GATE DA 2024 Question Paper with Solutions: 65 Questions, Answer Key & Section-wise Analysis

GATE DA 2024 previous year paper: 65 questions with answer key and detailed solutions, section-wise breakdown and free sample questions.

Paper breakdown

65 questions · 100 marks · 3 minutes. Programming, Data Structures and Algorithms: 13 · Probability and Statistics: 10 · Machine Learning: 10 · Linear Algebra: 6 · Artificial Intelligence: 6 · Quantitative Aptitude: 5 · Calculus and Optimization: 5 · Database Management and Warehousing: 4 · Spatial Aptitude: 3 · Verbal Aptitude: 2 · Analytical Aptitude: 1

Free sample questions from GATE DA 2024 Question Paper

Question 1 · Programming, Data Structures and Algorithms · 2024 MCQ

Consider performing depth-first search (DFS) on an undirected and unweighted
graph

G

starting at vertex

s

. For any vertex

u

G

d [u]

is the length of the shortest
path from

s

u

. Let

(u, v)

be an edge in

G

such that

d [u] < d [v]

. If the edge

(u, v)

is explored first in the direction from

u

v

during the above DFS, then

(u, v)

becomes a ______ edge.

A.
tree
B.
cross
C.
back
D.
gray

Correct Answer:

Step-by-Step Solution

Insight: In an undirected graph, DFS only produces tree edges and back edges. If $d [u] < d [v]$ and we explore $(u, v)$ from $u$ to $v$ (meaning $v$ is unvisited), it must be a tree edge.

Exam route: Recall the undirected DFS rule: no cross or forward edges exist. Since $v$ is unvisited when explored from $u$ , it's a tree edge by definition.

Learning route:

The graph is undirected and unweighted. $d [u]$ is the shortest path distance from $s$ to $u$ .
In undirected DFS, every edge is either a tree edge or a back edge. (Forward and cross edges are impossible.)
"Explored first in the direction from $u$ to $v$ " means when $u$ examines $v$ , $v$ is unvisited, so DFS traverses to $v$ .
By definition, an edge to an unvisited vertex is a tree edge.
The condition $d [u] < d [v]$ is consistent: if $(u, v)$ is a tree edge, $v$ is a child of $u$ , so $d [v] \leq d [u] + 1$ . Since $d [u] < d [v]$ , we have $d [v] = d [u] + 1$ .
Could it be a back edge? If $(u, v)$ were a back edge, $v$ would be an ancestor of $u$ , meaning $v$ was visited before $u$ . But the problem says the edge is explored from $u$ to $v$ first, implying $v$ was unvisited. Contradiction.

Answer: tree (A).

Question 2 · Programming, Data Structures and Algorithms · 2024 MCQ

Match the items in Column 1 with the items in Column 2 in the following table:

Column 1	Column 2
(p) First In First Out	(i) Stacks
(q) Lookup Operation	(ii) Queues
(r) Last In First Out	(iii) Hash Tables

A.
(p) − (ii), (q) − (iii), (r) − (i)
B.
(p) − (ii), (q) − (i), (r) − (iii)
C.
(p) − (i), (q) − (ii), (r) − (iii)
D.
(p) − (i), (q) − (iii), (r) − (ii)

Correct Answer:

Step-by-Step Solution

Key idea: This is an ADT property matching question, recognizable because it asks to pair fundamental access patterns (FIFO, LIFO, Lookup) with their corresponding data structures.

Step 1: Analyze "First In First Out" (p). This is the defining property of a Queue, where the first element added is the first to be removed. So, (p) matches with (ii).

Step 2: Analyze "Last In First Out" (r). This is the defining property of a Stack, where the most recently added element is the first to be removed. So, (r) matches with (i).

Step 3: Analyze "Lookup Operation" (q). Hash Tables are specifically designed to provide fast, average-case $O (1)$ key-based lookup operations. So, (q) matches with (iii).

Step 4: Combine the matches: (p) - (ii), (q) - (iii), (r) - (i).

Answer: Option A

Question 3 · Programming, Data Structures and Algorithms · 2024 MCQ

Consider performing uniform hashing on an open address hash table with load
factor

α = \frac{n}{m} < 1

, where

n

elements are stored in the table with

m

slots. The
expected number of probes in an unsuccessful search is at most

\frac{1}{1 - α}

.
Inserting an element in this hash table requires at most ______ probes, on average.

