GATE DA Maxima and Minima Previous Year Questions

GATE DA 2025 | Question 49

MSQ · 2 marks

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

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

Explanation

Answer: (C), (D)

On a closed interval the extreme values occur at critical points or at the endpoints, so compare all of them.

\(f'(x) = x^{2} + 7x + 10 = (x + 2)(x + 5)\), so there is a local maximum at \(x = -5\) and a local minimum at \(x = -2\). The values are \(f(-8) = \frac{239}{6} \approx 39.83\), \(f(-5) = \frac{187}{3} \approx 62.33\), \(f(-2) = \frac{347}{6} \approx 57.83\) and \(f(0) = \frac{133}{2} = 66.5\).

The maximum is \(\frac{133}{2}\) at the endpoint \(x = 0\), not at \(x = -5\): C holds and A fails.
The minimum is at the endpoint \(x = -8\), not at \(x = -2\): B fails.
\(f'(x) = x^{2} + 7x + 10\) is an upward parabola with its vertex at \(x = -\frac{7}{2}\), which lies inside \([-8, 0]\): D holds.

Answer: C, D

GATE DA 2025 | Question 51

MSQ · 2 marks

Let \(f : \mathbb{R} \to \mathbb{R}\) be a twice-differentiable function and suppose its second derivative satisfies \(f''(x) > 0\) for all \(x \in \mathbb{R}\). Which of the following statements is/are ALWAYS correct?

\(f\) has a local minima There does not exist \(x\) and \(y\), \(x \neq y\), such that \(f'(x) = f'(y) = 0\) \(f\) has at most one global minimum \(f\) has at most one local minimum

Explanation

Answer: (B), (C), (D)

Since \(f''(x) > 0\) everywhere, \(f'\) is strictly increasing, so it can be zero at most once; \(f\) is strictly convex.

A fails: \(f(x) = e^{x}\) has \(f''(x) = e^{x} > 0\), yet \(f'(x) = e^{x}\) is never 0, so this \(f\) has no minimum at all.
B holds: a strictly increasing \(f'\) cannot be 0 at two different points.
D holds: at a local minimum of a differentiable function \(f' = 0\), and there is at most one such point.
C holds: every global minimum is also a local minimum, so there is at most one.

Answer: B, C, D

GATE DA 2024 | Question 15

MCQ · 1 mark · −0.33

For any twice differentiable function \(f: \mathbb{R} \to \mathbb{R}\), if at some \(x^{*} \in \mathbb{R}\), \(f'(x^{*}) = 0\) and \(f''(x^{*}) > 0\), then the function \(f\) necessarily has a ______ at \(x = x^{*}\).

Note: \(\mathbb{R}\) denotes the set of real numbers.

local minimum global minimum local maximum global maximum

Explanation

Answer: (A)

This is the second-derivative test. \(f'(x^{*}) = 0\) makes \(x^{*}\) a stationary point, and \(f''(x^{*}) > 0\) means \(f'\) is increasing through \(x^{*}\): \(f' < 0\) just to the left and \(f' > 0\) just to the right. So \(f\) falls and then rises near \(x^{*}\), giving a local minimum; a maximum is impossible.

The test only looks near \(x^{*}\), so a global minimum is not guaranteed: \(f(x) = x^{3} - 3x\) has \(f'(1) = 0\) and \(f''(1) = 6 > 0\), yet \(f(x) \to -\infty\) as \(x \to -\infty\).

Answer: A

GATE DA 2024 | Question 50

MSQ · 2 marks

Consider the function \(f: \mathbb{R} \to \mathbb{R}\) where \(\mathbb{R}\) is the set of all real numbers.

\[ f(x) = \frac{x^{4}}{4} - \frac{2x^{3}}{3} - \frac{3x^{2}}{2} + 1 \]

Which of the following statements is/are TRUE?

\(x = 0\) is a local maximum of \(f\) \(x = 3\) is a local minimum of \(f\) \(x = -1\) is a local maximum of \(f\) \(x = 0\) is a local minimum of \(f\)

Explanation

Answer: (A), (B)

Differentiate and factor:

\[ f'(x) = x^{3} - 2x^{2} - 3x = x(x - 3)(x + 1) \]

so the critical points are \(x = -1, 0, 3\). Test each with the second derivative \(f''(x) = 3x^{2} - 4x - 3\):

\(f''(0) = -3 < 0\): a local maximum at \(x = 0\), so A holds and D fails.
\(f''(3) = 27 - 12 - 3 = 12 > 0\): a local minimum at \(x = 3\), so B holds.
\(f''(-1) = 3 + 4 - 3 = 4 > 0\): \(x = -1\) is a local minimum, not a maximum, so C fails.

Answer: A, B