Calculus and Optimization PYQs for GATE DA

GATE DA Calculus and Optimization: 2 units and 5 chapters, weightage from 13 previous year questions across 3 papers, a study order by exam weight and 511 practice questions.

About Calculus and Optimization Previous Year Questions (PYQs)

13 previous year questions from Calculus and Optimization in GATE DA, grouped by chapter with the exam year, answer key and step-by-step solution for each.

GATE DA Calculus and Optimization Unit-wise Weightage from Past Papers

We counted every GATE DA Calculus and Optimization previous year question in our bank (13 questions from 3 papers) and grouped them by unit.

Unit	Chapters	PYQs	Share of section	Avg per paper
Calculus	4	12	92%	4
Optimization	1	1	8%	0.3

Suggested Calculus and Optimization Study Order for GATE DA

Calculus: 92% of past Calculus and Optimization questions, about 4 per paper.
Optimization: 8% of past Calculus and Optimization questions, about 0.3 per paper.

Start where the marks are. Units at the top of this list have appeared most often in past GATE DA papers.

Units in GATE DA Calculus and Optimization

All Calculus and Optimization chapters

One Solved Question from Each Calculus and Optimization Chapter

Question 1 · Sequences, Series and Limits · 2026 NAT

The value of $\sum_{i = 0}^{\infty} \sum_{j = 1}^{\infty} 2^{- i} 3^{- j}$ is ______________ . (Answer in integer)

Correct Answer:

1.00

Step-by-Step Solution

Insight: The summand $2^{- i} 3^{- j}$ is a product of a function of $i$ alone and a function of $j$ alone, so the double sum separates into the product of two independent geometric series.

Exam route: Split $\sum_{i} \sum_{j} 2^{- i} 3^{- j} = (\sum_{i = 0}^{\infty} 2^{- i}) (\sum_{j = 1}^{\infty} 3^{- j})$ . Evaluate each geometric series using $S = \frac{a}{1 - r}$ , then multiply.

Learning route:

This is a separable double summation question, recognisable because the general term $a_{i, j} = 2^{- i} \cdot 3^{- j}$ factors cleanly into $f (i) \cdot g (j)$ .

Step 1 — Apply the separation rule (Fubini for positive terms):

$i = 0 \sum \infty j = 1 \sum \infty 2^{- i} 3^{- j} = (i = 0 \sum \infty (\frac{1}{2})^{i}) \times (j = 1 \sum \infty (\frac{1}{3})^{j})$

Step 2 — First geometric series ( $i$ starts at $0$ , ratio $r = \frac{1}{2}$ ):

$S_{1} = \frac{1}{1 - \frac{1}{2}} = 2$

Step 3 — Second geometric series ( $j$ starts at $1$ , ratio $r = \frac{1}{3}$ , first term $a = \frac{1}{3}$ ):

$S_{2} = \frac{\frac{1}{3}}{1 - \frac{1}{3}} = \frac{\frac{1}{3}}{\frac{2}{3}} = \frac{1}{2}$

Step 4 — Multiply:

$S = 2 \times \frac{1}{2} = 1$

Trap check: A common mistake is to start the $j$ -sum at $0$ instead of $1$ , which would give $S_{2} = \frac{1}{1 - 1/3} = \frac{3}{2}$ and a wrong total of $3$ . Always read the lower limit carefully.

Verification: Evaluate the inner sum first: $\sum_{j = 1}^{\infty} 2^{- i} 3^{- j} = 2^{- i} \cdot \frac{1}{2}$ . Then $\sum_{i = 0}^{\infty} 2^{- i} \cdot \frac{1}{2} = \frac{1}{2} \cdot 2 = 1$ . Confirmed.

The answer is $1$ .

Question 2 · Continuity and Differentiability · 2025 MSQ

Consider two functions $f : R \to R$ and $g : R \to (1, \infty)$ . Both functions are differentiable at a point $c$ . Which of the following functions is/are ALWAYS differentiable at $c$ ? The symbol $\cdot$ denotes product and the symbol $\circ$ denotes composition of functions.

A.
$f \pm g$
B.
$f \cdot g$
C.
$\frac{f}{g}$
D.
$f \circ g + g \circ f$

Correct Answer:

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

Step-by-Step Solution

Insight: Algebraic operations ( $\pm, \cdot, /$ ) only require differentiability at the point $c$ , but composition requires differentiability of the outer function at the inner function's value.

Exam route: Check if the outer function is guaranteed to be differentiable at the required point. For $f \circ g$ , $f$ must be diff at $g (c)$ , which is not guaranteed if $f$ is only diff at $c$ .

