Numerical Computation & Estimation, Algorithms, Matrices, Logic, Calculus… Test for GATE DA: 24 Questions with Solutions & Analysis

Attempt the Numerical Computation & Estimation, Algorithms, Matrices, Logic, Calculus… test for GATE DA: 24 exam-level questions, detailed solutions and performance analysis. First questions free.

Paper breakdown

24 questions · 36 marks · 1.08 minutes. Programming, Data Structures and Algorithms: 4 · Probability and Statistics: 4 · Machine Learning: 4 · Linear Algebra: 3 · Quantitative Aptitude: 2 · Database Management and Warehousing: 2 · Calculus and Optimization: 2 · Artificial Intelligence: 2 · Analytical Aptitude: 1

Free sample questions from Numerical Computation & Estimation, Algorithms, Matrices, Logic, Calculus…

Question 1 · Programming, Data Structures and Algorithms · 2025 MSQ

Suppose that insertion sort is applied to the array $[1, 3, 5, 7, 9, 11, x, 15, 13]$ and it takes exactly two swaps to sort the array. Select all possible values of $x$ .

A.
$10$
B.
$12$
C.
$14$
D.
$16$

Correct Answer:

["C"]

Step-by-Step Solution

Insight: This is the classic 'unknown element in a nearly sorted array' pattern recognisable because a sorted or nearly sorted array contains one unknown $x$ and the total number of swaps (inversions) is given. The pivot is the split method: partition the array around $x$ and count inversions in each piece.

Exam route:

Split $[1, 3, 5, 7, 9, 11, x, 15, 13]$ into Left $= [1, 3, 5, 7, 9, 11]$ , $x$ , Right $= [15, 13]$ .
Count inversions within Left: 0 (perfectly sorted).
Count inversions within Right: 1 (pair (15,13)).
Count inversions involving $x$ :

With Left: since all Left elements $\leq 11$ , if $x \geq 11$ then 0 inversions with Left; if $x < 11$ , then inversions with Left $= ∣ {l \in Left : l > x} ∣$ .
With Right $= [15, 13]$ :
If $x \geq 15$ : 0 inversions with Right.
If $13 \leq x < 15$ : 1 inversion (15 > x).
If $11 < x < 13$ : 2 inversions (15 > x and 13 > x).
If $x \leq 11$ : 2 inversions with Right, plus inversions with Left.

Total inversions $= 0 + 1 + I_{Left > x} + I_{x > Right} = 2$ .

So $I_{Left > x} + I_{x > Right} = 1$ .

Case analysis:

$x \geq 15$ : $I_{x > Right} = 0$ , $I_{Left > x} = 0$ (since all Left $\leq 11 < 15$ ). Sum = 0. No.
$13 \leq x < 15$ : $I_{x > Right} = 1$ , $I_{Left > x} = 0$ (since all Left $\leq 11 < 13$ ). Sum = 1. ✓
$11 < x < 13$ (i.e. $x = 12$ ): $I_{x > Right} = 2$ , $I_{Left > x} = 0$ . Sum = 2. No.
$x \leq 11$ : $I_{x > Right} = 2$ , $I_{Left > x} \geq 1$ . Sum $\geq 3$ . No.

So $x \in [13, 14]$ . Among the options ${10, 12, 14, 16}$ , only 14 is valid.

Learning route:

The split method works because inversions decompose additively:

$I_{total} = I_{Left} + I_{Right} + I_{Left > x} + I_{x > Right} .$

Here $I_{Left} = 0$ and $I_{Right} = 1$ , so we need $I_{Left > x} + I_{x > Right} = 1$ .

Since all Left elements are $\leq 11$ , any $x \geq 13$ gives $I_{Left > x} = 0$ . Then we need $I_{x > Right} = 1$ , which means exactly one of ${15, 13}$ is greater than $x$ . This happens iff $13 \leq x < 15$ , i.e. $x \in {13, 14}$ .

Verification: for $x = 14$ , inversions are (15,13) and (15,14) — total 2 ✓. For $x = 10$ , inversions are (15,13), (11,10), (15,10), (13,10) — total 4 ✗.

Answer: C (only 14).

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

Consider sorting the following array of integers in ascending order using an in-place
Quicksort algorithm that uses the last element as the pivot.

60708090100

The minimum number of swaps performed during this Quicksort is ______.

Correct Answer:

Step-by-Step Solution

Key idea: This is a Quicksort trace problem, recognizable by the request to count the exact number of swaps for a specific array and pivot choice. We must simulate the standard Lomuto partition scheme.

Step 1: Identify the partition scheme.

The problem specifies an "in-place Quicksort algorithm that uses the last element as the pivot". This corresponds to the standard Lomuto partition scheme.

Step 2: Trace the first partition on the array $[60, 70, 80, 90, 100]$ .

