Evaluate
$ I = \int_{0}^{\pi/2} \frac{\sin^2 x}{1+\sin x \cos x}dx $
| Column A | Column B |
|---|---|
| A. $ \frac{d}{dx}(\cos^{-1}x) $ | I. $ -\frac{1}{1+x^2} $ |
| B. $ \frac{d}{dx}(\cot^{-1}x) $ | II. $ \frac{1}{|x|\sqrt{x^2-1}} $ |
| C. $ \frac{d}{dx}(\cos ec^{-1}x) $ | III. $ -\frac{1}{\sqrt{1-x^2}} $ |
| D. $ \frac{d}{dx}(\sec^{-1}x) $ | IV. $ -\frac{1}{|x|\sqrt{x^2-1}}$ |
| Column A | Statement | Column B | Expression |
|---|---|---|---|
| A | Neither A nor B | I | $(A \cap \overline{B}) \cup (\overline{A} \cap B)$ |
| B | At least one of A, B or C | II | $\overline{A} \cap \overline{B}$ |
| C | Exactly one of A and B | III | $A \cup B \cup C$ |
| D | All three A, B, C | IV | $A \cap B \cap C$ |
| Column A | Column B |
|---|---|
| A. $(\sqrt{2}+1)+1+(\sqrt{2}-1)+\ldots \infty$ | IV. $\frac{4+3\sqrt{2}}{2}$ |
| B. $ \frac{1}{2}+\frac{1}{3^2}+\frac{1}{2^3}+\frac{1}{3^4}+\frac{1}{2^5}+\frac{1}{3^6}+\ldots \infty$ | I. $\frac{19}{24}$ |
| C. $6^{1/2}\times 6^{1/4}\times 6^{1/8}\ldots \infty$ | II. $6$ |
| D. $8+4\sqrt{2}+4+\ldots \infty$ | III. $8(2+\sqrt{2})$ |
| Column A | Column B |
|---|---|
| A. BFS | I. Shortest Path |
| B. Sparse graph | II. Adjacency list representation |
| C. Collision Resolution | III. Open Addressing |
| D. Link list | IV. Self Referential structure |
| Column A | Column B |
|---|---|
| A. AND | I. Output 1 only if both inputs are 1 |
| B. OR | II. Output 1 if any input is 1 |
| C. NAND | III. Output 0 only if both inputs are 1 |
| D. NOR | IV. Output 1 only if both inputs are 0 |
| Column A | Column B |
|---|---|
| A. Counter register | I. Read by host to get input |
| B. Status register | II. Written by host to sent output |
| C. Data in register | III. Contain bit that can be read by host |
| D. Data out register | IV. Can be written by host to start command |
| Column A | Column B |
|---|---|
| A. 4 : 17 :: 7 : ? | I. 18 |
| B. 720 : 840 :: 60 : ? | II. 82 |
| C. 225 : 256 :: 289 : ? | III. 324 |
| D. 21 : 3 :: 574 : ? | IV. 50 |
Options
(a) (A)-I, (B)-II, (C)-III, (D)-IVMatch the following:
| Column A | Column B |
|---|---|
| Decoder | I. Select input and output of many |
| Multiplexer | II. convert binary input into 2n output |
| Encoder | III. convert 2n input into binary output |
| Demultiplexer | IV. route one input to several output |
Options:
| (a) | Decoder – II, Multiplexer – I, Encoder – III, Demultiplexer – IV |
| (b) | Decoder – III, Multiplexer – I, Encoder – II, Demultiplexer – IV |
| (c) | Decoder – II, Multiplexer – III, Encoder – I, Demultiplexer – IV |
| (d) | Decoder – III, Multiplexer – IV, Encoder – II, Demultiplexer – I |
Assertion: 1/3, −1/2, 3/4, −9/8 form an AP
Reason: The constant sequence is the only sequence which is both AP as well as GP.
Options:
(a) Assertion is true, Reason is true, and Reason is the correct explanation of the Assertion.
(b) Assertion is true, Reason is true, but Reason is not the correct explanation of the Assertion.
(c) Assertion is true, but Reason is false.
(d) Assertion is false, but Reason is true.
Matching
| Column A | Column B | |
|---|---|---|
| A. Taxonomy | I. Biography | |
| B. Pedology | II. History | |
| C. Man | III. Social | |
| D. Nation | IV. Classification |
Options:
(a) A–IV, B–I, C–III, D–II
(b) A–III, B–I, C–IV, D–II
(c) A–IV, B–III, C–I, D–II
(d) A–II, B–IV, C–I, D–III
Taxonomy → Classification → IV
Pedology → Study of soil / earth (related to earth history) → II
Man → Social → III
Nation → Biography / historical account → I
Correct Answer: (a) A–IV, B–I, C–III, D–II
Half adder has
(a) two input two output
(b) one input two output
(c) two input one output
(d) one input one output
A Half Adder is a combinational circuit used to add two binary bits.
