Calculus Short Notes 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 Short Notes

Quick revision sheets for Calculus in GATE CS. Every chapter is condensed into key formulas, shortcuts and common traps so you can revise 3 chapters fast before the exam.

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 NAT

Consider the function $h : R \to R$ defined by

$h (x) = {\frac{e ^{a x} - 1 - s i n ( b x )}{x ^{2}} 2 if x \neq = 0 if x = 0$

where $a$ and $b$ are positive real numbers. If $h (x)$ is continuous at $x = 0$ , then the value of $a + b$ is _________.

Correct Answer:

4.00

Step-by-Step Solution

Key idea: This is a comparison question involving a hidden singularity at the boundary $x = 0$ . For $h (x)$ to be continuous at $x = 0$ , the limit of the rational piece as $x \to 0$ must exactly equal the defined value $h (0) = 2$ .

Step 1: Set up the continuity condition.

$lim_{x \to 0} \frac{e ^{a x} - 1 - s i n ( b x )}{x ^{2}} = 2$ .

Step 2: Analyze the limit using Taylor series expansions around $x = 0$ .

$e^{a x} = 1 + a x + \frac{( a x ) ^{2}}{2 !} + \dots = 1 + a x + \frac{a ^{2}}{2} x^{2} + O (x^{3})$

$sin (b x) = b x - \frac{( b x ) ^{3}}{3 !} + \dots = b x + O (x^{3})$

Step 3: Substitute the expansions into the numerator.

Numerator $= (1 + a x + \frac{a ^{2}}{2} x^{2}) - 1 - (b x) + O (x^{3})$

$= (a - b) x + \frac{a ^{2}}{2} x^{2} + O (x^{3})$

Step 4: Evaluate the limit.

$lim_{x \to 0} \frac{( a - b ) x + \frac{a ^{2}}{2} x ^{2} + O ( x ^{3} )}{x ^{2}} = lim_{x \to 0} (\frac{a - b}{x} + \frac{a ^{2}}{2} + O (x))$

Step 5: Apply the boundary condition for the limit to exist.

For the limit to be finite, the coefficient of the $\frac{1}{x}$ term must be zero. Otherwise, the limit would be $\pm \infty$ .

Therefore, $a - b = 0 ⟹ a = b$ .

Step 6: Solve for $a$ and $b$ .

With $a = b$ , the limit simplifies to $\frac{a ^{2}}{2}$ .

We are given that the limit equals $2$ :

$\frac{a ^{2}}{2} = 2 ⟹ a^{2} = 4$ .

Since $a$ is a positive real number, $a = 2$ .

Because $b = a$ , we have $b = 2$ .

Step 7: Find $a + b$ .

$a + b = 2 + 2 = 4$ .

Answer: 4.00

Question 2 · Differentiability and Optimization MCQ

Direction: The following question contains two statements — an Assertion (A) and a Reason (R).

Assertion (A): If the function $f (x) = {a x^{2} + b x + 3 x^{3} - x + 1 x < 2 x \geq 2$ is differentiable everywhere, then $a = 3$ and $b = - 1$ .

Reason (R): For $f (x)$ to be differentiable at $x = 2$ , the left-hand derivative must equal the right-hand derivative, which yields the equation $4 a + b = 11$ .

Select the correct answer using the codes given below:

A.
Both A and R are true and R is the correct explanation of A.
B.
Both A and R are true but R is NOT the correct explanation of A.
C.
A is true but R is false.
D.
A is false but R is true.

Correct Answer:

Step-by-Step Solution

Key idea: This is an assertion-reason question testing the algorithmic steps for finding unknown constants in a differentiable piecewise function.

Step 1: Check Reason (R). For $f (x)$ to be differentiable at $x = 2$ , we must have $f^{'} (2^{-}) = f^{'} (2^{+})$ .

Left piece: $g (x) = a x^{2} + b x + 3 ⟹ g^{'} (x) = 2 a x + b ⟹ g^{'} (2) = 4 a + b$ .