Pivot $x = 100$ .
Initialize $i = p - 1 = - 1$ .
Loop $j$ from $0$ to $3$:
$j = 0$ : $A [0] = 60 \leq 100$ . Increment $i$ to $0$. Swap $A [0]$ with $A [0]$ . (1 swap)
$j = 1$ : $A [1] = 70 \leq 100$ . Increment $i$ to $1$. Swap $A [1]$ with $A [1]$ . (1 swap)
$j = 2$ : $A [2] = 80 \leq 100$ . Increment $i$ to $2$. Swap $A [2]$ with $A [2]$ . (1 swap)
$j = 3$ : $A [3] = 90 \leq 100$ . Increment $i$ to $3$. Swap $A [3]$ with $A [3]$ . (1 swap)
End of loop. Swap $A [i + 1]$ with $A [r]$ , which is Swap $A [4]$ with $A [4]$ . (1 swap)

Total swaps in this step = $5$.

Step 3: Analyze the recursive calls.

The pivot $100$ is now in its correct final position. The left subarray is $[60, 70, 80, 90]$ (size 4), and the right subarray is empty.

Step 4: Repeat the process for the remaining subarrays.

For size 4 ( $[60, 70, 80, 90]$ ), pivot is $90$. By the same logic, it requires $4$ swaps.
For size 3 ( $[60, 70, 80]$ ), pivot is $80$. Requires $3$ swaps.
For size 2 ( $[60, 70]$ ), pivot is $70$. Requires $2$ swaps.
For size 1 ( $[60]$ ), pivot is $60$. Requires $1$ swap.

Step 5: Calculate the total number of swaps.

Total swaps = $5 + 4 + 3 + 2 + 1 = 15$ .

Answer: 15

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

Consider the following sorting algorithms:

i

Bubble sort

ii

Insertion sort

iii

Selection sort
Which ONE among the following choices of sorting algorithms sorts the numbers
in the array [4, 3, 2, 1, 5] in increasing order after exactly two passes over the array?

A.
$i$ only
B.
$iii$ only
C.
$i$ and $iii$ only
D.
$ii$ and $iii$ only

Correct Answer:

Step-by-Step Solution

Insight: This is a pass-state question recognisable because it asks which algorithm(s) fully sort a specific array in exactly two passes. The pivot is the invariant each algorithm guarantees after $k$ passes: Bubble locks the $k$ largest at the right, Selection locks the $k$ smallest at the left, Insertion sorts the first $k + 1$ elements.

Exam route:

Apply each algorithm for 2 passes on $[4, 3, 2, 1, 5]$ .
Bubble pass 1: $(4, 3) \to$ swap, $(3, 2) \to$ swap, $(2, 1) \to$ swap, $(1, 5) \to$ no swap → $[3, 2, 1, 4, 5]$ .

Bubble pass 2: $(3, 2) \to$ swap, $(2, 1) \to$ swap, $(1, 4) \to$ no, $(4, 5) \to$ no → $[2, 1, 3, 4, 5]$ . Not sorted.

Insertion pass 1 (key=3): shift 4 → $[3, 4, 2, 1, 5]$ .

Insertion pass 2 (key=2): shift 4, shift 3 → $[2, 3, 4, 1, 5]$ . Not sorted.

Selection pass 1: min of $[4, 3, 2, 1, 5]$ is 1, swap with index 0 → $[1, 3, 2, 4, 5]$ .

Selection pass 2: min of $[3, 2, 4, 5]$ is 2, swap with index 1 → $[1, 2, 3, 4, 5]$ . Sorted!

Only Selection (iii) sorts in exactly 2 passes. Answer: B.

Learning route:

The array $[4, 3, 2, 1, 5]$ has a special structure: the first four elements are reverse-sorted and the last is already the maximum.

Bubble Sort bubbles the largest element to the end in pass 1, but the remaining four elements are still reverse-sorted and need 3 more passes. After 2 passes, the array is $[2, 1, 3, 4, 5]$ — not sorted.
Insertion Sort sorts a growing prefix. After 2 passes, only the first 3 elements are sorted; the 1 at index 3 is still out of place.
Selection Sort places the smallest element at the front each pass. Pass 1 places 1, pass 2 places 2. Since the remaining elements $[3, 4, 5]$ are already sorted, the array is fully sorted after 2 passes.

Verification: re-tracing confirms only Selection produces $[1, 2, 3, 4, 5]$ after exactly 2 passes.

Answer: B.

Question 4 · Probability and Statistics · 2025 MCQ

Suppose $X$ and $Y$ are random variables. The conditional expectation of $X$ given $Y$ is denoted by $E [X ∣ Y]$ . Then $E [E [X ∣ Y]]$ equals

A.
$E [X ∣ Y]$
B.
$\frac{E [ X ]}{E [ Y ]}$
C.
$E [X]$
D.
$E [Y]$

Correct Answer:

Step-by-Step Solution

Insight: This is the exact statement of the Law of Iterated Expectations (Adam's Rule).

Exam route: $E [E [X ∣ Y]] = E [X]$ .

