Differentiability and Optimization PYQs for GATE CS

Solve 5+ Differentiability and Optimization previous year questions for GATE CS with answers and detailed solutions. Free sample questions below.

Chapter Roadmap: Differentiability and Optimization

Your journey through the calculus of change and optimization.

Differentiability Conditions for Piecewise Functions (Current)

Making sharp corners smooth. Matching limits and derivatives at joints.

Nondifferentiability of Maximum Functions

Analyzing

f (x) = max (g (x), h (x))

and finding where the 'winner' changes.

Local Extrema and Smoothness

First and second derivative tests, identifying peaks and valleys.

Mean Value Theorem and Derivative Bounds

Bounding function values using derivative limits over an interval.

4 Topics ~5 Core PYQs

Differentiability Conditions for Piecewise Functions

Why this matters: Piecewise functions often have 'joints' where the formula changes. These joints are natural candidates for sharp corners or breaks. To make the function differentiable everywhere, we must carefully stitch these pieces together.

What you'll learn here

The strict prerequisite of continuity before differentiability.
How to compute and equate Left-Hand and Right-Hand Derivatives.
A systematic method to find unknown constants in piecewise definitions.

Context: Differentiability & Optimization Topic 1 of 4

Differentiability and Optimization: Solved Questions with Step-by-Step Explanations (5 Problems)

Question 1 · Engineering Mathematics · 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 2 · Engineering Mathematics · 2025_Set1 NAT

Consider the given function

f (x)

f (x) = {a x + b for x l t; 1 \x^{3} + x^{2} + 1 for x \geq 1

If the function is differentiable everywhere, the value of

b

must be ________.
(rounded off to one decimal place)

Correct Answer:

-2.0

Step-by-Step Solution

Key idea: This is a piecewise differentiability constants question. The function has a joint at $x = 1$ with two unknowns $a, b$ . Differentiability everywhere forces two conditions at the joint: continuity (values match) and smoothness (derivatives match). Two conditions determine two unknowns.

Step 1: Enforce continuity at $x = 1$ .

Differentiability implies continuity, so the left and right limits must agree:

$x \to 1^{-} lim f (x) = a (1) + b = a + b$

$x \to 1^{+} lim f (x) = 1^{3} + 1^{2} + 1 = 3$

Continuity condition: $a + b = 3$ .

Step 2: Enforce derivative matching at $x = 1$ .

Left branch derivative: $\frac{d}{d x} (a x + b) = a$ , so the left derivative at $1$ is $a$ .

Right branch derivative: $\frac{d}{d x} (x^{3} + x^{2} + 1) = 3 x^{2} + 2 x$ , and at $x = 1$ this is $3 (1)^{2} + 2 (1) = 5$ .

Differentiability condition: $a = 5$ .

Step 3: Solve the system.

From $a = 5$ and $a + b = 3$ :

$5 + b = 3 ⟹ b = 3 - 5 = - 2$

Step 4: State the answer to one decimal place as required.

$b = - 2.0$

Answer: -2.0

Common trap: skipping the continuity condition and trying to use only the derivative match. That gives $a = 5$ but leaves $b$ undetermined. Continuity is a prerequisite of differentiability and always supplies one of the equations.

Question 3 · Engineering Mathematics · 2024_Set1 MCQ

Let $f : R \to R$ be a function such that $f (x) = max {x, x^{3}}$ , $x \in R$ , where $R$ is the set of all real numbers. The set of all points where $f (x)$ is NOT differentiable is

A.
${- 1, 1, 2}$
B.
${- 2, - 1, 1}$
C.
${0, 1}$
D.
${- 1, 0, 1}$

Correct Answer:

Step-by-Step Solution

Key idea: This is a nondifferentiability of a max function question. The function $f (x) = max {x, x^{3}}$ can only fail to be differentiable at points where the two pieces are equal, i.e. where $x = x^{3}$ . At such a point, $f$ is differentiable only if the two derivatives also match; otherwise there is a corner.

Step 1: Find the candidate points by solving $x = x^{3}$ .

$x^{3} - x = 0 ⟹ x (x^{2} - 1) = 0 ⟹ x (x - 1) (x + 1) = 0$

Candidates: $x = - 1, 0, 1$ . Everywhere else, one of $x$ or $x^{3}$ strictly dominates, so $f$ equals a smooth function locally and is differentiable.

Step 2: Check the derivative-match condition at each candidate.

Let $g (x) = x$ with $g^{'} (x) = 1$ , and $h (x) = x^{3}$ with $h^{'} (x) = 3 x^{2}$ .

At $x = - 1$ : $g^{'} (- 1) = 1$ , $h^{'} (- 1) = 3 (- 1)^{2} = 3$ . Since $1 \neq = 3$ , there is a corner. NOT differentiable.

At $x = 0$ : $g^{'} (0) = 1$ , $h^{'} (0) = 3 (0)^{2} = 0$ . Since $1 \neq = 0$ , there is a corner. NOT differentiable.

At $x = 1$ : $g^{'} (1) = 1$ , $h^{'} (1) = 3 (1)^{2} = 3$ . Since $1 \neq = 3$ , there is a corner. NOT differentiable.

Step 3: Collect the nondifferentiable points.

All three intersection points produce corners, so the set is ${- 1, 0, 1}$ .

Answer: Option D, ${- 1, 0, 1}$ .