A.
$ln (\frac{1}{1 - α})$
B.
$\frac{1}{1 - α}$
C.
$1 + \frac{α}{2}$
D.
$\frac{1}{1 + α}$

Correct Answer:

Step-by-Step Solution

Key idea: This is a probe complexity question, recognizable because it asks for the expected number of probes for a specific hash table operation under the uniform hashing assumption.

Step 1: Recall the mechanics of insertion in open addressing. To insert a new element, the algorithm must find an empty slot in the hash table.

Step 2: Relate insertion to search operations. Finding an empty slot is logically identical to an unsuccessful search, which probes sequentially until it finds an empty slot (indicating the key is not present).

Step 3: Apply the uniform hashing formula. The problem states that the expected number of probes for an unsuccessful search is at most $\frac{1}{1 - α}$ .

Step 4: Conclude the complexity. Since the insertion process requires exactly the same probing sequence as an unsuccessful search, its expected number of probes is also $\frac{1}{1 - α}$ .

Answer: B

Question 4 · Probability and Statistics · 2024 MCQ

The probability of a boy or a girl being born is 1/2. For a family having only
three children, what is the probability of having two girls and one boy?

A.
$\frac{3}{8}$
B.
$\frac{1}{8}$
C.
$\frac{1}{4}$
D.
$\frac{1}{2}$

Correct Answer:

Step-by-Step Solution

Insight: This is a classic binomial probability problem where the probability of success (girl) is 0.5, and we want exactly 2 successes in 3 trials.

Exam route: 3 children, exactly 2 girls. Number of arrangements = 3C2 = 3. Total outcomes = 2^3 = 8. Probability = 3/8.

Learning route: The sample space for a family of 3 children has $2^{3} = 8$ equally likely outcomes (since prob of boy/girl is 1/2). We want exactly 2 girls and 1 boy. The favorable outcomes are GGB, GBG, and BGG. There are 3 such outcomes. The probability is the ratio of favorable to total outcomes: $3/8$ . Alternatively, using the binomial formula: $(2 3) (1/2)^{2} (1/2)^{1} = 3 \times 1/8 = 3/8$ .

Question 5 · Probability and Statistics · 2024 MCQ

Consider the following statements:
(i) The mean and variance of a Poisson random variable are equal.
(ii) For a standard normal random variable, the mean is zero and the
variance is one.
Which ONE of the following options is correct?

A.
Both (i) and (ii) are true
B.
(i) is true and (ii) is false
C.
(ii) is true and (i) is false
D.
Both (i) and (ii) are false

Question 6 · Probability and Statistics · 2024 MCQ

Three fair coins are tossed independently.

T

is the event that two or more tosses
result in heads.

S

is the event that two or more tosses result in tails.
What is the probability of the event

T \cap S

A.
0
B.
0.5
C.
0.25
D.
1

Correct Answer:

Step-by-Step Solution

Insight: In 3 tosses, you cannot have 2 or more heads AND 2 or more tails simultaneously because that requires at least 4 tosses.

Exam route: T requires >= 2 heads, S requires >= 2 tails. Total coins = 3. Max heads + Max tails = 3. The intersection is empty. Probability = 0.

Learning route: The sample space for 3 fair coins has 8 equally likely outcomes. Event T (>= 2 heads) consists of {HHH, HHT, HTH, THH}. Event S (>= 2 tails) consists of {TTT, TTH, THT, HTT}. The intersection T ∩ S requires an outcome to have at least 2 heads and at least 2 tails, which means at least 4 coins. Since we only have 3 coins, T ∩ S = ∅. Therefore, P(T ∩ S) = 0.

Question 7 · Machine Learning · 2024 MCQ

Consider the dataset with six datapoints:

{(x_{1}, y_{1}), (x_{2}, y_{2}), \dots, (x_{6}, y_{6})}

,
where

x_{1} = [10]

x_{2} = [01]

x_{3} = [0 - 1]

x_{4} = [- 1 0]

x_{5} = [22]

x_{6} = [- 2 - 2]

and the labels are given by

y_{1} = y_{2} = y_{5} = 1

, and

y_{3} = y_{4} = y_{6} = - 1

. A hard
margin linear support vector machine is trained on the above dataset.
Which ONE of the following sets is a possible set of support vectors?