Learning route: This is a fundamental property question, recognizable because it asks for the expectation of a conditional expectation.

Step 1: Identify the concept. The expression $E [E [X ∣ Y]]$ represents the expected value of the conditional expectation of $X$ given $Y$ .

Step 2: Apply the Law of Iterated Expectations (also known as Adam's Rule or the Tower Property). This law states that the unconditional expectation of $X$ is equal to the expectation of its conditional expectation given any random variable $Y$ .

$E [E [X ∣ Y]] = E [X]$

Step 3: Intuition. $E [X ∣ Y]$ is a random variable that represents our best prediction of $X$ given the information $Y$ . If we average these predictions over all possible values of $Y$ , we get back our overall, unconditional best prediction of $X$ , which is simply $E [X]$ .

The correct option is (C).

Question 5 · Probability and Statistics · 2026 MSQ

Suppose a random variable

Z

follows

N or ma l (μ = 0, σ^{2} = 1)

distribution with probability density function

g (z)

and cumulative distribution function

G (z)

.
Another random variable

Y

follows

t_{1}

distribution with probability density function

h (y)

and cumulative distribution function

H (y)

. Let

c

be the positive real number for which

g (c) = h (c)

.
Which of the following statements is/are correct?

A.
$G (0) = H (0)$
B.
$G (c) < H (c)$
C.
$G (- c) < H (- c)$
D.
$g (0) = h (0)$

Correct Answer:

["A","C"]

Step-by-Step Solution

Insight: Both distributions are symmetric around zero, so their CDFs at zero are exactly 0.5. The normal distribution has a higher peak, meaning it accumulates area faster near the center, so $G (c) > H (c)$ . By symmetry, the left-tail inequality reverses.

Exam route: $G (0) = H (0) = 0.5$ . Since $g (0) > h (0)$ and $g (c) = h (c)$ , the normal PDF is strictly greater on $(0, c)$ , so $G (c) > H (c)$ . Using symmetry, $G (- c) = 1 - G (c)$ and $H (- c) = 1 - H (c)$ , which gives $G (- c) < H (- c)$ .

Learning route: This is a distribution comparison question, recognizable because it asks to compare the CDFs of a normal and a $t$ -distribution at specific points.

Step 1: Evaluate at zero. Both $Z$ and $Y$ are symmetric around zero. Therefore, the area to the left of zero is exactly half the total probability. $G (0) = 0.5$ and $H (0) = 0.5$ . Thus, $G (0) = H (0)$ . Statement (A) is correct.

Step 2: Compare the peaks. The standard normal PDF at zero is $g (0) = \frac{1}{2 π} \approx 0.399$ . The $t_{1}$ (Cauchy) PDF at zero is $h (0) = \frac{1}{π} \approx 0.318$ . Since $g (0) > h (0)$ , Statement (D) is incorrect.

Step 3: Compare CDFs at $c$ . We are given $g (c) = h (c)$ for some $c > 0$ . Because the normal distribution has a higher peak and lighter tails, its PDF must be strictly greater than the $t$ -distribution PDF for all $x \in (0, c)$ . That is, $g (x) > h (x)$ for $0 < x < c$ .

Step 4: Integrate to find the CDFs. The CDF at $c$ is the area from $- \infty$ to $c$ . Since both are symmetric, $G (c) = 0.5 + \int_{0}^{c} g (x) d x$ and $H (c) = 0.5 + \int_{0}^{c} h (x) d x$ . Because $g (x) > h (x)$ on this interval, the integral for $G$ is strictly larger. Thus, $G (c) > H (c)$ . Statement (B) is incorrect.

Step 5: Use symmetry for $- c$ . By the symmetry of both distributions, $G (- c) = 1 - G (c)$ and $H (- c) = 1 - H (c)$ . Since $G (c) > H (c)$ , subtracting both from 1 reverses the inequality: $1 - G (c) < 1 - H (c)$ , which means $G (- c) < H (- c)$ . Statement (C) is correct.

Question 6 · Probability and Statistics · 2024 NAT

Consider a joint probability density function of two random variables X and Y

f_{X, Y} (x, y) = {2 x y, 0, 0 < x < 2, 0 < y < x o t h er w i se

Then,

E [Y ∣ X = 1.5]

is ______.

Correct Answer:

1.00

Step-by-Step Solution

Insight: The joint density factors into a function of $x$ and a function of $y ∣ x$ , making the conditional distribution of $Y ∣ X$ easy to identify and integrate.

Exam route: $f_{X} (x) = \int_{0}^{x} 2 x y d y = x^{3}$ . $f (y ∣ x) = \frac{2 x y}{x ^{3}} = \frac{2 y}{x ^{2}}$ . $E [Y ∣ X = x] = \int_{0}^{x} y \frac{2 y}{x ^{2}} d y = \frac{2 x}{3}$ . At $x = 1.5$ , it is 1.

