GATE 2024 DA Q.15: solution and answer

Official answer: ASource: Official answer key, GATE 2024 DA (IISc Bengaluru)

1. Since $f^{\prime }(x^{\ast })=0$, the point $x^{\ast }$ is a stationary (critical) point of $f$.

2. Because $f^{\prime \prime }(x^{\ast })={\lim }_{x\to x^{\ast }}\frac{f^{\prime }(x)-f^{\prime }(x^{\ast })}{x-x^{\ast }}={\lim }_{x\to x^{\ast }}\frac{f^{\prime }(x)}{x-x^{\ast }}>0$, the quotient is positive for $x$ close to $x^{\ast }$. So $f^{\prime }(x)<0$ just to the left of $x^{\ast }$ and $f^{\prime }(x)>0$ just to the right. Hence $f$ decreases and then increases, and $x^{\ast }$ is a strict local minimum (the second-derivative test).

3. It cannot be a local maximum, since $f$ would have to stop increasing after $x^{\ast }$, but $f^{\prime }>0$ immediately to the right.

4. The conclusion is local only, because $f^{\prime \prime }(x^{\ast })>0$ gives no information about $f$ far from $x^{\ast }$. To show that a global minimum is not guaranteed, take $f(x)=x^3-3x$ and $x^{\ast }=1$. Then $f^{\prime }(x)=3x^2-3$, so $f^{\prime }(1)=0$, and $f^{\prime \prime }(x)=6x$, so $f^{\prime \prime }(1)=6>0$. Yet $f(1)=-2$ while $f(-3)=-27+9=-18<-2$ (indeed $f(x)\to -\infty$ as $x\to -\infty$). So $x=1$ is a local minimum but not a global one. This also rules out a global maximum, since a global maximum would be a local maximum, which was excluded in step 3.

5. So $f$ necessarily has a local minimum at $x^{\ast }$, which is option A.