Inputs: A, B → 2 inputs
Outputs: Sum (S) and Carry (C) → 2 outputs
Sum = A ⊕ B
Carry = A · B
| Column A | Column B | |
|---|---|---|
| (A) FCFS | I. Give CPU according to highest priority | |
| (B) SJF | II. Gives a CPU to process which come first | |
| (C) Priority Scheduling | III. Gives CPU to process with shortest burst time | |
| (D) RR | IV. Time sharing |
Options:
(a) (A)-II, (B)-III, (C)-I, (D)-IV
(b) (A)-I, (B)-II, (C)-III, (D)-IV
(c) (A)-II, (B)-I, (C)-III, (D)-IV
(d) (A)-IV, (B)-III, (C)-II, (D)-I
FCFS → process which comes first → II
SJF → shortest burst time → III
Priority Scheduling → highest priority → I
Round Robin (RR) → time sharing → IV
Find the missing ?
M, N, O, L, R, J, V, ?
Options:
(a) A
(b) E
(c) F
(d) H
Split the series into two parts:
1st, 3rd, 5th, 7th terms: M, O, R, V
Positions: 13, 15, 18, 22
Pattern: +2, +3, +4
2nd, 4th, 6th, 8th terms: N, L, J, ?
Positions: 14, 12, 10, 8
Pattern: −2, −2, −2
Position 8 corresponds to H.
Correct Answer: (d) H
Assertion: Binary search tree property allow to print all keys in binary search tree in sorted order by simple recursive algorithm called inorder tree walk.
Reason: The inorder tree walk algorithm print the key root of subtree between printing value in its left subtree and printing those in its right subtree.
Options:
(a) Assertion is true, Reason is true, and Reason is the correct explanation of the Assertion.
(b) Assertion is true, Reason is true, but Reason is not the correct explanation of the Assertion.
(c) Assertion is true, but Reason is false.
(d) Assertion is false, but Reason is true.
In a Binary Search Tree (BST), all keys in the left subtree are smaller than the root and all keys in the right subtree are larger than the root.
Inorder traversal follows the order:
Left Subtree → Root → Right Subtree
Because of the BST property, this traversal prints the elements in sorted order.
The reason correctly explains the inorder traversal process where the root is printed between its left and right subtree.
Correct Answer: (a) Assertion is true, Reason is true, and Reason is the correct explanation of the Assertion.
Assertion: A(BA) and (AB)A are symmetric matrices.
Reason: AB is symmetric matrix, if matrix multiplication of A with B is commutation.
Options:
(a) Assertion is true, Reason is true, and Reason is the correct explanation of the Assertion.
(b) Assertion is true, Reason is true, but Reason is not the correct explanation of the Assertion.
(c) Assertion is true, but Reason is false.
(d) Assertion is false, but Reason is true.
A(BA) = (AB)A = ABA
If A and B are symmetric matrices, then
(ABA)T = ATBTAT
Since A and B are symmetric,
AT = A and BT = B
Therefore
(ABA)T = ABA
Hence A(BA) and (AB)A are symmetric matrices.
Also, AB is symmetric if AB = BA (i.e., A and B commute).
Correct Answer: (a) Assertion is true, Reason is true, and Reason is the correct explanation of the Assertion.
Arrange the following steps of retrieving data from memory unit.
(A) Memory Address Register send address to memory chip
(B) The address is placed on address bus
(C) The control unit sends signal
(D) Data is retrieved and place into the Memory Data Register
Options:
(a) D, C, A, B
(b) A, B, C, D
(c) A, C, D, B
(d) B, D, A, C
1. Memory Address Register (MAR) sends the address → (A)
2. Address is placed on the address bus → (B)
3. Control unit sends the read signal → (C)
4. Data is fetched and stored in Memory Data Register (MDR) → (D)
Correct Order: A → B → C → D
Correct Answer: (b) A, B, C, D
| Column A | Column B | |
|---|---|---|
| (A) Minimum Spanning Tree | I. Height balanced | |
| (B) Topological Sort | II. Stack | |
| (C) Recursion | III. Kruskal's algorithm | |
| (D) AVL Tree | IV. DFS |
Options:
(a) (A)-I, (B)-II, (C)-III, (D)-IV
(b) (A)-II, (B)-I, (C)-III, (D)-IV
(c) (A)-IV, (B)-III, (C)-II, (D)-I
(d) (A)-III, (B)-IV, (C)-II, (D)-I
Minimum Spanning Tree → Kruskal's algorithm → III
Topological Sort → DFS based algorithm → IV
Recursion → uses stack → II
AVL Tree → height balanced tree → I
Correct Answer: (d) (A)-III, (B)-IV, (C)-II, (D)-I
(a) A is diagonal matrix
(b) A is zero matrix
(c) A is scalar matrix
(d) A is square matrix
If a matrix A is symmetric, then
AT = A
If a matrix A is skew symmetric, then
AT = −A
For both conditions to hold:
A = −A
⇒ 2A = 0
⇒ A = 0
Therefore, the matrix must be a zero matrix.
