GATE 2025 DA Q.49: solution and answer

GATE 2025 · DAQ.49 · 2 marks · MSQ

Consider the function $f (x) = \frac{x ^{3}}{3} + \frac{7}{2} x^{2} + 10 x + \frac{133}{2}$ , $x \in [- 8, 0]$ . Which of the following statements is/are correct?

(A)
The maximum value of $f$ is attained at $x = - 5$
(B)
The minimum value of $f$ is attained at $x = - 2$
(C)
The maximum value of $f$ is $\frac{133}{2}$
(D)
The minimum value of the derivative of $f$ is attained at $x = - \frac{7}{2}$

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Answer and solution

ANSWER

The correct statements are (C) and (D).

Official answer: C, DSource: Official answer key, GATE 2025 DA (IIT Roorkee)

Differentiate: $f^{'} (x) = x^{2} + 7 x + 10 = (x + 2) (x + 5)$ , so the critical points are $x = - 5$ and $x = - 2$ , both inside $[- 8, 0]$ .
Sign of $f^{'}$ : positive on $[- 8, - 5)$ , negative on $(- 5, - 2)$ , positive on $(- 2, 0]$ . So $x = - 5$ is a local maximum and $x = - 2$ is a local minimum.
On a closed interval the absolute extremes occur at a critical point or at an endpoint, so compare $f (- 8)$ , $f (- 5)$ , $f (- 2)$ and $f (0)$ .
$f (- 8) = - \frac{512}{3} + 224 - 80 + \frac{133}{2} = \frac{239}{6} \approx 39.83$ .
$f (- 5) = - \frac{125}{3} + \frac{175}{2} - 50 + \frac{133}{2} = \frac{187}{3} \approx 62.33$ .
$f (- 2) = - \frac{8}{3} + 14 - 20 + \frac{133}{2} = \frac{347}{6} \approx 57.83$ , and $f (0) = \frac{133}{2} = 66.5$ .
The largest value is $f (0) = \frac{133}{2}$ , attained at the endpoint $x = 0$ , not at $x = - 5$ (where $f \approx 62.33$ ). So (A) is false and (C) is true.
The smallest value is $f (- 8) = \frac{239}{6}$ , attained at the endpoint $x = - 8$ , not at $x = - 2$ (where $f \approx 57.83$ ). So (B) is false.
For (D), $f^{'} (x) = x^{2} + 7 x + 10$ is an upward-opening parabola, so it is minimised where its derivative $f^{''} (x) = 2 x + 7$ is zero, i.e. $x = - \frac{7}{2}$ . This lies in $[- 8, 0]$ , so (D) is true.
Correct options: (C) and (D).

KEY IDEA

On a closed interval the absolute maximum and minimum are found by comparing the function at the critical points and at both endpoints. A local maximum or minimum is not automatically the absolute one. The derivative's own minimum comes from setting the second derivative to zero.

WATCH OUT FOR

Stopping at the critical points $x = - 5$ and $x = - 2$ and calling them the maximum and minimum, which gives (A) and (B). This ignores the endpoints $x = 0$ and $x = - 8$ , where the function is actually largest and smallest.

From the GATE 2025 DA paper. Every question of this paper · Download the full paper · All 5 Maxima & minima questions, with notes

GATE 2025 DA Q.49: solution and answer

Answer and solution

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