Learning route: This is a conditional expectation from a joint PDF question, recognizable because it gives $f_{X, Y} (x, y)$ and asks for $E [Y ∣ X = x]$ .

Step 1: Find the marginal density of $X$ . Integrate the joint density over the valid range of $y$ . The support is $0 < y < x$ , so the limits for $y$ are $0$ to $x$ .

$f_{X} (x) = \int_{0}^{x} 2 x y d y = 2 x [\frac{y ^{2}}{2}]_{0}^{x} = x^{3} for 0 < x < 2$

Step 2: Find the conditional density of $Y$ given $X = x$ . Divide the joint density by the marginal density.

$f_{Y ∣ X} (y ∣ x) = \frac{f _{X, Y} ( x , y )}{f _{X} ( x )} = \frac{2 x y}{x ^{3}} = \frac{2 y}{x ^{2}} for 0 < y < x$

Step 3: Calculate the conditional expectation $E [Y ∣ X = x]$ . Integrate $y$ times the conditional density over the support of $y$ .

$E [Y ∣ X = x] = \int_{0}^{x} y (\frac{2 y}{x ^{2}}) d y = \frac{2}{x ^{2}} \int_{0}^{x} y^{2} d y = \frac{2}{x ^{2}} [\frac{y ^{3}}{3}]_{0}^{x} = \frac{2 x}{3}$

Step 4: Evaluate at $x = 1.5$ .

$E [Y ∣ X = 1.5] = \frac{2 ( 1.5 )}{3} = \frac{3}{3} = 1$

Question 7 · Machine Learning · 2024 MCQ

Given a dataset with

K

binary-valued attributes (where

K > 2

) for a two-class
classification task, the number of parameters to be estimated for learning a naïve
Bayes classifier is

A.
$2^{K} + 1$
B.
$2 K + 1$
C.
$2^{K + 1} + 1$
D.
$K^{2} + 1$

Correct Answer:

Step-by-Step Solution

Insight: The number of independent parameters in Naive Bayes is the sum of independent prior probabilities and independent conditional probabilities, leveraging the fact that probabilities for a given variable must sum to 1.

Exam route: For 2 classes, priors need $2 - 1 = 1$ parameter. For $K$ binary attributes, each needs $2 - 1 = 1$ parameter per class. Total conditionals = $K \times 1 \times 2 = 2 K$ . Total = $2 K + 1$ .

Learning route:

Step 1: Prior Probabilities. There are $C = 2$ classes. Since $P (Y = 0) + P (Y = 1) = 1$ , we only need to estimate $C - 1 = 1$ independent parameter (e.g., $P (Y = 1)$ ).

Step 2: Conditional Probabilities. There are $K$ binary-valued attributes. For each attribute $X_{i}$ , and for each of the 2 classes, we need to estimate $P (X_{i} = 1∣ Y = c)$ . The probability $P (X_{i} = 0∣ Y = c)$ is simply $1 - P (X_{i} = 1∣ Y = c)$ . Thus, each attribute requires $V_{i} - 1 = 2 - 1 = 1$ independent parameter per class.

Step 3: Total conditional parameters = $K attributes \times 2 classes \times 1 parameter = 2 K$ .

Step 4: Total parameters to estimate = $1 (prior) + 2 K (conditionals) = 2 K + 1$ .

Answer: B

Question 8 · Machine Learning · 2025 MCQ

Consider the neural network shown in the figure with
inputs:

u, v

weights:

a, b, c, d, e, f

output:

y

R

denotes the ReLU function,

R (x) = max (0, x)

.

Given

u = 2, v = 3,

a = 1, b = 1, c = 1, d = - 1, e = 4, f = - 1,

which one of the following is correct?

A.
$\frac{\partial y}{\partial a} = 8, \frac{\partial y}{\partial f} = 0$
B.
$\frac{\partial y}{\partial a} = 1, \frac{\partial y}{\partial f} = 0$
C.
$\frac{\partial y}{\partial a} = 1, \frac{\partial y}{\partial f} = - 1$
D.
$\frac{\partial y}{\partial a} = 2, \frac{\partial y}{\partial f} = - 1$

Correct Answer:

Step-by-Step Solution

Insight: This is a backpropagation through ReLU problem, requiring careful evaluation of the forward pass to determine which gradients are blocked (zero) and which pass through.

Exam route: Compute forward pass values first. If a pre-activation is $\leq 0$ , its gradient is 0. If $> 0$ , the gradient passes through unchanged.

Learning route:

Step 1: Forward pass for Hidden Layer 1.

$z_{1} = a \cdot u + b \cdot v = 1 (2) + 1 (3) = 5$ . Since $5 > 0$ , $h_{1} = R (5) = 5$ .

$z_{2} = c \cdot u + d \cdot v = 1 (2) + (- 1) (3) = - 1$ . Since $- 1 \leq 0$ , $h_{2} = R (- 1) = 0$ .

Step 2: Forward pass for Output Layer.

