Calculus Short Notes 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 Short Notes

Quick revision sheets for Calculus in GATE DA. Every chapter is condensed into key formulas, shortcuts and common traps so you can revise 4 chapters fast before the exam.

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 NAT

Consider the function $f (x) = \sum_{n = 0}^{\infty} (x - x^{2})^{n}$ defined for $x \in [0, 1]$ . The maximum value of $f (x)$ in this interval is ______________. (Round off to two decimal places)

Correct Answer:

1.33

Step-by-Step Solution

Key idea: The function $f (x)$ is an infinite geometric series with common ratio $r (x) = x - x^{2}$ . To maximize $f (x) = \frac{1}{1 - r ( x )}$ , we must maximize $r (x)$ within the valid domain.

Step 1: Identify the geometric series. The sum is $f (x) = \frac{1}{1 - ( x - x ^{2} )}$ , provided $∣ x - x^{2} ∣ < 1$ .

Step 2: Analyze the ratio $r (x) = x - x^{2}$ . This is a downward-opening parabola with roots at $x = 0$ and $x = 1$ . Its maximum occurs at the vertex $x = 1/2$ .

Step 3: Calculate the maximum ratio. $r (1/2) = 1/2 - (1/2)^{2} = 1/4$ . Since $1/4 < 1$ , the series converges at the vertex.

Step 4: Maximize $f (x)$ . Since $f (x) = \frac{1}{1 - r ( x )}$ is an increasing function of $r (x)$ for $r < 1$ , the maximum of $f (x)$ occurs when $r (x)$ is maximized.

Step 5: Calculate the maximum value. $f (1/2) = \frac{1}{1 - 1/4} = \frac{1}{3/4} = 4/3 \approx 1.33$ .

Answer: 1.33

Question 2 · Continuity and Differentiability NAT

Let $f (x) = ∣ x^{3} - 2 x^{2} + x ∣$ . Find the number of points in the interval $[- 1, 2]$ at which $f (x)$ is NOT differentiable.

Correct Answer:

1.00

Step-by-Step Solution

Key idea: The absolute value function $∣ g (x) ∣$ is non-differentiable at $x = c$ if and only if $g (c) = 0$ and $g^{'} (c) \neq = 0$ . This requires casework for each root.

Step 1: Let $g (x) = x^{3} - 2 x^{2} + x$ . Factor it: $g (x) = x (x^{2} - 2 x + 1) = x (x - 1)^{2}$ .

Step 2: The roots of $g (x)$ are $x = 0$ and $x = 1$ . Both lie in the interval $[- 1, 2]$ . $f (x)$ can only be non-differentiable at these roots.

Step 3: Check the derivative $g^{'} (x) = 3 x^{2} - 4 x + 1$ at each root.

Step 4: At $x = 0$ , $g^{'} (0) = 1 \neq = 0$ . Since $g (0) = 0$ and $g^{'} (0) \neq = 0$ , the graph of $g (x)$ crosses the x-axis sharply. Thus, $f (x) = ∣ g (x) ∣$ has a corner and is NOT differentiable at $x = 0$ .

Step 5: At $x = 1$ , $g^{'} (1) = 3 (1)^{2} - 4 (1) + 1 = 0$ . Since the derivative is zero, $g (x)$ is tangent to the x-axis at $x = 1$ . Near $x = 1$ , $g (x) \approx (x - 1)^{2}$ , so $f (x) = ∣ g (x) ∣ \approx (x - 1)^{2}$ , which IS differentiable at $x = 1$ (with derivative 0).

Conclusion: There is exactly 1 point ( $x = 0$ ) where $f (x)$ is not differentiable.

Answer: 1

Question 3 · Differentiation and Higher Order Derivatives MCQ

Let $f (x) = 2^{3 x} + 3^{2 x}$ , $x \in R$ . What is the value of $f^{'} (0)$ ?

A.
\ln 72
B.
0
C.
\ln 6
D.
\ln 108

Correct Answer:

Step-by-Step Solution

Key idea: This is an exponential derivative question, recognisable because the variable $x$ is in the exponent, not the base. The rule is $\frac{d}{d x} a^{u (x)} = a^{u (x)} ln (a) \cdot u^{'} (x)$ .

Step 1: Differentiate the first term $2^{3 x}$ .

Here $a = 2$ and $u (x) = 3 x$ , so $u^{'} (x) = 3$ .

$\frac{d}{d x} 2^{3 x} = 2^{3 x} \cdot ln 2 \cdot 3 = 3 \cdot 2^{3 x} ln 2$ .

Step 2: Differentiate the second term $3^{2 x}$ .

Here $a = 3$ and $u (x) = 2 x$ , so $u^{'} (x) = 2$ .

$\frac{d}{d x} 3^{2 x} = 3^{2 x} \cdot ln 3 \cdot 2 = 2 \cdot 3^{2 x} ln 3$ .

Step 3: Combine and evaluate at $x = 0$ .

$f^{'} (x) = 3 \cdot 2^{3 x} ln 2 + 2 \cdot 3^{2 x} ln 3$ .

$f^{'} (0) = 3 \cdot 2^{0} ln 2 + 2 \cdot 3^{0} ln 3 = 3 ln 2 + 2 ln 3$ .

Step 4: Simplify using log properties.

$3 ln 2 = ln 2^{3} = ln 8$ and $2 ln 3 = ln 3^{2} = ln 9$ .

$f^{'} (0) = ln 8 + ln 9 = ln (8 \times 9) = ln 72$ .

Answer: \ln 72

Question 4 · Maxima, Minima and Applications of Derivatives NAT

Let $f : R \to R$ be a twice differentiable function having exactly $5$ distinct critical points. Among these critical points, $f^{''} (x) > 0$ at exactly $2$ points, $f^{''} (x) < 0$ at exactly $2$ points, and $f^{''} (x) = 0$ at exactly $1$ point. How many of the following statements are necessarily true?

I. $f$ has at least $2$ local minima.

II. $f$ has at least $2$ local maxima.

III. The critical point where $f^{''} (x) = 0$ is neither a local maximum nor a local minimum.

IV. $f$ has at least $4$ local extrema.

Correct Answer:

Step-by-Step Solution

Key idea: This question tests your understanding of necessary vs. sufficient conditions for local extrema. The second derivative test gives sufficient conditions, not necessary ones. Step 1: Analyze what we know. -

f

has 5 critical points where

f^{'} (x) = 0

. - At 2 of these points,

f^{''} (x) > 0

. - At 2 of these points,

f^{''} (x) < 0

. - At 1 point,

f^{''} (x) = 0

. Step 2: Evaluate each statement. Statement I: $f$ has at least 2 local minima. At the 2 points where

f^{''} (x) > 0

, the second derivative test guarantees local minima (sufficient condition). So

f

has at least 2 local minima. TRUE. Statement II: $f$ has at least 2 local maxima. At the 2 points where

f^{''} (x) < 0

, the second derivative test guarantees local maxima (sufficient condition). So

f

has at least 2 local maxima. TRUE. Statement III: The critical point where $f^{''} (x) = 0$ is neither a local maximum nor a local minimum. This is not necessarily true. Example:

f (x) = x^{4}

has

f^{''} (0) = 0

but

x = 0

is a local minimum. Example:

f (x) = - x^{4}

has

f^{''} (0) = 0

but

x = 0

is a local maximum. Example:

f (x) = x^{3}

has

f^{''} (0) = 0

and

x = 0

is neither. So the point where

f^{''} (x) = 0

could be a local extremum or not. NOT NECESSARILY TRUE. Statement IV: $f$ has at least 4 local extrema. From statements I and II,

f

has at least 2 local minima and at least 2 local maxima, giving at least 4 local extrema total. TRUE. Step 3: Count the true statements. Statements I, II, and IV are necessarily true. That's 3 statements. Answer: 3.

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Calculus Short Notes for GATE DA

About Calculus Short Notes

Calculus Weightage in GATE DA

Calculus Chapter Matrix

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One Solved Question from Each Calculus Chapter

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