A.
${x_{1}, x_{2}, x_{5}}$
B.
${x_{3}, x_{4}, x_{5}}$
C.
${x_{4}, x_{5}}$
D.
${x_{1}, x_{2}, x_{3}, x_{4}}$

Question 8 · Machine Learning · 2024 MCQ

Match the items in Column 1 with the items in Column 2 in the following table:

Column 1	Column 2
(p) Principal Component Analysis	(i) Discriminative Model
(q) Naïve Bayes Classification	(ii) Dimensionality Reduction
(r) Logistic Regression	(iii) Generative Model

A.
(p) − (iii), (q) − (i), (r) − (ii)
B.
(p) − (ii), (q) − (i), (r) − (iii)
C.
(p) − (ii), (q) − (iii), (r) − (i)
D.
(p) − (iii), (q) − (ii), (r) − (i)

Question 9 · Machine Learning · 2024 MCQ

Euclidean distance based

k

-means clustering algorithm was run on a dataset of 100
points with

k = 3

. If the points

[11]

and

[- 1 1]

are both part of cluster 3, then which
ONE of the following points is necessarily also part of cluster 3?

A.
$[00]$
B.
$[02]$
C.
$[20]$
D.
$[01]$

Question 10 · Linear Algebra · 2024 MCQ

Consider the matrix

M = [23 - 1 1]

.
Which ONE of the following statements is TRUE?

A.
The eigenvalues of $M$ are non-negative and real.
B.
The eigenvalues of $M$ are complex conjugate pairs.
C.
One eigenvalue of $M$ is positive and real, and another eigenvalue of $M$ is zero.
D.
One eigenvalue of $M$ is non-negative and real, and another eigenvalue of $M$ is negative and real.

Correct Answer:

Step-by-Step Solution

Insight: For a $2 \times 2$ matrix, the characteristic equation is $λ^{2} - Tr (M) λ + det (M) = 0$ . The discriminant determines the nature of the eigenvalues.

Exam route: $Tr (M) = 2 + 1 = 3$ , $det (M) = 2 (1) - (- 1) (3) = 5$ . Equation: $λ^{2} - 3 λ + 5 = 0$ . Discriminant $Δ = 9 - 20 = - 11 < 0$ . So eigenvalues are a complex conjugate pair.

Learning route:

Compute trace: $Tr (M) = 2 + 1 = 3$ .
Compute determinant: $det (M) = (2) (1) - (- 1) (3) = 2 + 3 = 5$ .
Characteristic equation: $λ^{2} - 3 λ + 5 = 0$ .
Discriminant: $Δ = (- 3)^{2} - 4 (1) (5) = 9 - 20 = - 11$ .
Since $Δ < 0$ , the roots are complex conjugates: $λ = \frac{3 \pm i 11}{2}$ .
This matches option B: "The eigenvalues of $M$ are complex conjugate pairs."
Note: $M$ is a real matrix but not symmetric, so there is no guarantee of real eigenvalues. This is the key insight — real entries do not imply real eigenvalues.

Question 11 · Linear Algebra · 2024 NAT

Consider the

3 \times 3

matrix

M = 134213336

.
The determinant of

(M^{2} + 12 M)

is ______.

Correct Answer:

0.00

Step-by-Step Solution

Insight: The matrix polynomial $M^{2} + 12 M$ factors as $M (M + 12 I)$ . If $det (M) = 0$ , the whole product has determinant $0$ without any need to square the matrix or compute the second factor.

Exam route:

Factor the expression: $M^{2} + 12 M = M (M + 12 I)$ .
Use multiplicativity: $det (M^{2} + 12 M) = det (M) \cdot det (M + 12 I)$ .
Inspect $M$ for linear dependence. Row 3 $= (4, 3, 6) = (1, 2, 3) + (3, 1, 3) = R_{1} + R_{2}$ .
Therefore $det (M) = 0$ .
The product is $0 \times det (M + 12 I) = 0$ .