$z_{3} = e \cdot h_{1} + f \cdot h_{2} = 4 (5) + (- 1) (0) = 20$ . Since $20 > 0$ , $y = R (20) = 20$ .

Step 3: Backward pass for $\frac{\partial y}{\partial a}$ .

$\frac{\partial y}{\partial a} = \frac{\partial y}{\partial z _{3}} \cdot \frac{\partial z _{3}}{\partial h _{1}} \cdot \frac{\partial h _{1}}{\partial z _{1}} \cdot \frac{\partial z _{1}}{\partial a}$ .

$\frac{\partial y}{\partial z _{3}} = 1$ (since $z_{3} > 0$ ).

$\frac{\partial z _{3}}{\partial h _{1}} = e = 4$ .

$\frac{\partial h _{1}}{\partial z _{1}} = 1$ (since $z_{1} > 0$ ).

$\frac{\partial z _{1}}{\partial a} = u = 2$ .

$\frac{\partial y}{\partial a} = 1 \cdot 4 \cdot 1 \cdot 2 = 8$ .

Step 4: Backward pass for $\frac{\partial y}{\partial f}$ .

$\frac{\partial y}{\partial f} = \frac{\partial y}{\partial z _{3}} \cdot \frac{\partial z _{3}}{\partial f}$ .

$\frac{\partial z _{3}}{\partial f} = h_{2} = 0$ .

$\frac{\partial y}{\partial f} = 1 \cdot 0 = 0$ .

Answer: A.

Question 9 · Machine Learning · 2025 MSQ

A.
ReLU is continuous everywhere
B.
ReLU is differentiable everywhere
C.
ReLU is not differentiable at $x = 0$
D.
ReLU( $x$ ) = ReLU( $a x$ ), for all $a \in R$

Correct Answer:

["A","C"]

Step-by-Step Solution

Insight: This is a conceptual question testing the mathematical properties of the ReLU activation function, specifically its continuity and differentiability.

Exam route: Recall the piecewise definition of ReLU. Check continuity at $x = 0$ . Check left and right derivatives at $x = f or a l l$ a \in \mathbb{R}$?

Let $x = 1, a = - 1$ . $R (1) = 1$ . $R (- 1) = 0$ . $1 \neq = 0$ . (Option D is False)

Answer: A, C.

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 · 2026 MCQ

Let

M = (cos θ s in θ - s in θ cos θ)

be a

2 \times 2

matrix, where

θ = \frac{2 π}{5}

, and

I_{2} = (1001)

.
Which of the following options is equal to

M^{2026}

A.
$M^{2}$
B.
$M$
C.
$M^{- 1}$
D.
$I_{2}$

Correct Answer:

Step-by-Step Solution

Insight: $M = R (2 π /5)$ is a rotation by $2 π /5$ , so $M^{5} = I$ . Reduce the exponent $2026$ modulo $5$.

Exam route: $2026 = 5 \times 405 + 1$ , so $M^{2026} = (M^{5})^{405} \cdot M^{1} = I^{405} \cdot M = M$ .

Learning route:

Identify $M$ as the standard counter-clockwise rotation matrix $R (θ)$ with $θ = \frac{2 π}{5}$ .
A rotation by $θ$ raised to the $n$ -th power is a rotation by $n θ$ : $M^{n} = R (n θ)$ .
Compute $n θ = 2026 \times \frac{2 π}{5} = \frac{4052 π}{5}$ .
Reduce modulo $2 π$ : $\frac{4052}{5} = 810 + \frac{2}{5}$ , so $\frac{4052 π}{5} = 810 π + \frac{2 π}{5} = 405 (2 π) + \frac{2 π}{5}$ .
The effective angle is $\frac{2 π}{5} = θ$ , so $M^{2026} = R (θ) = M$ .
Alternatively, $M^{5} = R (5 \cdot 2 π /5) = R (2 π) = I$ . Then $2026 mod 5 = 1$ , so $M^{2026} = M^{1} = M$ .

Question 12 · Linear Algebra · 2026 MCQ

Let

γ_{1}, γ_{2}, γ_{3}

be the eigenvalues of the matrix

100 0 cos t - s in t 0 s in t cos t

, where

t \in [- π, π]

is in radians.

Which one of the following options lists all the possible values of

t

satisfying

γ_{1} + γ_{2} + γ_{3} = 1 + 2

A.
${\frac{π}{3}, - \frac{π}{4}}$
B.
${\frac{π}{4}, - \frac{π}{3}}$
C.
${\frac{π}{4}, - \frac{π}{4}}$
D.
${\frac{π}{3}, - \frac{π}{3}}$

Correct Answer:

Step-by-Step Solution

Insight: The matrix is block-diagonal — a $1 \times 1$ block $[1]$ and a $2 \times 2$ rotation block — so the eigenvalues are the union of the blocks' eigenvalues.