Learning route:

We are given $f$ and $g$ are differentiable at $c$ . This does not imply they are differentiable everywhere.
Option A ( $f \pm g$ ): Sum/difference rule requires $f, g$ diff at $c$ . Always true.
Option B ( $f \cdot g$ ): Product rule requires $f, g$ diff at $c$ . Always true.
Option C ( $f / g$ ): Quotient rule requires $f, g$ diff at $c$ and $g (c) \neq = 0$ . Since $g : R \to (1, \infty)$ , $g (c) > 1 \neq = 0$ . Always true.
Option D ( $f \circ g + g \circ f$ ): Chain rule for $f \circ g$ requires $f$ to be differentiable at $g (c)$ . Since $f$ is only guaranteed to be differentiable at $c$ , this fails if $g (c) \neq = c$ . Similarly, $g \circ f$ requires $g$ to be diff at $f (c)$ , which fails if $f (c) \neq = c$ . Not always true.

Correct options are A, B, C.

Question 3 · Differentiation and Higher Order Derivatives · 2025 MCQ

Let $f (x) = \frac{e ^{x} - e ^{- x}}{2}$ , $x \in R$ . Let $f^{(k)} (a)$ denote the $k^{t h}$ derivative of $f$ evaluated at $a$ . What is the value of $f^{(10)} (0)$ ? (Note: ! denotes factorial)

A.
$0$
B.
$1$
C.
$\frac{1}{10 !}$
D.
$\frac{2}{10 !}$

Correct Answer:

Step-by-Step Solution

Insight: The function is the hyperbolic sine, $sinh (x)$ , whose derivatives cycle every two steps.

Exam route: Recognize $f (x) = sinh (x)$ . The $n$ -th derivative of $sinh (x)$ is $sinh (x)$ for even $n$ and $cosh (x)$ for odd $n$ . Since 10 is even, $f^{(10)} (x) = sinh (x)$ . Evaluate at $x = 0$ to get $sinh (0) = 0$ .

Learning route:

Write $f (x) = \frac{1}{2} e^{x} - \frac{1}{2} e^{- x}$ .

Differentiate 10 times using the rule $\frac{d ^{n}}{d x ^{n}} e^{a x} = a^{n} e^{a x}$ .

$f^{(10)} (x) = \frac{1}{2} (1)^{10} e^{x} - \frac{1}{2} (- 1)^{10} e^{- x} = \frac{1}{2} e^{x} - \frac{1}{2} e^{- x} = f (x)$ .

Substitute $x = 0$ : $f^{(10)} (0) = f (0) = \frac{e ^{0} - e ^{0}}{2} = 0$ .

Question 4 · Maxima, Minima and Applications of Derivatives · 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 5 · Optimization: Local and Global Extrema · 2025 MSQ

A.
$f$ has a local minima
B.
There does not exist $x$ and $y$ , $x \neq = y$ , such that $f^{'} (x) = f^{'} (y) = 0$
C.
$f$ has at most one global minimum
D.
$f$ has at most one local minimum

Correct Answer:

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

Step-by-Step Solution

Insight: Strict convexity means the function curves strictly upwards, preventing multiple flat spots or multiple dips, but not guaranteeing a minimum exists.

Exam route: Use the definition of strict convexity. (A) is false because $e^{x}$ is strictly convex but has no minimum. (B) is true because the derivative is strictly increasing. (C) and (D) are true because strict convexity forbids multiple minima.

Learning route:

Define strict convexity: $f (t x + (1 - t) y) < t f (x) + (1 - t) f (y)$ for $t \in (0, 1)$ and $x \neq = y$ .
Tempting wrong path: "Convex means bowl-shaped, so it must have a bottom." This leads to selecting (A). This breaks at the counterexample $f (x) = e^{x}$ , which is strictly convex but strictly increasing, so it never flattens out. Generalization: Strict convexity guarantees the shape, but not the existence of an extremum unless the function is coercive.
Check (A): False, as shown by $f (x) = e^{x}$ .
Check (B): If $f$ is strictly convex and differentiable at $x < y$ , then $f^{'} (x) < f^{'} (y)$ . Thus, $f^{'}$ can be zero at most once. There do not exist $x \neq = y$ with $f^{'} (x) = f^{'} (y) = 0$ . (B) is true.
Check (C): Suppose two global minima $x \neq = y$ exist with $f (x) = f (y) = m$ . Then $f ((x + y) /2) < (f (x) + f (y)) /2 = m$ , contradicting $m$ is the global minimum. (C) is true.
Check (D): For convex functions, any local minimum is a global minimum. Since there is at most one global minimum, there is at most one local minimum. (D) is true.

Verification: For $f (x) = e^{x}$ , $f^{'} (x) = e^{x} \neq = 0$ (B holds), no global min (C holds), no local min (D holds), no local min (A fails).

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Calculus and Optimization PYQs for GATE DA

About Calculus and Optimization Previous Year Questions (PYQs)

GATE DA Calculus and Optimization Unit-wise Weightage from Past Papers

Suggested Calculus and Optimization Study Order for GATE DA

Units in GATE DA Calculus and Optimization

All Calculus and Optimization chapters

One Solved Question from Each Calculus and Optimization Chapter

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution