official answer key, GATE 2024 EE (IISc Bengaluru)

1. Write it piecewise. $\max (0,t)=0$ for $t<0$ and $\max (0,t)=t$ for $t\ge 0$, so $$f(t)=\{\begin{array}{ll}0,& t<0\\ t^2,& t\ge 0\end{array}$$

2. Away from $t=0$ each piece is a polynomial, so $f^{\prime }(t)=0$ for $t<0$ and $f^{\prime }(t)=2t$ for $t>0$.

3. At $t=0$, use the definition $f^{\prime }(0)={\lim }_{h\to 0}\frac{f(h)-f(0)}{h}$ with $f(0)=0$. The right-hand limit is ${\lim }_{h\to 0^{+}}\frac{h^2}{h}={\lim }_{h\to 0^{+}}h=0$ and the left-hand limit is ${\lim }_{h\to 0^{-}}\frac{0}{h}=0$. They agree, so $f^{\prime }(0)=0$ and $f$ is differentiable at every $t$. That makes (A) false.

4. So $$f^{\prime }(t)=\{\begin{array}{ll}0,& t<0\\ 2t,& t\ge 0\end{array}=2\max (0,t).$$ Since ${\lim }_{t\to 0^{-}}{f}^{\prime }(t)=0$, ${\lim }_{t\to 0^{+}}{f}^{\prime }(t)={\lim }_{t\to 0^{+}}2t=0$ and $f^{\prime }(0)=0$, $f^{\prime }$ is continuous at $0$ (and it is clearly continuous elsewhere). That makes (B) true and (C) false.

5. Is $f^{\prime }$ itself differentiable at $0$? Right derivative: ${\lim }_{h\to 0^{+}}\frac{f^{\prime }(h)-f^{\prime }(0)}{h}={\lim }_{h\to 0^{+}}\frac{2h-0}{h}=2$. Left derivative: ${\lim }_{h\to 0^{-}}\frac{0-0}{h}=0$. Since $2\ne 0$, $f^{\prime \prime }(0)$ does not exist: $f^{\prime }=2\max (0,t)$ is a ramp with a corner at the origin.

6. So $f^{\prime }$ is not differentiable at $t=0$, and (D) is false. $f$ is continuously differentiable but not twice differentiable. Only (B) is correct.