Exam route: Eigenvalues of the $1 \times 1$ block: ${1}$ . Eigenvalues of the $2 \times 2$ block $[cos t - sin t sin t cos t]$ : solve $(cos t - λ)^{2} + sin^{2} t = 0 \Rightarrow λ = cos t \pm i sin t = e^{\pm i t}$ . Sum of all three eigenvalues (the trace) is $1 + 2 cos t$ . Set $1 + 2 cos t = 1 + 2 \Rightarrow cos t = \frac{1}{2}$ . For $t \in [- π, π]$ , this gives $t = \pm \frac{π}{4}$ .

Learning route:

Identify the block structure: $A = diag (1, R (- t))$ where $R (- t)$ is a clockwise rotation by $t$ .
The spectrum of a block-diagonal matrix is the union of the spectra of its blocks.
Block 1 contributes $γ_{1} = 1$ .
Block 2 has characteristic polynomial $λ^{2} - 2 cos t \cdot λ + 1 = 0$ , giving $γ_{2} = e^{i t}, γ_{3} = e^{- i t}$ .
Sum: $γ_{1} + γ_{2} + γ_{3} = 1 + e^{i t} + e^{- i t} = 1 + 2 cos t$ .
Equation: $1 + 2 cos t = 1 + 2 \Rightarrow cos t = \frac{2}{2}$ .
In $[- π, π]$ , $cos t = \frac{2}{2}$ has exactly two solutions: $t = \frac{π}{4}$ and $t = - \frac{π}{4}$ .
The set of all such $t$ is ${\frac{π}{4}, - \frac{π}{4}}$ , matching option C.

Question 13 · 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 14 · Quantitative Aptitude · 2026 NAT

The number of bijections $f (\cdot)$ from the set $S = {1, 2, 3, 4}$ to itself such that $f (f (n)) = n$ , for all $n \in S$ , is __________ . (Answer in integer)

Correct Answer:

10.00

Step-by-Step Solution

Insight: The condition $f (f (n)) = n$ defines an involution, meaning the permutation consists entirely of 1-cycles (fixed points) and 2-cycles (swaps).

Exam route: Use the involution recurrence $I (n) = I (n - 1) + (n - 1) I (n - 2)$ . With $I (1) = 1$ and $I (2) = 2$ , we get $I (3) = 2 + 2 (1) = 4$ , and $I (4) = 4 + 3 (2) = 10$ .

Learning route:

We classify the bijections by their cycle structure for $n = 4$ :

Zero swaps (4 fixed points): There is exactly $(0 4) = 1$ way.
One swap (2 fixed points): Choose 2 elements to swap out of 4. This is $(2 4) = 6$ ways.
Two swaps (0 fixed points): Choose 2 elements for the first swap ( $(2 4) = 6$ ), and the remaining 2 form the second swap ( $(2 2) = 1$ ). Since the two swaps are indistinguishable, we divide by $2!$ . This gives $\frac{6 \times 1}{2} = 3$ ways.

Total involutions = $1 + 6 + 3 = 10$ .

Common Trap: Forgetting to divide by $2!$ in the two-swap case leads to $6 \times 1 = 6$ ways, incorrectly totaling $1 + 6 + 6 = 13$ .

Verification: The recurrence $I (4) = I (3) + 3 I (2) = 4 + 3 (2) = 10$ perfectly matches the manual enumeration.

Question 15 · Database Management and Warehousing · 2026 NAT

In a relational database, a B+ Tree Index is to be constructed for a relation on a key field. In a B+ Tree, a Node Pointer points to a sub-tree and a Data Record Pointer points to a block of database records.

Let, Node size = 4096 bytes, Node Pointer size = 10 bytes, Search Key Field size = 11 bytes and Data Record Pointer size = 12 bytes.

The maximum number of Node Pointers that can be present in a non-leaf node of the B+ Tree is ________ . (Answer in integer)

Correct Answer:

195

Step-by-Step Solution

Insight: The maximum fanout of a non-leaf node is constrained by the fixed block size, ensuring all pointers and keys fit within a single node.

Exam route: Use the non-leaf fanout inequality: $p \times P_{t r} + (p - 1) \times V \leq B$ . Substitute $P_{t r} = 10$ , $V = 11$ , $B = 4096$ , and solve for integer $p$ .

Learning route:

Identify node components: A non-leaf node contains $p$ node pointers and $p - 1$ search keys. Data record pointers are only present in leaf nodes.
Set up the size inequality: Total size = $p \times (Node Pointer Size) + (p - 1) \times (Search Key Size) \leq Node Size$ .
Substitute given values: $p \times 10 + (p - 1) \times 11 \leq 4096$ .
Simplify the inequality: $10 p + 11 p - 11 \leq 4096 ⟹ 21 p - 11 \leq 4096$ .
Solve for $p$ : $21 p \leq 4107 ⟹ p \leq 4107/21 \approx 195.57$ .
Apply integer constraint: The maximum number of pointers $p$ must be an integer, so we take the floor, yielding $p = 195$ .

