GATE 2025 EC Q.25: solution and answer

GATE 2025 · ECQ.25 · 1 mark · MSQ

Consider the function $f : R \to R$ , defined as
$f (x) = 2 x^{3} - 3 x^{2} - 12 x + 1.$
Which of the following statements is/are correct?

(Here, $R$ is the set of real numbers.)

(A)
$f$ has no global maximizer
(B)
$f$ has no global minimizer
(C)
$x = - 1$ is a local minimizer of $f$
(D)
$x = 2$ is a local maximizer of $f$

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Answer and solution

ANSWER

(A) and (B) are correct. $f$ has a local maximum at $x = - 1$ and a local minimum at $x = 2$ , but as a cubic on all of $R$ it is unbounded above and below, so it has no global maximizer and no global minimizer.

Official answer: A, BSource: Official answer key, GATE 2025 EC (IIT Roorkee)

Find the stationary points. $f^{'} (x) = 6 x^{2} - 6 x - 12 = 6 (x^{2} - x - 2) = 6 (x - 2) (x + 1)$ , so the critical points are $x = - 1$ and $x = 2$ .
Second derivative: $f^{''} (x) = 12 x - 6$ .
At $x = - 1$ : $f^{''} (- 1) = - 18 < 0$ , so $x = - 1$ is a local maximizer, with $f (- 1) = - 2 - 3 + 12 + 1 = 8$ . So (C) is false.
At $x = 2$ : $f^{''} (2) = 18 > 0$ , so $x = 2$ is a local minimizer, with $f (2) = 16 - 12 - 24 + 1 = - 19$ . So (D) is false.
Behaviour on all of $R$ : $f$ is a cubic with positive leading coefficient, so $f (x) \to + \infty$ as $x \to + \infty$ and $f (x) \to - \infty$ as $x \to - \infty$ .
$f$ is unbounded above, so no point is a global maximizer. For example $f (4) = 128 - 48 - 48 + 1 = 33 > 8$ , so the local maximum at $x = - 1$ is not global. So (A) is true.
$f$ is unbounded below, so no point is a global minimizer. For example $f (- 3) = - 54 - 27 + 36 + 1 = - 44 < - 19$ , so the local minimum at $x = 2$ is not global. So (B) is true.
Correct options: (A) and (B).

KEY IDEA

A local extremum is global only if the function never goes past it anywhere else in the domain. A polynomial of odd degree on all of $R$ goes to $+ \infty$ at one end and $- \infty$ at the other, so it never has a global maximizer or a global minimizer, whatever its local extrema are. To classify the stationary points of $f^{'} (x) = 6 (x - 2) (x + 1)$ , use the sign of $f^{''} (x) = 12 x - 6$ .

WATCH OUT FOR

Swapping the local classifications. For a cubic with positive leading coefficient the smaller critical point ( $x = - 1$ ) is the local maximum and the larger one ( $x = 2$ ) is the local minimum. Another mistake is to call a local extremum global without checking the ends of the domain: on a closed interval the cubic would have global extrema, but on $R$ it has none.

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GATE 2025 EC Q.25: solution and answer

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