Learning route:

The determinant is multiplicative: $det (A B) = det (A) det (B)$ . Since $M$ and $M + 12 I$ are both polynomials in the same matrix $M$ , they commute, and the factorisation $M^{2} + 12 M = M (M + 12 I)$ is valid at the matrix level. Hence

$det (M^{2} + 12 M) = det (M) \cdot det (M + 12 I) .$

Compute $det (M)$ by inspection. The rows of $M$ are

$R_{1} = (1, 2, 3), R_{2} = (3, 1, 3), R_{3} = (4, 3, 6) .$

Observe that $R_{1} + R_{2} = (1 + 3, 2 + 1, 3 + 3) = (4, 3, 6) = R_{3}$ . The rows are linearly dependent, so $det (M) = 0$ .

(Equivalently, expanding along the first row: $det (M) = 1 (1 \cdot 6 - 3 \cdot 3) - 2 (3 \cdot 6 - 3 \cdot 4) + 3 (3 \cdot 3 - 1 \cdot 4) = 1 (- 3) - 2 (6) + 3 (5) = - 3 - 12 + 15 = 0$ .)

Since one factor has determinant $0$ , the entire product has determinant $0$ , regardless of the value of $det (M + 12 I)$ .

The common wrong path is to compute $M^{2}$ explicitly (a $3 \times 3$ matrix multiplication), then add $12 M$ , then expand a $3 \times 3$ determinant. That is legal but wasteful and invites arithmetic errors. The factorisation shortcut collapses the work to a single dependency check.

Verification: $det (M) = 0$ is confirmed by two independent routes (row dependence and cofactor expansion). Any product containing a singular factor is singular, so $det (M (M + 12 I)) = 0$ is exact.

Question 12 · Linear Algebra · 2024 MSQ

Select all choices that are subspaces of

R^{3}

.
Note:

R

denotes the set of real numbers.

A.
${x = [x_{1} \x_{2} \x_{3}] \in R^{3} : x = α 110 + β 100, α, β \in R}$
B.
${x = [x_{1} \x_{2} \x_{3}] \in R^{3} : x = α^{2} 120 + β^{2} 101, α, β \in R}$
C.
${x = [x_{1} \x_{2} \x_{3}] \in R^{3} : 5 x_{1} + 2 x_{3} = 0, 4 x_{1} - 2 x_{2} + 3 x_{3} = 0}$
D.
${x = [x_{1} \x_{2} \x_{3}] \in R^{3} : 5 x_{1} + 2 x_{3} + 4 = 0}$

Correct Answer:

["A","C"]

Step-by-Step Solution

Insight: Subspaces must be closed under all scalar multiplications (including negative) and contain the origin. Spans of vectors and solutions to homogeneous linear equations are always subspaces.

Exam route: Check each option against the subspace criteria. A is a span (subspace). B uses squared parameters (only non-negative scalars, fails closure). C is a homogeneous system (subspace). D is a non-homogeneous system (fails zero vector).

Learning route:

Option A: This is the span of two vectors, $span {v_{1}, v_{2}}$ . The span of any set of vectors is always a subspace because it is closed under addition and scalar multiplication by definition.
Option B: The coefficients are $α^{2}$ and $β^{2}$ . Since squares of real numbers are always non-negative ( $α^{2} \geq 0$ ), we can only form non-negative linear combinations. If we multiply a vector in this set by $- 1$ , we cannot guarantee it remains in the set. It fails closure under scalar multiplication.
Option C: This is the solution set to a system of homogeneous linear equations ( $A x = 0$ ). The null space of any matrix is always a subspace. It contains the zero vector and is closed under addition and scalar multiplication.
Option D: The equation is $5 x_{1} + 2 x_{3} = - 4$ . To be a subspace, it must contain the zero vector $(0, 0, 0)$ . Substituting gives $5 (0) + 2 (0) = 0 \neq = - 4$ . It fails the zero vector test.

Question 13 · Artificial Intelligence · 2024 MCQ

Let

h_{1}

and

h_{2}

be two admissible heuristics used in

A^{*}

search.
Which ONE of the following expressions is always an admissible heuristic?

A.
$h_{1} + h_{2}$
B.
$h_{1} \times h_{2}$
C.
$h_{1} / h_{2}, (h_{2} \neq = 0)$
D.
$∣ h_{1} - h_{2} ∣$

Question 14 · Artificial Intelligence · 2024 MCQ

Consider five random variables

U, V, W, X,

and

Y

whose joint distribution
satisfies:

P (U, V, W, X, Y) = P (U) P (V) P (W ∣ U, V) P (X ∣ W) P (Y ∣ W)

Which ONE of the following statements is FALSE?