Question 16 · Database Management and Warehousing · 2026 MCQ

Consider a B+ Tree where the maximum number of key values in each leaf node is 2 and the maximum number of pointers in each non-leaf node is 3. Let the content of the B+ Tree be as shown in the figure.

Which of the following options denotes the key value(s) stored in the root node after inserting a key value 3 in the given B+ Tree?

A.
5
B.
8
C.
3 and 5
D.
3, 5 and 8

Correct Answer:

Step-by-Step Solution

Insight: Leaf overflow triggers a split, promoting a separator to the parent. If the parent exceeds its pointer capacity, it recursively splits, pushing the middle key up to a new root.

Exam route: Insert 3 into [1, 5] $\to$ [1, 3, 5]. Split leaf: left gets [1, 3], right gets [5]. Promote separator to parent. Root [5, 8] becomes [3, 5, 8] (4 pointers), which overflows (max 3 pointers). Split root: middle key 5 moves to new root.

Learning route:

Locate insertion point: Key 3 is $\leq 5$ , so it goes down the first pointer into the leaf [1, 5].
Leaf overflow: The leaf becomes [1, 3, 5], exceeding the max capacity of 2 keys.
Leaf split: Distribute entries evenly. Left leaf gets $⌈ 3/2 ⌉ = 2$ keys: [1, 3]. Right leaf gets the remaining 1 key: [5]. The separator key is promoted to the parent.
Parent overflow: The root originally had keys [5, 8] and 3 pointers. Adding a new pointer and key makes it 4 pointers and 3 keys (e.g., [3, 5, 8]), exceeding the max pointer limit of 3.
Root split: The root splits into two nodes. The middle key (5) is pushed up alone to form a new root. The left child keeps [3] and the right child keeps [8].
Final state: The new root contains only the single key 5.

Question 17 · 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 18 · Calculus and Optimization · 2024 MSQ

Consider the function

f : R \to R

where

R

is the set of all real numbers.

f (x) = \frac{x ^{4}}{4} - \frac{2 x ^{3}}{3} - \frac{3 x ^{2}}{2} + 1

Which of the following statements is/are TRUE?

A.
$x = 0$ is a local maximum of f
B.
$x = 3$ is a local minimum of f
C.
$x = - 1$ is a local maximum of f
D.
$x = 0$ is a local minimum of f

Correct Answer:

["A","B"]

Step-by-Step Solution

Insight: Standard polynomial extrema classification using the Second Derivative Test.

Exam route: $f^{'} (x) = x^{3} - 2 x^{2} - 3 x = x (x^{2} - 2 x - 3) = x (x - 3) (x + 1)$ . Critical points: $x = 0, 3, - 1$ . $f^{''} (x) = 3 x^{2} - 4 x - 3$ .

Evaluate $f^{''} (x)$ at each critical point:

$f^{''} (0) = - 3 < 0 ⟹$ local maximum at $x = 0$ . (Option A is TRUE, D is FALSE)
$f^{''} (3) = 27 - 12 - 3 = 12 > 0 ⟹$ local minimum at $x = 3$ . (Option B is TRUE)
$f^{''} (- 1) = 3 + 4 - 3 = 4 > 0 ⟹$ local minimum at $x = - 1$ . (Option C is FALSE)

Learning route: The Second Derivative Test is the fastest way to classify critical points for polynomials. Always factor $f^{'} (x)$ completely to avoid missing roots. A positive second derivative means concave up (local min), negative means concave down (local max).

Question 19 · Artificial Intelligence · 2026 MSQ

Sentence

X

is said to entail Sentence

Y

if whenever

X

T R U E

Y

also must hold

T R U E

.
Which of the following statements is/are correct if

X

entails

Y

A.
$X \Rightarrow Y$
B.
$X \land \neg Y$ is $F A L S E$
C.
if $X$ then $Y$
D.
if $Y$ then $X$

Correct Answer:

["A","B","C"]

Step-by-Step Solution

Insight: Entailment $X ⊨ Y$ means the material implication $X ⟹ Y$ is valid, which is logically equivalent to $X \land \neg Y$ being unsatisfiable.

Exam route: If $X ⊨ Y$ , then $X ⟹ Y$ is a tautology (Option A). This means its negation, $X \land \neg Y$ , is always false (Option B). "If X then Y" is the natural language equivalent (Option C). The converse "if Y then X" (Option D) is not guaranteed.

Learning route:

Step 1: Recall the definition of semantic entailment: $X ⊨ Y$ means that in every possible model where $X$ is true, $Y$ is also true.

Step 2: Evaluate Option A. This condition is exactly the definition of the material implication $X ⟹ Y$ being valid (true in all models). Thus, A is correct.

Step 3: Evaluate Option B. By propositional equivalence, $X ⟹ Y \equiv \neg (X \land \neg Y)$ . If the implication is valid (always true), its negation $X \land \neg Y$ must be false in all models (unsatisfiable). Thus, B is correct.

