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Maxima, Minima and Applications of Derivatives PYQs for GATE DA

Solve 4+ Maxima, Minima and Applications of Derivatives previous year questions for GATE DA with answers and detailed solutions. Free sample questions below.

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Question 1

2026 PYQ

Level 3: Exam Standard

Let f (x) = x^{3} - 3 x^{2} + 2 be a function defined on (- 1, 3] . Which of the following statements is/are correct?

Question 2

2025 PYQ

Level 3: Exam Standard

Question 3

2024 PYQ

Level 3: Exam Standard

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

Question 4

2024 PYQ

Level 3: Exam Standard

Consider the function f : R \to R where R is the set of all real numbers. f (x) = \frac{x ^{4}}{4} - \frac{2 x ^{3}}{3} - \frac{3 x ^{2}}{2} + 1 Which of the following statements is/are TRUE?

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Maxima, Minima and Applications of Derivatives PYQs for GATE DA

Solve 4+ Maxima, Minima and Applications of Derivatives previous year questions for GATE DA with answers and detailed solutions. Free sample questions below.

Chapter Roadmap: Maxima, Minima and Applications

Chapter Journey

Critical Points & Second Derivative Test

Current Topic • Foundation

Polynomial Extrema & Interval Analysis

Global Max/Min, Boundary Checks

Goal for this topic: Master the identification of critical points and use the second derivative to classify them as local maxima, minima, or saddle points.

The Hero Concept: What is a Critical Point?

Intuition: The Flat Spots

A critical point of a function $f (x)$ occurs at $x = c$ if:

$f^{'} (c) = 0$ (The tangent is horizontal)
OR $f^{'} (c)$ does not exist (Sharp corner or vertical tangent)

f^{'} (c) = 0 or f^{'} (c) is undefined

Why care?
Local maxima (peaks) and local minima (valleys) can only occur at critical points. If the slope is not zero and exists, you are still going up or down.

Maxima, Minima and Applications of Derivatives: Solved Questions with Step-by-Step Explanations (4 Problems)

Question 1 · Calculus and Optimization · 2026 MSQ

Let

f (x) = x^{3} - 3 x^{2} + 2

be a function defined on

(- 1, 3]

.
Which of the following statements is/are correct?

A.
$f (x)$ has exactly two roots in $[- 0.9, 0]$ .
B.
$f (x)$ has a minimum at 2 only.
C.
$f (x)$ has maximum at 0 only.
D.
$f (x)$ has a root at 1.

Correct Answer:

["B","D"]

Step-by-Step Solution

Insight: Half-open interval extrema and root counting require careful boundary analysis.

Exam route: $f (x) = x^{3} - 3 x^{2} + 2$ on $(- 1, 3]$ . $f^{'} (x) = 3 x^{2} - 6 x = 3 x (x - 2)$ . Critical points: $x = 0, 2$ .

Evaluate $f$ : $f (0) = 2$ , $f (2) = - 2$ , $f (3) = 2$ . Limit as $x \to - 1^{+}$ is $f (- 1) = - 2$ .

Absolute max is $2$ (attained at $x = 0$ and $x = 3$ ). Absolute min is $- 2$ (attained at $x = 2$ ; $x = - 1$ is excluded).

Check options:

A: On $[- 0.9, 0]$ , $f^{'} (x) > 0$ (except at 0), so $f$ is strictly increasing. Exactly ONE root. FALSE.
B: Min is $- 2$ , attained only at $x = 2$ since $x = - 1$ is excluded. TRUE.
C: Max is $2$ , attained at both $x = 0$ and $x = 3$ . "At 0 only" is FALSE.
D: $f (1) = 1 - 3 + 2 = 0$ . TRUE.

Learning route: For half-open intervals, the Extreme Value Theorem doesn't guarantee extrema are attained. Always check if the supremum/infimum occurs at an excluded boundary. If it does, the function approaches it but never reaches it, meaning no actual max/min exists at that boundary.

Question 2 · Calculus and Optimization · 2025 MSQ

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 $x = - \frac{7}{2}$

Correct Answer:

["A","B","C","D"]

Step-by-Step Solution

Defect: The question statement is missing. Reconstructed from options: $f (x) = x^{3} + \frac{21}{2} x^{2} + 30 x + 79$ on $[- 5, - 2]$ .

Insight: Reconstruct the cubic using the derivative's vertex and endpoint values to verify all claims.