A.
$Y$ is conditionally independent of $V$ given $W$
B.
$X$ is conditionally independent of $U$ given $W$
C.
$U$ and $V$ are conditionally independent given $W$
D.
$Y$ and $X$ are conditionally independent given $W$

Question 15 · Artificial Intelligence · 2024 MCQ

Consider the following statement:
In adversarial search,

α

–

β

pruning can be applied to game trees of any depth where

α

is the (m) value choice we have formed so far at any choice point along the
path for the MAX player and

β

is the (n) value choice we have formed so far
at any choice point along the path for the MIN player.
Which ONE of the following choices of (m) and (n) makes the above statement
valid?

A.
(m) = highest, (n) = highest
B.
(m) = lowest, (n) = highest
C.
(m) = highest, (n) = lowest
D.
(m) = lowest, (n) = lowest

Question 16 · Quantitative Aptitude · 2024 MCQ

How many 4-digit positive integers divisible by 3 can be formed using only the
digits {1, 3,4, 6, 7}, such that no digit appears more than once in a number?

A.
24
B.
48
C.
72
D.
12

Correct Answer:

Step-by-Step Solution

Insight: A number is divisible by 3 if the sum of its digits is divisible by 3. The sum of all 5 available digits is 21. To form a 4-digit number, we must drop exactly one digit, and that dropped digit must be a multiple of 3.

Exam route: The multiples of 3 in ${1, 3, 4, 6, 7}$ are 3 and 6.

Case 1: Drop 3. Digits are ${1, 4, 6, 7}$ . All are non-zero, so $4! = 24$ ways.

Case 2: Drop 6. Digits are ${1, 3, 4, 7}$ . All are non-zero, so $4! = 24$ ways.

Total = $24 + 24 = 48$ .

Learning route:

Check the sum of the given set: $1 + 3 + 4 + 6 + 7 = 21$ .
Since 21 is divisible by 3, removing a digit $x$ leaves a sum of $21 - x$ .
For the remaining 4 digits to be divisible by 3, $21 - x$ must be a multiple of 3, which means $x$ itself must be a multiple of 3.
The available multiples of 3 in the set are 3 and 6. Thus, we have exactly two valid subsets of 4 digits: ${1, 4, 6, 7}$ and ${1, 3, 4, 7}$ .
Neither subset contains the digit 0, so there are no leading-zero constraints to worry about. Each subset can form $4! = 24$ valid 4-digit numbers.
Total valid numbers = $24 + 24 = 48$ .

Common Trap: Forgetting that 0 is not in the set and unnecessarily subtracting leading zero cases, or failing to realize that dropping a non-multiple of 3 ruins the divisibility.

Verification: Both subsets sum to a multiple of 3 (18 and 15), and both yield 24 permutations. $24 + 24 = 48$ .

Question 17 · Quantitative Aptitude · 2024 MCQ

The sum of the following infinite series is

2 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{8} + \frac{1}{9} + \frac{1}{16} + \frac{1}{27} + \dots

A.
$\frac{11}{3}$
B.
$\frac{7}{2}$
C.
$\frac{13}{4}$
D.
$\frac{9}{2}$

Correct Answer:

Step-by-Step Solution

Insight: The series is a mix of a constant, a geometric series with ratio 1/2, and another with ratio 1/3.

Exam route: Group the terms by their denominators' patterns. Sum the two infinite geometric series separately using $S = a / (1 - r)$ and add to the initial constant.

Learning route:

The given series is $2 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{8} + \frac{1}{9} + \frac{1}{16} + \frac{1}{27} + \dots$

Observe the denominators after the first term: 2, 4, 8, 16... are powers of 2. 3, 9, 27... are powers of 3.

We can split the series into three parts:

$S = 2 + (\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots) + (\frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \dots)$

The first bracket is a geometric series with $a = 1/2$ and $r = 1/2$ . Its sum is $\frac{1/2}{1 - 1/2} = 1$ .

The second bracket is a geometric series with $a = 1/3$ and $r = 1/3$ . Its sum is $\frac{1/3}{1 - 1/3} = \frac{1/3}{2/3} = 1/2$ .

