Maxima, Minima and Applications of Derivatives Practice Questions for GATE DA

Solve 118+ Maxima, Minima and Applications of Derivatives practice 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 (10 Problems)

Question 1 · Calculus and Optimization MCQ

Let $f (x) = e^{x} - 2 x$ , which has a critical point at $x = ln 2$ . Using the Second Derivative Test, what does the test determine about $x = ln 2$ ?

A.
Local minimum
B.
Local maximum
C.
Inflection point
D.
Test fails

Correct Answer:

Step-by-Step Solution

Key idea: observation-type application of the second derivative test: sign of $f^{''}$ at the critical point classifies it.

Step 1: $f^{'} (x) = e^{x} - 2$ , and indeed $f^{'} (ln 2) = 2 - 2 = 0$ .

Step 2: $f^{''} (x) = e^{x}$ , so $f^{''} (ln 2) = e^{l n 2} = 2 > 0$ .

Step 3: Positive second derivative ⇒ concave up ⇒ local minimum by the test.

Note: the question asks only what the TEST determines; no boundary/global claim is needed.

Answer: A

Question 2 · Calculus and Optimization MCQ

Let $f (x) = - x^{2} + 6 x - 5$ . The function has a critical point at $x = 3$ . What does the Second Derivative Test conclude about $x = 3$ ?

A.
Local minimum
B.
Inflection point
C.
Local maximum
D.
Test is inconclusive

Correct Answer:

Step-by-Step Solution

Key idea: observation-type application of the second derivative test — sign of $f^{''}$ at the critical point classifies it.

Step 1: Compute $f^{'} (x) = - 2 x + 6$ . Verify: $f^{'} (3) = - 6 + 6 = 0$ . Confirmed critical point.

Step 2: Compute $f^{''} (x) = - 2$ .

Step 3: Evaluate: $f^{''} (3) = - 2 < 0$ .

Step 4: Negative second derivative ⇒ concave down ⇒ local maximum by the test.

Answer: C

Question 3 · Calculus and Optimization MCQ

For $f (x) = \frac{x ^{4}}{4} - \frac{2 x ^{3}}{3} - \frac{3 x ^{2}}{2} + 1$ , consider the assertion: " $x = 3$ is a local minimum" and the reason: " $f^{''} (3) = 12 > 0$ ".

A.
Both assertion and reason are true, and the reason is the correct explanation
B.
Both are true, but the reason is not the correct explanation
C.
Assertion is true, reason is false
D.
Assertion is false, reason is true

Correct Answer:

Step-by-Step Solution

Key idea: construction-by-substitution to verify both the assertion and the reason, then check their logical link.

Step 1: Recall $f^{'} (x) = x^{3} - 2 x^{2} - 3 x$ and $f^{''} (x) = 3 x^{2} - 4 x - 3$ .

Step 2: Verify the reason: $f^{''} (3) = 3 (3)^{2} - 4 (3) - 3 = 27 - 12 - 3 = 12$ . Since $12 > 0$ , the reason is TRUE.

Step 3: Verify the assertion: Since $f^{'} (3) = 3^{3} - 2 (3)^{2} - 3 (3) = 27 - 18 - 9 = 0$ , $x = 3$ is a critical point. And since $f^{''} (3) = 12 > 0$ , the Second Derivative Test confirms it is a local minimum. The assertion is TRUE.

Step 4: Check the link: The reason ( $f^{''} (3) > 0$ ) is the exact condition of the Second Derivative Test that proves the assertion (local minimum). Thus, the reason is the correct explanation.

Answer: A

Question 4 · Calculus and Optimization MCQ

For $f (x) = x^{3} - 3 x$ , consider the two statements about $x = 1$ :

(I) $f^{'} (1) = 0$ .

(II) $x = 1$ is a critical point of $f$ .

Which conclusion is correct?

A.
Both I and II are true
B.
I is true, II is false
C.
I is false, II is true
D.
Both I and II are false

Correct Answer:

Step-by-Step Solution

Key idea: this is a casework check of the definition of a critical point ( $f^{'} (c) = 0$ or $f^{'} (c)$ undefined).

Step 1: Compute $f^{'} (x) = 3 x^{2} - 3$ .

Step 2: Evaluate at $x = 1$ : $f^{'} (1) = 3 (1)^{2} - 3 = 0$ . So statement I is TRUE.

Step 3: Since $f^{'} (1) = 0$ (and the derivative exists everywhere for a polynomial), $x = 1$ satisfies the definition of a critical point. Statement II is TRUE.

Step 4: Both statements hold → option A.

Answer: A

Question 5 · Calculus and Optimization MCQ

Consider $f (x) = x^{1/3}$ . Which statement correctly classifies $x = 0$ ?

A.
It is a critical point because $f^{'} (0)$ does not exist
B.
It is a critical point because $f^{'} (0) = 0$
C.
It is not a critical point because $f (0) = 0$
D.
It is not a critical point because $f$ is continuous at $0$

Correct Answer:

Step-by-Step Solution

Key idea: this is a casework check of the critical-point definition — two branches: $f^{'} (c) = 0$ OR $f^{'} (c)$ undefined.

Step 1: Compute $f^{'} (x) = \frac{1}{3} x^{- 2/3} = \frac{1}{3 x ^{2/3}}$ .

Step 2: At $x = 0$ , the denominator is zero, so $f^{'} (0)$ does not exist (vertical tangent).

Step 3: Since $f^{'} (0)$ is undefined, $x = 0$ satisfies the second branch of the critical-point definition. It is a critical point.

Step 4: Option A states this correctly. Option B incorrectly claims $f^{'} (0) = 0$ . Options C and D confuse function value or continuity with the derivative condition.