Exam route: $f^{'} (x) = 3 x^{2} + 21 x + 30$ . The vertex of this quadratic is at $x = - 21/6 = - 7/2$ , confirming D. Critical points of $f$ are $x = - 2, - 3$ . Evaluating on $[- 5, - 2]$ : $f (- 5) = 133/2$ , $f (- 2) = 53$ , $f (- 3) = 56.5$ . The absolute maximum is $133/2$ at $x = - 5$ (confirming A and C). The absolute minimum is $53$ at $x = - 2$ (confirming B). All options are true.

Learning route: When a question statement is missing, reverse-engineer the function from the options. The derivative's minimum at $x = - 7/2$ forces $f^{''} (x) = 6 a x + 2 b$ to have a root at $- 7/2$ . Choosing $a = 1$ gives $b = 21/2$ . Matching $f (- 5) = 133/2$ and $f^{'} (- 2) = 0$ uniquely determines the remaining coefficients. Always check both critical points and endpoints for absolute extrema on closed intervals.

Question 3 · Calculus and Optimization · 2024 MCQ

For any twice differentiable function

f : R \to R

, if at some

x^{*} \in R

f^{'} (x^{*}) = 0

and

f^{''} (x^{*}) > 0

, then the function

f

necessarily has a ______ at

x = x^{*}

.
Note:

R

denotes the set of real numbers.

A.
local minimum
B.
global minimum
C.
local maximum
D.
global maximum

Correct Answer:

Step-by-Step Solution

Insight: The Second Derivative Test only guarantees local extrema, not global.

Exam route: $f^{'} (x^{*}) = 0$ and $f^{''} (x^{*}) > 0$ means the function is concave up at $x^{*}$ . By the Second Derivative Test, this guarantees a local minimum at $x^{*}$ . It does not guarantee a global minimum because the function could go to $- \infty$ elsewhere (e.g., $f (x) = x^{3} /3 - x$ has a local min at $x = 1$ but no global min). Thus, "local minimum" is the only necessarily true statement.

Learning route: In optimization, local conditions (like $f^{''} > 0$ ) only describe the neighborhood of a point. Global extrema require analyzing the entire domain, especially for functions that are unbounded or have multiple critical points. Always distinguish between "local" and "global" in theoretical questions.

Question 4 · Calculus and Optimization · 2024 MSQ

Consider the function

f : R \to R

where

R

is the set of all real numbers.

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

Which of the following statements is/are TRUE?

A.
$x = 0$ is a local maximum of f
B.
$x = 3$ is a local minimum of f
C.
$x = - 1$ is a local maximum of f
D.
$x = 0$ is a local minimum of f

Correct Answer:

["A","B"]

Step-by-Step Solution

Insight: Standard polynomial extrema classification using the Second Derivative Test.

Exam route: $f^{'} (x) = x^{3} - 2 x^{2} - 3 x = x (x^{2} - 2 x - 3) = x (x - 3) (x + 1)$ . Critical points: $x = 0, 3, - 1$ . $f^{''} (x) = 3 x^{2} - 4 x - 3$ .

Evaluate $f^{''} (x)$ at each critical point:

$f^{''} (0) = - 3 < 0 ⟹$ local maximum at $x = 0$ . (Option A is TRUE, D is FALSE)
$f^{''} (3) = 27 - 12 - 3 = 12 > 0 ⟹$ local minimum at $x = 3$ . (Option B is TRUE)
$f^{''} (- 1) = 3 + 4 - 3 = 4 > 0 ⟹$ local minimum at $x = - 1$ . (Option C is FALSE)

Learning route: The Second Derivative Test is the fastest way to classify critical points for polynomials. Always factor $f^{'} (x)$ completely to avoid missing roots. A positive second derivative means concave up (local min), negative means concave down (local max).

Try a question

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Built around you, not around a syllabus PDF

Revision that hits your weak spots

Questions calibrated to the real exam

Notes written for recall, not for volume

One place for everything

Honest progress

Unlock the whole course

Maxima, Minima and Applications of Derivatives PYQs for GATE DA

Chapter Roadmap: Maxima, Minima and Applications

Chapter Journey

The Hero Concept: What is a Critical Point?

Intuition: The Flat Spots

Maxima, Minima and Applications of Derivatives: Solved Questions with Step-by-Step Explanations (4 Problems)

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

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