Total sum $S = 2 + 1 + 1/2 = 3.5 = 7/2$ .

Correct option is B.

Question 18 · Quantitative Aptitude · 2024 MCQ

In an election, the share of valid votes received by the four candidates A, B, C, and
D is represented by the pie chart shown. The total number of votes cast in the
election were 1,15,000, out of which 5,000 were invalid.

Based on the data provided, the total number of valid votes received by the
candidates B and C is

A.
45,000
B.
49,500
C.
51,750
D.
54,000

Correct Answer:

Step-by-Step Solution

Key idea: This is a pie chart interpretation question with a valid/invalid data trap, recognisable because it gives total votes, invalid votes, and percentages that apply only to valid votes.

Step 1: Calculate the total number of valid votes. Total votes = 1,15,000. Invalid votes = 5,000. Valid votes = 1,15,000 - 5,000 = 1,10,000.

Step 2: Identify the combined percentage share of candidates B and C from the pie chart. Share of B = 25%, Share of C = 20%. Combined share = 25% + 20% = 45%.

Step 3: Calculate the number of valid votes for B and C. 45% of 1,10,000 = 0.45 \times 1,10,000 = 49,500.

Answer: B

Question 19 · Calculus and Optimization · 2024 MCQ

For any twice differentiable function

f : R \to R

, if at some

x^{*} \in R

f^{'} (x^{*}) = 0

and

f^{''} (x^{*}) > 0

, then the function

f

necessarily has a ______ at

x = x^{*}

.
Note:

R

denotes the set of real numbers.

A.
local minimum
B.
global minimum
C.
local maximum
D.
global maximum

Correct Answer:

Step-by-Step Solution

Insight: The Second Derivative Test only guarantees local extrema, not global.

Exam route: $f^{'} (x^{*}) = 0$ and $f^{''} (x^{*}) > 0$ means the function is concave up at $x^{*}$ . By the Second Derivative Test, this guarantees a local minimum at $x^{*}$ . It does not guarantee a global minimum because the function could go to $- \infty$ elsewhere (e.g., $f (x) = x^{3} /3 - x$ has a local min at $x = 1$ but no global min). Thus, "local minimum" is the only necessarily true statement.

Learning route: In optimization, local conditions (like $f^{''} > 0$ ) only describe the neighborhood of a point. Global extrema require analyzing the entire domain, especially for functions that are unbounded or have multiple critical points. Always distinguish between "local" and "global" in theoretical questions.

Question 20 · Calculus and Optimization · 2024 NAT

Let

f : R \to R

be the function

f (x) = \frac{1}{1 + e ^{- x}}

.
The value of the derivative of

f

x

where

f (x) = 0.4

is ______
(rounded off to two decimal places).
Note:

R

denotes the set of real numbers.

Correct Answer:

0.24

Step-by-Step Solution

Insight: The derivative of the sigmoid function $f (x)$ can be expressed entirely in terms of $f (x)$ itself: $f^{'} (x) = f (x) (1 - f (x))$ .

Exam route: Use the identity $f^{'} (x) = f (x) (1 - f (x))$ . Substitute $f (x) = 0.4$ to get $0.4 \times 0.6 = 0.24$ .

Learning route:

Given $f (x) = (1 + e^{- x})^{- 1}$ .

Using the chain rule, $f^{'} (x) = - 1 (1 + e^{- x})^{- 2} \cdot (- e^{- x}) = \frac{e ^{- x}}{( 1 + e ^{- x} ) ^{2}}$ .

Rewrite this as $\frac{1}{1 + e ^{- x}} \cdot \frac{e ^{- x}}{1 + e ^{- x}}$ .

Notice that $\frac{e ^{- x}}{1 + e ^{- x}} = \frac{1 + e ^{- x} - 1}{1 + e ^{- x}} = 1 - \frac{1}{1 + e ^{- x}} = 1 - f (x)$ .

Thus, $f^{'} (x) = f (x) (1 - f (x))$ .

Given $f (x) = 0.4$ , we have $f^{'} (x) = 0.4 (1 - 0.4) = 0.4 \times 0.6 = 0.24$ .

Question 21 · 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 $a = 1/4, b = 0, c = 1$ .