Correct Answer: (b) A is zero matrix
(a) Monkey
(b) Stag
(c) Ass
(d) Sheep
Cub → young one of Tiger
Fawn → young one of Deer
Adult male deer is called Stag.
Correct Answer: (b) Stag
A. Deletion from Queue
B. Worst case search in binary search tree
C. Insertion in stack
D. Quick sort worst case
Options:
(a) C, D
(b) A, C
(c) A, B
(d) B, C
Deletion from Queue → O(1)
Worst case search in Binary Search Tree → O(n)
Insertion in Stack → O(1)
Quick Sort worst case → O(n2)
Same order complexity:
Deletion from Queue and Insertion in Stack → O(1)
Correct Answer: (b) A, C
A. Half adder
B. SR flip flop
C. Full adder
D. D flip flop
Options:
(a) A, C
(b) B, D
(c) B, C, D
(d) A, B, C, D
Half Adder → Combinational circuit
Full Adder → Combinational circuit
SR Flip Flop → Sequential circuit
D Flip Flop → Sequential circuit
Combinational circuits depend only on present input, while sequential circuits depend on present input and previous state.
Correct Answer: (a) A, C
Length or number of data element of Array can be obtained from index set by which of the following formulae.
(a) Lower bound − Upper bound − 1
(b) Lower bound − Upper bound + 1
(c) Upper bound − Lower bound + 1
(d) Upper bound + Lower bound − 1
Number of elements in an array is calculated using:
Length = Upper Bound − Lower Bound + 1
Example:
If Lower Bound = 0 and Upper Bound = 9
Length = 9 − 0 + 1 = 10
Correct Answer: (c) Upper bound − Lower bound + 1
Cooperating process required an interprocess communication mechanism. This work on which of the following fundamental model.
(A) Shared Memory
(B) Cascading termination
(C) Message passing
(D) Cooperating
Options:
(a) C, D
(b) A, C
(c) A, D
(d) B, D
Interprocess Communication (IPC) between cooperating processes is mainly based on two fundamental models:
Shared Memory Model
Message Passing Model
Shared memory allows processes to communicate by sharing a common memory space, while message passing allows processes to exchange information through messages.
Correct Answer: (b) A, C
Options:
(a) x2 − 3xy + y2 = 0
(b) x2 + 3xy − y2 = 0
(c) x2 + 4xy − y2 = 0
(d) x2 − 3xy − y2 = 0
Which of the following is not an application of stack.
(a) Tower of Hanoi
(b) evaluation of postfix expression
(c) Priority queue
(d) Recursion
Tower of Hanoi → solved using recursion and stack concept
Evaluation of postfix expression → uses stack
Recursion → implemented using call stack
Priority queue is implemented using heap or special queue structure, not stack.
Correct Answer: (c) Priority queue
Arrange the following steps of K-map simplification in correct order.
(A) Write minimized SOP / POS expression
(B) Plot truth table outputs (0's and 1's) on K-map
(C) Identify prime implicants and essential prime implicants
(D) Draw groups (loops) of adjacent 1's or 0's that are powers of 2
Options:
(a) A, B, C, D
(b) C, A, B, D
(c) B, C, A, D
(d) B, D, C, A
Step 1: Plot truth table outputs on K-map → (B)
Step 2: Draw groups (loops) of adjacent 1's or 0's → (D)
Step 3: Identify prime implicants and essential prime implicants → (C)
Step 4: Write minimized SOP / POS expression → (A)
Correct Order: B → D → C → A
Correct Answer: (d) B, D, C, A
Assertion (A): Copper is used to make electric wires.
Reason (R): Copper has very low electrical resistance.
Options:
(A) Both A and R are true and R is the correct explanation of A.
(B) Both A and R are true but R is NOT the correct explanation of A.
(C) A is true but R is false.
(D) A is false and R is true.
Copper is widely used in electric wires because it has very low electrical resistance (high conductivity), which allows electric current to flow easily with minimal energy loss.
Thus, the Assertion is true and the Reason is also true, and the Reason correctly explains the Assertion.
Correct Answer: (A) Both A and R are true and R is the correct explanation of A.
Assertion (A): When lightning strikes, the light appears shortly after the sound is heard.
Reason (R): The velocity of light is greater than the velocity of sound.
Options:
(1) A is true but R is false.
(2) Both A and R are true and R is the correct explanation of A.
(3) A is false but R is true.
(4) Both A and R are true but R is not the correct explanation of A.
In reality, when lightning strikes, we see the light before hearing the thunder because the speed of light is much greater than the speed of sound.
Therefore, the Assertion is false, while the Reason is true.
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