Common trap: Option C ( ${0, 1}$ ) misses $x = - 1$ . Students often forget that $(- 1)^{3} = - 1$ , so $x = - 1$ is also an intersection point. Option A and B contain points like $2$ or $- 2$ that are not intersections at all.

Question 4 · Engineering Mathematics · 2023 MSQ

Let

f (x) = x^{3} + 15 x^{2} - 33 x - 36

be a real-valued function.

Which of the following statements is/are TRUE?

A.
$f (x)$ does not have a local maximum.
B.
$f (x)$ has a local maximum.
C.
$f (x)$ does not have a local minimum.
D.
$f (x)$ has a local minimum.

Correct Answer:

["B","D"]

Step-by-Step Solution

Key idea: This is a polynomial local extrema question, recognisable because we are given a cubic polynomial and asked whether local maxima/minima exist. The standard tool is: find critical points via $f^{'} (x) = 0$ , then classify them.

Step 1: Compute the first derivative.

$f (x) = x^{3} + 15 x^{2} - 33 x - 36 ⟹ f^{'} (x) = 3 x^{2} + 30 x - 33$

Step 2: Find critical points by solving $f^{'} (x) = 0$ .

$3 x^{2} + 30 x - 33 = 0 ⟹ x^{2} + 10 x - 11 = 0$

Factorising:

$x^{2} + 10 x - 11 = (x + 11) (x - 1) = 0$

So the critical points are $x = - 11$ and $x = 1$ . Both are real and distinct, so both are candidates for extrema.

Step 3: Classify using the second derivative test.

$f^{''} (x) = 6 x + 30$

At $x = - 11$ : $f^{''} (- 11) = 6 (- 11) + 30 = - 66 + 30 = - 36 < 0$ .

Negative second derivative means concave down, so $f$ has a local maximum at $x = - 11$ .

At $x = 1$ : $f^{''} (1) = 6 (1) + 30 = 36 > 0$ .

Positive second derivative means concave up, so $f$ has a local minimum at $x = 1$ .

Step 4: Verify with the first derivative test as a cross-check.

$f^{'} (x) = 3 (x + 11) (x - 1)$ is an upward parabola, so $f^{'} > 0$ for $x < - 11$ , $f^{'} < 0$ on $(- 11, 1)$ , $f^{'} > 0$ for $x > 1$ .

Sign pattern $+ \to -$ at $x = - 11$ confirms local maximum; $- \to +$ at $x = 1$ confirms local minimum.

Step 5: Evaluate the four statements.

A ("no local maximum") is FALSE. B ("has a local maximum") is TRUE.

C ("no local minimum") is FALSE. D ("has a local minimum") is TRUE.

Answer: B and D.

Common trap: A cubic has no global maximum or minimum (it tends to $\pm \infty$ ), but it does have local extrema whenever $f^{'}$ has two distinct real roots. Do not confuse local with global.

Question 5 · Engineering Mathematics · 2021_Set2 NAT

Suppose that $f : R \to R$ is a continuous function on the interval $[- 3, 3]$ and a differentiable function in the interval $(- 3, 3)$ such that for every $x$ in the interval, $f^{'} (x) \leq 2$ . If $f (- 3) = 7$ , then $f (3)$ is at most __________.

Correct Answer:

Step-by-Step Solution

Key idea: This is a Mean Value Theorem bound question, recognisable because we are given a derivative bound $f^{'} (x) \leq 2$ , an endpoint value $f (- 3) = 7$ , and asked for the maximum possible value of $f (3)$ . The MVT converts a derivative bound into a function-value bound.

Step 1: State the MVT in inequality form.

Since $f$ is continuous on $[- 3, 3]$ and differentiable on $(- 3, 3)$ , Lagrange's MVT guarantees some $c \in (- 3, 3)$ with

$f^{'} (c) = \frac{f ( 3 ) - f ( - 3 )}{3 - ( - 3 )} = \frac{f ( 3 ) - 7}{6}$

Step 2: Apply the derivative bound.

We are told $f^{'} (x) \leq 2$ for every $x$ in the interval, so in particular $f^{'} (c) \leq 2$ . Therefore:

$\frac{f ( 3 ) - 7}{6} \leq 2$

Step 3: Solve for $f (3)$ .

Multiply both sides by $6$ (positive, so inequality direction is preserved):

$f (3) - 7 \leq 12 ⟹ f (3) \leq 19$

Step 4: Confirm the bound is tight.

The function $f (x) = 2 x + 13$ satisfies all conditions: it is continuous and differentiable everywhere, $f^{'} (x) = 2 \leq 2$ , and $f (- 3) = 6 + 13 = 19 - 12 = 7$ . For this function, $f (3) = 6 + 13 = 19$ . So the upper bound $19$ is actually attained.

Answer: 19

Common trap: The interval length is $3 - (- 3) = 6$ , not $3$ . Using length $3$ gives the wrong bound $f (3) \leq 7 + 6 = 13$ . Always compute $b - a$ carefully when the interval crosses zero.

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Differentiability and Optimization PYQs for GATE CS

Chapter Roadmap: Differentiability and Optimization

Chapter Roadmap: Differentiability and Optimization

Differentiability Conditions for Piecewise Functions

Differentiability Conditions for Piecewise Functions

What you'll learn here

Differentiability and Optimization: Solved Questions with Step-by-Step Explanations (5 Problems)

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

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