Step 4: Evaluate Option C. "If X then Y" is the standard English reading of the valid implication $X ⟹ Y$ . Thus, C is correct.

Step 5: Evaluate Option D. The converse $Y ⟹ X$ (or $Y ⊨ X$ ) is not logically implied by $X ⊨ Y$ . A true premise guaranteeing a true conclusion does not mean the conclusion guarantees the premise. Thus, D is incorrect.

Question 20 · Artificial Intelligence · 2024 MSQ

Let game(ball, rugby) be true if the ball is used in rugby and false otherwise.
Let shape(ball, round) be true if the ball is round and false otherwise.
Consider the following logical sentences:
s1:

\forall

ball

\neg

game(ball, rugby)

⟹

shape(ball, round)
s2:

\forall

ball

\neg

shape(ball, round)

⟹

game(ball, rugby)
s3:

\forall

ball game(ball, rugby)

⟹ \neg

shape(ball, round)
s4:

\forall

ball shape(ball, round)

⟹ \neg

game(ball, rugby)
Which of the following choices is/are logical representations of the assertion,
“All balls are round except balls used in rugby”?

A.
$s 1 \land s 3$
B.
$s 1 \land s 2$
C.
$s 2 \land s 3$
D.
$s 3 \land s 4$

Correct Answer:

["A","C"]

Step-by-Step Solution

Insight: "All A except B" is a biconditional-like statement meaning "If not B, then A" AND "If B, then not A".

Exam route: s1 is $\neg$ rugby $⟹$ round. s3 is rugby $⟹ \neg$ round. Together, s1 $\land$ s3 captures the full meaning. Since s2 is the contrapositive of s1, s2 $\land$ s3 is logically equivalent to s1 $\land$ s3.

Learning route:

Step 1: Parse the assertion "All balls are round except balls used in rugby". This means two things must hold simultaneously:

(1) If a ball is NOT used in rugby, it is round.

(2) If a ball IS used in rugby, it is NOT round.

Step 2: Translate (1) into FOL: $\forall ball (\neg game (ball, rugby) ⟹ shape (ball, round))$ . This matches sentence s1 exactly.

Step 3: Translate (2) into FOL: $\forall ball (game (ball, rugby) ⟹ \neg shape (ball, round))$ . This matches sentence s3 exactly.

Step 4: Check logical equivalences. Sentence s2 is $\forall ball (\neg shape ⟹ game)$ . This is the contrapositive of s1 ( $\neg Q ⟹ \neg P$ ), so s1 $\equiv$ s2. Sentence s4 is the contrapositive of s3, so s3 $\equiv$ s4.

Step 5: Evaluate the options. Option A (s1 $\land$ s3) directly combines the two required translations. Option C (s2 $\land$ s3) is also correct because substituting the logically equivalent s2 for s1 yields the exact same logical meaning. Options B and D are redundant, each only capturing half of the required meaning.

Question 21 · Analytical Aptitude · 2025 MCQ

Weight of a person can be expressed as a function of their age. The function usually varies from person to person. Suppose this function is identical for two brothers, and it monotonically increases till the age of

50

years and then it monotonically decreases. Let

a_{1}

and

a_{2}

(in years) denote the ages of the brothers and

a_{1} < a_{2}

.
Which one of the following statements is correct about their age on the day when they attain the same weight?

A.
$a_{1} < a_{2} < 50$
B.
$a_{1} < 50 < a_{2}$
C.
$50 < a_{1} < a_{2}$
D.
Either $a_{1} = 50$ or $a_{2} = 50$

Correct Answer:

Step-by-Step Solution

Key idea: This is a monotonicity-based logical deduction question, recognizable by the function's trend (increasing then decreasing) and the equality of outputs for two different inputs.

Step 1: Understand the function's behavior. The weight function $W (a)$ monotonically increases for $a \leq 50$ and monotonically decreases for $a > 50$ . This creates a single peak at $a = 50$ .

Step 2: Analyze the condition $W (a_{1}) = W (a_{2})$ with $a_{1} < a_{2}$ .

Step 3: Evaluate the positions of $a_{1}$ and $a_{2}$ relative to the peak.

If both $a_{1} < a_{2} \leq 50$ , they lie on the monotonically increasing segment. For a strictly monotonic function, $W (a_{1}) < W (a_{2})$ , so they cannot be equal.

If both $50 \leq a_{1} < a_{2}$ , they lie on the monotonically decreasing segment. Similarly, $W (a_{1}) > W (a_{2})$ , so they cannot be equal.

Step 4: The only way for $W (a_{1}) = W (a_{2})$ with $a_{1} \neq = a_{2}$ is if one age is on the increasing slope and the other is on the decreasing slope.

Step 5: Since $a_{1} < a_{2}$ , it must be that $a_{1} < 50$ and $a_{2} > 50$ .

Step 6: This matches the condition $a_{1} < 50 < a_{2}$ .

Answer: B

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