GATE 2024 EE Q.43: solution and answer

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

1. Since $f^{\prime \prime }(t)>0$ for every real $t$, the function $f^{\prime }(t)$ is strictly increasing on $(-\infty ,\infty )$, because a function with a positive derivative is strictly increasing. Also, $f$ is differentiable, so any local extremum $t_0$ must satisfy $f^{\prime }(t_0)=0$ (Fermat's theorem).

2. (A) Counter-example: $f(t)=e^t$ has $f^{\prime \prime }(t)=e^t>0$, but $f^{\prime }(t)=e^t\ne 0$ for every $t$. It has no stationary point and therefore no local minimum. So (A) is not always true.

3. (B) Suppose $t_1<t_2$ were two distinct local minima. Then $f^{\prime }(t_1)=f^{\prime }(t_2)=0$. But $f^{\prime }$ is strictly increasing, so $f^{\prime }(t_1)<f^{\prime }(t_2)$, which is a contradiction. So $f^{\prime }$ has at most one zero, and $f$ has at most one local minimum. (B) is always true.

4. (C) Suppose $t_0$ were a local maximum. Then $f^{\prime }(t_0)=0$. Because $f^{\prime }$ is strictly increasing, $f^{\prime }(t)<0$ for $t<t_0$ and $f^{\prime }(t)>0$ for $t>t_0$. So $f$ falls and then rises around $t_0$, which makes $t_0$ a strict local minimum, not a maximum. So no such $f$ can have a local maximum; $f(t)=e^t$ and $f(t)=t^2$ are examples. (C) is false.

5. (D) Counter-example: $f(t)=t^2-1$ has $f^{\prime \prime }(t)=2>0$, and its minimum value is $f(0)=-1<0$. Convexity controls the shape of the graph, not how high or low it sits, because adding any constant leaves $f^{\prime \prime }$ unchanged. So (D) is not always true.

6. Only (B) holds for every function with $f^{\prime \prime }(t)>0$ on $(-\infty ,\infty )$. Answer: (B)