Answer: A

Question 6 · Calculus and Optimization MCQ

How many critical points does the function $f (x) = ∣ x^{2} - 4∣$ have in the set of real numbers?

A.
1
B.
2
C.
3
D.
4

Correct Answer:

Step-by-Step Solution

Key idea: this is a casework check of the critical-point definition across the pieces of an absolute-value function.

Step 1: The expression inside the absolute value is zero at $x = 2$ and $x = - 2$ . At these points, the function has sharp corners, so $f^{'} (x)$ does not exist. These are two critical points.

Step 2: For $- 2 < x < 2$ , $f (x) = 4 - x^{2}$ , so $f^{'} (x) = - 2 x$ . Setting $f^{'} (x) = 0$ gives $x = 0$ . This is a third critical point.

Step 3: For $x > 2$ and $x < - 2$ , $f (x) = x^{2} - 4$ , so $f^{'} (x) = 2 x$ . Setting $f^{'} (x) = 0$ gives $x = 0$ , which is not in these intervals. No additional critical points.

Step 4: Total critical points: $x = - 2, 0, 2$ . The count is 3.

Answer: C

Question 7 · Calculus and Optimization MCQ

How many distinct critical points does the function $f (x) = x^{4} - 4 x^{3} + 6 x^{2}$ have in the set of real numbers?

A.
0
B.
1
C.
2
D.
3

Correct Answer:

Step-by-Step Solution

Key idea: Critical points occur where $f^{'} (x) = 0$ . For polynomials, we find the distinct real roots of the derivative.

Step 1: Find the first derivative.

$f^{'} (x) = 4 x^{3} - 12 x^{2} + 12 x$ .

Step 2: Factor out the greatest common factor.

$f^{'} (x) = 4 x (x^{2} - 3 x + 3)$ .

Step 3: Set each factor to zero and solve for real roots.

$4 x = 0 ⟹ x = 0$ .

For the quadratic $x^{2} - 3 x + 3 = 0$ , check the discriminant:

$Δ = b^{2} - 4 a c = (- 3)^{2} - 4 (1) (3) = 9 - 12 = - 3$ .

Since $Δ < 0$ , the quadratic has no real roots.

Step 4: Count the distinct real roots.

The only real root is $x = 0$ .

Answer: There is exactly 1 distinct critical point.

Question 8 · Calculus and Optimization MCQ

Given $f (x) = \frac{x ^{4}}{4} - \frac{2 x ^{3}}{3} - \frac{3 x ^{2}}{2} + 1$ with $f^{''} (x) = 3 x^{2} - 4 x - 3$ , construct the value of $f^{''} (- 1)$ and classify $x = - 1$ .

A.
$f^{''} (- 1) = 4 > 0$ , local minimum
B.
$f^{''} (- 1) = - 4 < 0$ , local maximum
C.
$f^{''} (- 1) = 2 > 0$ , local minimum
D.
$f^{''} (- 1) = 0$ , test fails

Correct Answer:

Step-by-Step Solution

Key idea: construction-by-substitution — build $f^{''} (- 1)$ term by term from the supplied formula, keeping signs intact.

Step 1: $f^{''} (- 1) = 3 (- 1)^{2} - 4 (- 1) - 3$ .

Step 2: Term values: $3 (1) = 3$ ; $- 4 (- 1) = + 4$ ; constant $- 3$ .

Step 3: Sum: $3 + 4 - 3 = 4 > 0$ ⇒ concave up ⇒ local minimum at $x = - 1$ .

Step 4: Only option A reports both the correct constructed value and the correct classification.

Answer: A

Question 9 · Calculus and Optimization MCQ

Given $f (x) = \frac{x ^{4}}{4} - \frac{2 x ^{3}}{3} - \frac{3 x ^{2}}{2} + 1$ with $f^{''} (x) = 3 x^{2} - 4 x - 3$ , construct the value of $f^{''} (0)$ and classify $x = 0$ .

A.
$f^{''} (0) = 3 > 0$ , local minimum
B.
$f^{''} (0) = 0$ , test fails
C.
$f^{''} (0) = - 3 < 0$ , local minimum
D.
$f^{''} (0) = - 3 < 0$ , local maximum

Correct Answer:

Step-by-Step Solution

Key idea: construction-by-substitution — build $f^{''} (0)$ term by term from the supplied formula.

Step 1: $f^{''} (0) = 3 (0)^{2} - 4 (0) - 3$ .

Step 2: Term values: $3 (0) = 0$ ; $- 4 (0) = 0$ ; constant $- 3$ .

Step 3: Sum: $0 + 0 - 3 = - 3 < 0$ .

Step 4: Negative second derivative ⇒ concave down ⇒ local maximum at $x = 0$ .

Step 5: Only option D reports both the correct constructed value and the correct classification.

Answer: D

Question 10 · Calculus and Optimization MCQ

Let $f (x) = - cos x$ . At $x = π$ , the function has a critical point. What does the Second Derivative Test conclude about $x = π$ ?

A.
Local minimum
B.
Local maximum
C.
Inflection point
D.
Test fails

Correct Answer:

Step-by-Step Solution

Key idea: observation-type application of the second derivative test to a trigonometric function.

Step 1: Compute $f^{'} (x) = sin x$ . Verify: $f^{'} (π) = sin (π) = 0$ . Confirmed critical point.

Step 2: Compute $f^{''} (x) = cos x$ .

Step 3: Evaluate: $f^{''} (π) = cos (π) = - 1$ .

Step 4: Since $f^{''} (π) = - 1 < 0$ , the function is concave down at $x = π$ . By the Second Derivative Test, this indicates a local maximum.

Answer: B

Try a question

Login to view the complete practice questions and solutions

Personalised first. High quality throughout.

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 Practice Questions 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 (10 Problems)

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

More practice questions in this unit