Question 22 · Database Management and Warehousing · 2024 MCQ

Consider a database that includes the following relations:
Defender(name, rating, side, goals)
Forward(name, rating, assists, goals)
Team(name, club, price)
Which ONE of the following relational algebra expressions checks that every name
occurring in Team appears in either Defender or Forward, where

ϕ

denotes the
empty set?

A.
$Π_{nam e} (Team) ∖ (Π_{nam e} (Defender) \cap Π_{nam e} (Forward)) = ϕ$
B.
$(Π_{nam e} (Defender) \cap Π_{nam e} (Forward)) ∖ Π_{nam e} (Team) = ϕ$
C.
$Π_{nam e} (Team) ∖ (Π_{nam e} (Defender) \cup Π_{nam e} (Forward)) = ϕ$
D.
$(Π_{nam e} (Defender) \cup Π_{nam e} (Forward)) ∖ Π_{nam e} (Team) = ϕ$

Question 23 · Database Management and Warehousing · 2024 NAT

Consider the following two tables named Raider and Team in a relational database
maintained by a Kabaddi league. The attribute ID in table Team references the
primary key of the Raider table, ID.

Raider
ID	Name	Raids	RaidPoints
1	Arjun	200	250
2	Ankush	190	219
3	Sunil	150	200
4	Reza	150	190
5	Pratham	175	220
6	Gopal	193	215

The SQL query described below is executed on this database:

SELECT *
FROM Raider, Team
WHERE Raider.ID=Team.ID AND City=“Jaipur” AND
RaidPoints > 200;

The number of rows returned by this query is ______.

Team
City	ID	BidPoints
Jaipur	2	200
Patna	3	195
Hyderabad	5	175
Jaipur	1	250
Patna	4	200
Jaipur	6	200

Question 24 · Database Management and Warehousing · 2024 MSQ

Given the relational schema

R = (U, V, W, X, Y, Z)

and the set of functional
dependencies:

{U \to V, U \to W, W X \to Y, W X \to Z, V \to X}

Which of the following functional dependencies can be derived from the above
set?

A.
$V W \to Y Z$
B.
$W X \to Y Z$
C.
$V W \to U$
D.
$V W \to Y$

Question 25 · Spatial Aptitude · 2024 MCQ

The 15 parts of the given figure are to be painted such that no two adjacent parts
with shared boundaries (excluding corners) have the same color. The minimum
number of colors required is

A.
4
B.
3
C.
5
D.
6

Question 26 · Spatial Aptitude · 2024 MCQ

Three different views of a dice are shown in the figure below.

The piece of paper that can be folded to make this dice is

Question 27 · Spatial Aptitude · 2024 MCQ

Visualize two identical right circular cones such that one is inverted over the other
and they share a common circular base. If a cutting plane passes through the vertices
of the assembled cones, what shape does the outer boundary of the
resulting cross-section make?

A.
A rhombus
B.
A triangle
C.
An ellipse
D.
A hexagon

Question 28 · Verbal Aptitude · 2024 MCQ

If ‘→’ denotes increasing order of intensity, then the meaning of the words
[sick → infirm → moribund] is analogous to [silly → _______ → daft].
Which one of the given options is appropriate to fill the blank?

A.
frown
B.
fawn
C.
vein
D.
vain

Correct Answer:

Step-by-Step Solution

Insight: This question tests semantic gradients, requiring a word that fits the increasing intensity of cognitive foolishness between "silly" and "daft", while strictly maintaining part-of-speech parallelism.

Exam route: Identify the progression in the first pair: sick (mild) → infirm (moderate) → moribund (extreme). Apply this to the second pair: silly (mild) → ? → daft (extreme). Evaluate options by part of speech. "Vain" is the only adjective among the choices, making it the only structurally viable bridge for a personal quality.

Learning route:

Analyze the source vector: "sick" (general unwellness) → "infirm" (significant weakness) → "moribund" (near death). This is a strict increasing intensity gradient of physical decline.
Analyze the target vector: "silly" (mild lack of sense) → [blank] → "daft" (extreme foolishness). We need an intermediate stage of cognitive inadequacy.
Evaluate options by grammatical category:

"frown": Noun/verb (facial expression). Fails part-of-speech match.
"fawn": Noun/verb (young deer / show affection). Fails part-of-speech match.
"vein": Noun (blood vessel / style). Fails part-of-speech match.
"vain": Adjective (having excessive pride or producing no result). In the context of this specific exam question, it is the only adjective provided, serving as the structural bridge for a personal quality.

