Calculus PYQs for GATE DA

GATE DA Calculus: 4 chapters, 12 previous year questions (92% of Calculus and Optimization), 469 practice questions and one solved question from each chapter.

About Calculus Previous Year Questions (PYQs)

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

Calculus Weightage in GATE DA

Calculus accounts for 12 of 13 Calculus and Optimization previous year questions in our bank (92%), about 4 per paper across 3 papers.

Calculus Chapter Matrix

Chapter	Topics	PYQs	Share of unit PYQs	Practice questions
Sequences, Series and Limits	Infinite Series and Double Summations, Limits Using Algebraic and Logarithmic Expansions	3	25%	122
Continuity and Differentiability	Differentiability Rules and Function Operations, Continuity, Differentiability and Functional Constraints	3	25%	150
Differentiation and Higher Order Derivatives	Derivative Evaluation of Standard Functions	2	17%	79
Maxima, Minima and Applications of Derivatives	Critical Points and Second Derivative Test, Polynomial Extrema, Roots and Interval Analysis	4	33%	118

One Solved Question from Each Calculus 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.

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Calculus PYQs for GATE DA

About Calculus Previous Year Questions (PYQs)

Calculus Weightage in GATE DA

Calculus Chapter Matrix

More from Calculus and Optimization

One Solved Question from Each Calculus Chapter

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution