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Continuity and Differentiability PYQs for GATE DA

Solve 3+ Continuity and Differentiability previous year questions for GATE DA with answers and detailed solutions. Free sample questions below.

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Question 1

2025 PYQ

Level 3: Exam Standard

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.

Question 2

2025 PYQ

Level 3: Exam Standard

Let f : R \to R be such that ∣ f (x) - f (y) ∣ \leq (x - y)^{2} for all x, y \in R . Then f (1) - f (0) = (Answer in integer)

Question 3

2024 PYQ

Level 3: Exam Standard

Let f : R \to R be a function. Note: R denotes the set of real numbers. f (x) = {- x, if x < - 2 \ax^{2} + b x + c, if x \in [- 2, 2] \x, if x > 2 Which ONE of the following choices gives the values of a, b, c that make the function f continuous and differentiable?

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Continuity and Differentiability PYQs for GATE DA

Solve 3+ Continuity and Differentiability previous year questions for GATE DA with answers and detailed solutions. Free sample questions below.

Chapter Roadmap: Continuity and Differentiability

Chapter Journey

1. Differentiability Rules & Operations

Algebra of derivatives: Sum, Product, Quotient, Chain Rule. Composition logic.

2. Continuity & Functional Constraints

Lipschitz conditions, piecewise continuity, necessary vs sufficient conditions.

Why this matters:
Before solving complex optimization problems, you must know if a function can be differentiated and how to compute it efficiently when functions are combined. This topic provides the toolkit.

The Algebra of Derivatives

Let $f$ and $g$ be differentiable at $x = c$ . The following operations preserve differentiability:

Sum / Difference

(f \pm g)^{'} (c) = f^{'} (c) \pm g^{'} (c)

Product Rule

(f \cdot g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)

Quotient Rule

(\frac{f}{g})^{'} (c) = \frac{f ^{'} ( c ) g ( c ) - f ( c ) g ^{'} ( c )}{[ g ( c ) ] ^{2}}

Condition:

g (c) \neq = 0

Key Insight: The set of differentiable functions forms an algebra. You can build complex differentiable functions from simple ones (polynomials, exponentials, trig) using these operations without returning to first principles.

Continuity and Differentiability: Solved Questions with Step-by-Step Explanations (3 Problems)

Question 1 · Calculus and Optimization · 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 2 · Calculus and Optimization · 2025 NAT

Let

f : R \to R

be such that

∣ f (x) - f (y) ∣ \leq (x - y)^{2}

for all

x, y \in R

. Then

f (1) - f (0) =

(Answer in integer)

Correct Answer:

0.00

Step-by-Step Solution

Insight: An inequality bounding the difference of function values by a power $> 1$ of the difference in inputs forces the derivative to be zero everywhere.

Exam route: Divide the given inequality by $∣ x - y ∣$ and take the limit as $x \to y$ to show $f^{'} (y) = 0$ , meaning $f$ is constant.

Learning route:

Given $∣ f (x) - f (y) ∣ \leq (x - y)^{2}$ for all $x, y$ .
For $x \neq = y$ , divide both sides by $∣ x - y ∣$ : $\frac{f ( x ) - f ( y )}{x - y} \leq ∣ x - y ∣$ .
Take the limit as $x \to y$ : The right side approaches 0.
By the Squeeze Theorem, $lim_{x \to y} \frac{f ( x ) - f ( y )}{x - y} = 0$ , which means $∣ f^{'} (y) ∣ = 0 ⟹ f^{'} (y) = 0$ .
Since $f^{'} (y) = 0$ for all $y \in R$ , $f (x)$ is a constant function.
Therefore, $f (1) - f (0) = 0$ .

Question 3 · Calculus and Optimization · 2024 MCQ

Let

f : R \to R

be a function. Note:

R

denotes the set of real numbers.

f (x) = {- x, if x < - 2 \ax^{2} + b x + c, if x \in [- 2, 2] \x, if x > 2

Which ONE of the following choices gives the values of

a, b, c

that make the
function

f

continuous and differentiable?

A.
$a = \frac{1}{4}, b = 0, c = 1$
B.
$a = \frac{1}{2}, b = 0, c = 0$
C.
$a = 0, b = 0, c = 0$
D.
$a = 1, b = 1, c = - 4$

Correct Answer:

Step-by-Step Solution

Insight: Piecewise smoothness requires matching both function values (continuity) and slopes (differentiability) at every boundary point.

Exam route: Set up 4 equations: 2 for continuity at $x = - 2, 2$ and 2 for differentiability at $x = - 2, 2$ . Solve the linear system for $a, b, c$ .

Learning route:

Continuity at $x = - 2$ : $lim_{x \to - 2^{-}} (- x) = f (- 2) ⟹ 2 = 4 a - 2 b + c$ .
Continuity at $x = 2$ : $lim_{x \to 2^{+}} (x) = f (2) ⟹ 2 = 4 a + 2 b + c$ .
Differentiability at $x = - 2$ : LHD = $lim_{x \to - 2^{-}} (- 1) = - 1$ . RHD = $lim_{x \to - 2^{+}} (2 a x + b) = - 4 a + b$ . So $- 4 a + b = - 1$ .
Differentiability at $x = 2$ : LHD = $lim_{x \to 2^{-}} (2 a x + b) = 4 a + b$ . RHD = $lim_{x \to 2^{+}} (1) = 1$ . So $4 a + b = 1$ .
Solve the system: Adding the derivative equations gives $2 b = 0 ⟹ b = 0$ . Subtracting gives $8 a = 2 ⟹ a = 1/4$ .
Substitute $a = 1/4, b = 0$ into the first continuity equation: $4 (1/4) - 0 + c = 2 ⟹ 1 + c = 2 ⟹ c = 1$ .

The values are $a = 1/4, b = 0, c = 1$ .

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Notes written for recall, not for volume

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Unlock the whole course

Continuity and Differentiability PYQs for GATE DA

Chapter Roadmap: Continuity and Differentiability

Chapter Journey

1. Differentiability Rules & Operations

2. Continuity & Functional Constraints

The Algebra of Derivatives

The Algebra of Derivatives

Continuity and Differentiability: Solved Questions with Step-by-Step Explanations (3 Problems)

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

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