Conclusion: "vain" is the only grammatically and structurally appropriate option to complete the analogy.

Question 29 · Verbal Aptitude · 2024 MCQ

Thousands of years ago, some people began dairy farming. This coincided with a
number of mutations in a particular gene that resulted in these people developing
the ability to digest dairy milk.
Based on the given passage, which of the following can be inferred?

A.
All human beings can digest dairy milk.
B.
No human being can digest dairy milk.
C.
Digestion of dairy milk is essential for human beings.
D.
In human beings, digestion of dairy milk resulted from a mutated gene.

Correct Answer:

Step-by-Step Solution

Insight: The passage explicitly links the ability to digest dairy milk to a genetic mutation in "some people" using the causal phrase "resulted in", making Option D a direct paraphrase of the stated causality.

Exam route: Scan options for extreme modifiers ("All", "No", "essential"). Eliminate A, B, and C immediately as they overstate or invent claims. Option D perfectly matches the passage's explicit causal link without adding outside scope.

Learning route:

Analyze the passage: It states two facts: (a) some people began dairy farming, and (b) this coincided with gene mutations that resulted in these people developing the ability to digest dairy milk.
Identify the causal link: The phrase "resulted in" explicitly establishes causation between the mutated gene and the ability to digest dairy milk for that specific group.
Evaluate Option A: Uses the extreme modifier "All", but the passage explicitly limits the scope to "some people". This is a classic extreme modifier trap.
Evaluate Option B: Uses the extreme modifier "No", which directly contradicts the passage's statement that some people developed the ability.
Evaluate Option C: Introduces the concept of "essential", which is outside the scope of the passage. The passage describes a historical development, not a biological necessity for all humans.
Evaluate Option D: Accurately reflects the passage's explicit statement that the digestion ability resulted from a mutated gene.

Question 30 · Analytical Aptitude · 2024 MSQ

Let

x

and

y

be two propositions. Which of the following statements is a tautology
/are tautologies?

A.
$(\neg x \land y) ⟹ (y ⟹ x)$
B.
$(x \land \neg y) ⟹ (\neg x ⟹ y)$
C.
$(\neg x \land y) ⟹ (\neg x ⟹ y)$
D.
$(x \land \neg y) ⟹ (y ⟹ x)$

Correct Answer:

["B","C","D"]

Step-by-Step Solution

Insight: Convert each implication to disjunction form and simplify. A tautology simplifies to $T$ .

Exam route: For each option, replace $A \to B$ with $\neg A \lor B$ , then simplify using De Morgan's and absorption. If the result is $T$ , it's a tautology.

Learning route:

Step 1: Recall $A \to B \equiv \neg A \lor B$ .

Step 2: Evaluate option A: $(\neg x \land y) \to (y \to x)$ .

$= \neg (\neg x \land y) \lor (\neg y \lor x)$

$= (x \lor \neg y) \lor (\neg y \lor x)$

$= x \lor \neg y$

This is NOT always true (false when $x = F, y = T$ ). So A is not a tautology.

Step 3: Evaluate option B: $(x \land \neg y) \to (\neg x \to y)$ .

$= \neg (x \land \neg y) \lor (x \lor y)$

$= (\neg x \lor y) \lor (x \lor y)$

$= \neg x \lor y \lor x \lor y$

$= (\neg x \lor x) \lor (y \lor y)$

$= T \lor y$

$= T$

This IS a tautology.

Step 4: Evaluate option C: $(\neg x \land y) \to (\neg x \to y)$ .

$= \neg (\neg x \land y) \lor (x \lor y)$

$= (x \lor \neg y) \lor (x \lor y)$

$= x \lor \neg y \lor x \lor y$

$= x \lor (\neg y \lor y)$

$= x \lor T$

$= T$

This IS a tautology.

Step 5: Evaluate option D: $(x \land \neg y) \to (y \to x)$ .

$= \neg (x \land \neg y) \lor (\neg y \lor x)$

$= (\neg x \lor y) \lor (\neg y \lor x)$

$= (\neg x \lor x) \lor (y \lor \neg y)$

$= T \lor T$

$= T$

This IS a tautology.

Answer: B, C, D are tautologies.

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