Right piece: $h (x) = x^{3} - x + 1 ⟹ h^{'} (x) = 3 x^{2} - 1 ⟹ h^{'} (2) = 3 (4) - 1 = 11$ .

Equating them gives $4 a + b = 11$ . Thus, Reason (R) is TRUE.

Step 2: Check Assertion (A). Differentiability requires continuity first.

Continuity at $x = 2$ : $g (2) = h (2) ⟹ 4 a + 2 b + 3 = 8 - 2 + 1 ⟹ 4 a + 2 b + 3 = 7 ⟹ 4 a + 2 b = 4 ⟹ 2 a + b = 2$ .

Now we have a system:

$4 a + b = 11$
$2 a + b = 2$

Subtracting (2) from (1): $2 a = 9 ⟹ a = 4.5$ .

Substituting $a = 4.5$ into (2): $2 (4.5) + b = 2 ⟹ 9 + b = 2 ⟹ b = - 7$ .

The assertion claims $a = 3$ and $b = - 1$ , which is incorrect. Thus, Assertion (A) is FALSE.

Conclusion: A is false, but R is true.

Answer: D

Question 3 · Integral Calculus MCQ

Let $h (a) = \int_{- a}^{a} (x^{3} + x^{2} - 2 x - 1) d x$ for $a \in [0, 3]$ . The maximum value of $h (a)$ is

A.
12
B.
$- \frac{4}{3}$
C.
0
D.
18

Correct Answer:

Step-by-Step Solution

Key idea: This is an even-odd decomposition followed by optimisation. First use symmetry to eliminate odd terms, then find the maximum of the resulting function on $[0, 3]$ .

Exam route: Separate even and odd parts of the integrand, drop the odd terms, integrate the even part, then check critical points and boundaries.

Step 1: Decompose the integrand into even and odd parts.

Odd terms: $x^{3}$ and $- 2 x$ (odd powers of $x$ ).

Even terms: $x^{2}$ and $- 1$ (even powers of $x$ , including $x^{0}$ ).

Step 2: Over the symmetric interval $[- a, a]$ , the integral of every odd function is $0$ . So:

$h (a) = \int_{- a}^{a} (x^{2} - 1) d x = 2 \int_{0}^{a} (x^{2} - 1) d x$

Step 3: Evaluate the integral:

$h (a) = 2 [\frac{x ^{3}}{3} - x]_{0}^{a} = 2 (\frac{a ^{3}}{3} - a) = \frac{2 a ^{3}}{3} - 2 a$

Step 4: Find critical points by differentiating:

$h^{'} (a) = 2 a^{2} - 2 = 0 ⟹ a^{2} = 1 ⟹ a = 1 (since a \geq 0)$

Step 5: Classify the critical point. $h^{''} (a) = 4 a$ . At $a = 1$ : $h^{''} (1) = 4 > 0$ , so $a = 1$ is a local minimum.

Step 6: Evaluate $h (a)$ at the critical point and both boundaries:

$h (0) = 0$

$h (1) = \frac{2}{3} - 2 = - \frac{4}{3}$

$h (3) = \frac{2 \cdot 27}{3} - 6 = 18 - 6 = 12$

Step 7: The maximum value on $[0, 3]$ is $12$ , occurring at the boundary $a = 3$ .

Wrong path (Option B): Stopping at the critical point $a = 1$ and reporting $h (1) = - \frac{4}{3}$ as the maximum. This is actually a local minimum. Always check the boundaries.

Wrong path (Option C): Evaluating only at $a = 0$ and getting $0$ .

Wrong path (Option D): Forgetting the $- 2 a$ term and computing only $\frac{2 a ^{3}}{3}$ at $a = 3$ , giving $18$ .

Generalisation: When optimising a function on a closed interval, the global maximum can occur at a boundary even when interior critical points exist.

Verification: $h (3) = \int_{- 3}^{3} (x^{2} - 1) d x = 2 \int_{0}^{3} (x^{2} - 1) d x = 2 [\frac{x ^{3}}{3} - x]_{0}^{3} = 2 (9 - 3) = 12$ .

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Calculus Short Notes for GATE CS

About Calculus Short Notes

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