Calculus PYQs for GATE CS

GATE CS Calculus: 3 chapters, 13 previous year questions (12% of Engineering Mathematics), 206 practice questions and one solved question from each chapter.

About Calculus Previous Year Questions (PYQs)

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

Calculus Weightage in GATE CS

Calculus accounts for 13 of 105 Engineering Mathematics previous year questions in our bank (12%), about 1.3 per paper across 10 papers.

Calculus Chapter Matrix

Chapter	Topics	PYQs	Share of unit PYQs	Practice questions
Limits and Continuity	Evaluation of Indeterminate Limits, Continuity of Piecewise Functions, Discontinuities of Digit-Defined Functions	4	31%	63
Differentiability and Optimization	Differentiability Conditions for Piecewise Functions, Nondifferentiability of Maximum Functions, Local Extrema and Smoothness, Mean Value Theorem and Derivative Bounds	5	38%	80
Integral Calculus	Integration by Parts and Integral Equations, Symmetry Properties of Definite Integrals, Multiple Integrals and Symmetry	4	31%	63

One Solved Question from Each Calculus Chapter

Question 1 · Limits and Continuity · 2026_Set2 NAT

Consider a function

f : (0, 1) \to {0, 1}

defined as follows.

For a real number

r \in (0, 1)

f (r) = 1

if the second digit after the decimal point in

r

is one of the four digits 2, 3, 6 and 7. Otherwise,

f (r)

is equal to 0.

The number of points in

(0, 1)

at which

f

is discontinuous is ___________. (answer in integer)

Correct Answer:

Step-by-Step Solution

Key idea: This is a digit-defined function continuity question. The function value depends on the second decimal digit of $r$ . Discontinuities occur only at points where the second digit changes AND the function value changes.

Step 1: Understand the second digit $d_{2} (r)$ . For $r \in (0, 1)$ , write $r = 0. d_{1} d_{2} d_{3} \dots$ . The second digit $d_{2}$ is constant on each interval $[k /100, (k + 1) /100)$ for $k = 0, 1, \dots, 99$ . Specifically, $d_{2} (r) = k mod 10$ for $r \in [k /100, (k + 1) /100)$ .

Step 2: Identify where $f$ can be discontinuous. Since $f$ is constant on the interior of each interval $[k /100, (k + 1) /100)$ , it is continuous there. Discontinuities can only occur at the boundary points $x = k /100$ for $k = 1, 2, \dots, 99$ .

Step 3: Check continuity at each boundary $x = k /100$ . At such a point:

$f (x)$ depends on $d_{2} (x) = k mod 10$ .
Left limit: as $r \to x^{-}$ , $r$ is in $[(k - 1) /100, k /100)$ , so $d_{2} (r) = (k - 1) mod 10$ .
Right limit: as $r \to x^{+}$ , $r$ is in $[k /100, (k + 1) /100)$ , so $d_{2} (r) = k mod 10$ .

Step 4: $f$ is discontinuous at $x = k /100$ if and only if $f ((k - 1) mod 10) \neq = f (k mod 10)$ . Let's compute $f (d)$ for $d = 0, 1, \dots, 9$ :

$f (0) = 0, f (1) = 0, f (2) = 1, f (3) = 1, f (4) = 0, f (5) = 0, f (6) = 1, f (7) = 1, f (8) = 0, f (9) = 0$ .

Step 5: Find transitions where $f$ changes value:

$d = 1 \to 2$ : $f (1) = 0 \to f (2) = 1$ . Change! Occurs at $k \equiv 2 (mod 10)$ : $k = 2, 12, 22, \dots, 92$ (10 values).
$d = 3 \to 4$ : $f (3) = 1 \to f (4) = 0$ . Change! Occurs at $k \equiv 4 (mod 10)$ : $k = 4, 14, 24, \dots, 94$ (10 values).
$d = 5 \to 6$ : $f (5) = 0 \to f (6) = 1$ . Change! Occurs at $k \equiv 6 (mod 10)$ : $k = 6, 16, 26, \dots, 96$ (10 values).
$d = 7 \to 8$ : $f (7) = 1 \to f (8) = 0$ . Change! Occurs at $k \equiv 8 (mod 10)$ : $k = 8, 18, 28, \dots, 98$ (10 values).

Other transitions ( $0 \to 1$ , $2 \to 3$ , $4 \to 5$ , $6 \to 7$ , $8 \to 9$ , $9 \to 0$ ) do not change $f$ , so no discontinuity.

Step 6: Total discontinuities = $4 \times 10 = 40$ .

Answer: 40

Question 2 · Differentiability and Optimization · 2026_Set1 MSQ

Let

f : R \to R

be defined as follows:

f (x) = (\frac{∣ x ∣}{2} - x) (x - \frac{∣ x ∣}{2})

Which of the following statements is/are true?

A.
$f$ has a local maximum
B.
$f$ has a local minimum
C.
$f^{'}$ is continuous over $R$
D.
$f^{'}$ is not differentiable over $R$

Correct Answer:

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

Step-by-Step Solution

Insight: The function is the negative of a perfect square, $f (x) = - u^{2}$ , which immediately reveals its global maximum at $u = 0$ (i.e., $x = 0$ ) and allows easy piecewise differentiation.

Exam route:

Simplify $f (x) = - (\frac{∣ x ∣}{2} - x)^{2}$ . Since a square is $\geq 0$ , $f (x) \leq 0$ . At $x = 0$ , $f (0) = 0$ , so $x = 0$ is a local (and global) maximum. No local minimum exists. (A is true, B is false).
Write $f (x)$ piecewise: $- x^{2} /4$ for $x \geq 0$ , and $- 9 x^{2} /4$ for $x < 0$ .
Differentiate: $f^{'} (x) = - x /2$ for $x > 0$ , and $- 9 x /2$ for $x < 0$ . At $x = 0$ , both left and right limits of the difference quotient are 0, so $f^{'} (0) = 0$ . Since $lim_{x \to 0} f^{'} (x) = 0 = f^{'} (0)$ , $f^{'}$ is continuous. (C is true).
Check differentiability of $f^{'}$ at $x = 0$ : left derivative of $f^{'}$ is $- 9/2$ , right derivative is $- 1/2$ . They are unequal, so $f^{'}$ is not differentiable at $x = 0$ . (D is true).

Learning route:

Step 1: Notice the algebraic structure. The two factors are negatives of each other. Let $u = \frac{∣ x ∣}{2} - x$ . Then $f (x) = u (- u) = - u^{2}$ .

Step 2: Analyze extrema. Since $u^{2} \geq 0$ for all real $x$ , $f (x) = - u^{2} \leq 0$ . The maximum possible value is 0, which occurs when $u = 0 ⟹ \frac{∣ x ∣}{2} = x ⟹ x = 0$ . Thus, $f (0) = 0$ is a local (and global) maximum. The function strictly decreases as $x$ moves away from 0 in either direction, so there is no local minimum.

Step 3: Analyze $f (x)$ piecewise to find derivatives.

For $x \geq 0$ , $∣ x ∣ = x$ , so $u = - x /2 ⟹ f (x) = - x^{2} /4$ .

For $x < 0$ , $∣ x ∣ = - x$ , so $u = - 3 x /2 ⟹ f (x) = - 9 x^{2} /4$ .

Step 4: Find $f^{'} (x)$ .

For $x > 0$ , $f^{'} (x) = - x /2$ .

For $x < 0$ , $f^{'} (x) = - 9 x /2$ .

At $x = 0$ , use the limit definition: $lim_{h \to 0} \frac{f ( h ) - f ( 0 )}{h}$ .

From the right: $lim_{h \to 0^{+}} \frac{- h ^{2} /4}{h} = 0$ .

From the left: $lim_{h \to 0^{-}} \frac{- 9 h ^{2} /4}{h} = 0$ .

Thus, $f^{'} (0) = 0$ .

Step 5: Check continuity of $f^{'}$ .

$lim_{x \to 0^{+}} f^{'} (x) = 0$ and $lim_{x \to 0^{-}} f^{'} (x) = 0$ . Both equal $f^{'} (0) = 0$ . So $f^{'}$ is continuous on $R$ .

Step 6: Check differentiability of $f^{'}$ .

We need $f^{''} (0)$ .

From the right: $lim_{h \to 0^{+}} \frac{f ^{'} ( h ) - f ^{'} ( 0 )}{h} = lim_{h \to 0^{+}} \frac{- h /2}{h} = - 1/2$ .

From the left: $lim_{h \to 0^{-}} \frac{f ^{'} ( h ) - f ^{'} ( 0 )}{h} = lim_{h \to 0^{-}} \frac{- 9 h /2}{h} = - 9/2$ .

Since $- 1/2 \neq = - 9/2$ , $f^{''} (0)$ does not exist. Thus, $f^{'}$ is not differentiable over $R$ .

Question 3 · Integral Calculus · 2026_Set2 MSQ

For a real number $a$ , let $I (a) = \int_{- 1}^{1} (3 x^{2} - a x + 1) d x$ . Which of the following statements is/are true?

A.
The value of $I (a)$ is independent of the value of $a$
B.
The value of $I (a)$ can vary with the value of $a$
C.
There exists $a \in (- \infty, + \infty)$ such that $I (a)$ is a positive real number
D.
There exists $a \in (- \infty, + \infty)$ such that $I (a)$ is a negative real number

Correct Answer:

["A","C"]

Step-by-Step Solution

Key idea: Split the integral using linearity and apply the Even-Odd function rules over the symmetric limits $[- 1, 1]$ to eliminate the parameter-dependent term.

Step 1: Write the integral as $I (a) = \int_{- 1}^{1} (3 x^{2} - a x + 1) d x$ .

Step 2: Use the linearity of integration to split it into three separate integrals:

$I (a) = \int_{- 1}^{1} 3 x^{2} d x - \int_{- 1}^{1} a x d x + \int_{- 1}^{1} 1 d x$ .

Step 3: Evaluate the middle term. The function $g (x) = a x$ is an odd function because $g (- x) = a (- x) = - a x = - g (x)$ . The integral of any odd function over symmetric limits $[- 1, 1]$ is exactly 0. So, $\int_{- 1}^{1} a x d x = 0$ .

Step 4: Evaluate the remaining terms. Both $3 x^{2}$ and $1$ are even functions.

$\int_{- 1}^{1} 3 x^{2} d x = 2 \int_{0}^{1} 3 x^{2} d x = 2 [x^{3}]_{0}^{1} = 2 (1 - 0) = 2$ .

$\int_{- 1}^{1} 1 d x = 2 \int_{0}^{1} 1 d x = 2 [x]_{0}^{1} = 2 (1 - 0) = 2$ .

Step 5: Combine the results: $I (a) = 2 - 0 + 2 = 4$ .

Step 6: Analyze the options based on $I (a) = 4$ .

The value is always 4, regardless of $a$ . Thus, it is independent of $a$ (Option A is true, Option B is false).
Since 4 is a positive real number, there exists an $a$ (in fact, all real $a$ ) that makes $I (a)$ positive (Option C is true, Option D is false).

Answer: Options A and C.

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Calculus PYQs for GATE CS

About Calculus Previous Year Questions (PYQs)

Calculus Weightage in GATE CS

Calculus Chapter Matrix

More from Engineering Mathematics

One Solved Question from Each Calculus Chapter

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution