CUET PG Computer Science Mock Test - 4 - CUET PG MCQ

The correct answer is O(n).

Explanation

• The time complexity of finding the maximum number in an unsorted array of n elements is O(n).
• This is because you need to check each element of the array to ensure you have found the maximum value.
• In the worst-case scenario, every element must be compared, leading to a linear time complexity of O(n).

Other Related Points

• O(1): This represents constant time complexity, which means the operation takes the same amount of time regardless of the size of the input. This is not applicable here because we need to scan all elements in the array.
• O(log n): This represents logarithmic time complexity, which usually applies to operations in sorted arrays or binary trees. It is not applicable for finding the maximum in an unsorted array.
• O(n^2): This represents quadratic time complexity, which is typically seen in algorithms like bubble sort or selection sort. It is much higher than necessary for finding the maximum in an unsorted array.

Multiplexer:

• Combining one or more signals and transmitting on a single channel
• It consists more than one input line and only one output line and more than one select line

2 X 1 MUX

Decoder:

• It consists of n input lines and maximum of 2ⁿ output lines.
• Outputs of the decoder are nothing but the min terms of n input variables.

2 X 4 Decoder

Encoder:

• Reverse operation of decoder.
• It consists of 2ⁿ input lines and n output lines.
• Enable signal is optional to represent.

2² × 2 Encoder

Demultiplexer:

• Reverse operation of multiplexer.
• It consists of n selection lines and maximum of 2ⁿ outputs and input will be connected to one of these outputs based on the values of selection lines.

1 × 2 DEMUX

Hence option 3 is the correct answer.

The correct answer is -128, 0

Explanation

The least values of a signed and unsigned 8-bit data type variables are:

• -128 for a signed variable
• 0 for an unsigned variable

Explanation:

• Signed variables: With 8 bits, a signed variable can represent both positive and negative numbers. The leftmost bit is used for the sign (0 for positive, 1 for negative). Leaving 7 bits for the value, it can range from -2^(7-1) (which is -128) to 2^(7-1) - 1 (which is 127).
• Unsigned variables: Since unsigned variables only represent non-negative numbers, all 8 bits are available for the value. This allows it to range from 0 (all bits 0) to 2^8 - 1 (which is 255).

Given:
Arithmetic progression: -10, -7, -4, ...
Sum of first k terms (S_k) = 104
Formula used:
S_k = (k/2)[2a + (k - 1)d]
where a is the first term and d is the common difference.
Calculation:
a = -10
d = -7 - (-10) = 3
Now, substitute the values into the sum formula:
⇒ 104 = (k/2)[2(-10) + (k - 1)3]
⇒ 104 = (k/2)[-20 + 3k - 3]
⇒ 104 = (k/2)[3k - 23]
⇒ 208 = k(3k - 23)
⇒ 3k² - 23k - 208 = 0
⇒ 3k² - 39k + 16k - 208 = 0
⇒ 3k(k - 13) + 16(k - 13) = 0
⇒ (3k + 16)(k - 13)
⇒ k = 13, or -16/3(Negative not possible)
The value of (k + 9) / (k - 5) is
⇒ (13 + 9) / (13 - 5) = 22/8 = 11/4.
Hence the correct answer is 11/4.

Calculation
Parabola x² = 6y and the ellipse (x² / 4) + y² = 1 will satisfy each other
⇒ (6y / 4) + y² = 1
⇒ (3y / 2) + y² = 1
⇒ 3y + 2y² = 2
⇒ 2y² + 3y = 2
⇒ 2y² + 3y - 2 = 0
⇒ (y + 2)(2y - 1) = 0
⇒ (y + 2) = 0 or (2y - 1) = 0
Hence Option(2) is the correct answer.

Given:
First term (a) = 3
Common difference (d) = 5
Number of terms (n) = 11
Formula Used:
Sum of the first n terms of an A.P. (Sₙ) = n/2 × (2a + (n-1)d)
Calculation:
S₁₁ = 11/2 × (2 × 3 + (11-1) × 5)
⇒ S₁₁ = 11/2 × (6 + 50)
⇒ S₁₁ = 11/2 × 56
⇒ S₁₁ = 11 × 28
⇒ S₁₁ = 308
The sum of the first 11 terms is 308.
Answer: 3) 30

The logic followed here is:

Logic: Synonyms

Here,

Summon : Call → Call is a synonym of Summon.

Similarly,

Disperse : Spread → Spread is a synonym of Disperse.

Hence, the correct answer is "Option 3".

Concept -
(1) |kA| = kn |A|
(2) |adj A| = |A|n–1
Explanation -
We have,
= 2(4 – 1) – 1(2 – 0) + 0
= 6 – 2 = 4
So, |2A| = 2³ |A| = 8 × 4 = 32
Now, |adj (adj (adj 2A))| = |2A|^(n-1)³
= (32)^(2³) = 32⁸
⇒ 16ⁿ = (32)⁸ = 2⁸ × 16⁸
⇒ 16ⁿ = 16²⁺⁸ ⇒ n = 10
Hence Option (4) is correct.

Equation of normal at (ct, c/ct) is yt - xt³ - c + ct⁴ = 0.
If it passes through any point say Q having its coordinates as t', i.e., (ct', c/t').
The coordinates must satisfy the equation.
Hence, we get (c/t') * t - ct't³ - c + ct⁴ = 0.
⇒ t - t't²t³ - t' + t't⁴ = 0.
On factorizing, we get, (t't³ + 1)(t - t') = 0.
Hence, t't³ + 1 = 0 or t't³ = -1 as t ≠ t'.

Binary search algorithm:
Binary search algorithm is used to find an element in an already sorted array.
STEPS 1:
It finds the middle element of the array and compare the element with the element to be searched, if it matches then return true.
STEPS 2:
If not, then divide the array into two halves in which if element to be search is less than middle element then search takes place in left part otherwise search in right half.
STEP 3:
Repeat this process, until you get the element.
Explanation:
For worst case 52
Worst Case: Search 50 in below give array

midde index = (0 + 9) / 2 = 4 ∴ a[4] = 35
50 > 32
midde index = (5 + 9) / 2 = 7 ∴ a[7] = 63
50 < 63
midde index = (5 + 6) / 2 = 8 ∴ a[5] = 50
matched
T(n) = O(log n)Also, for average case:
T(n) = O(log n)

For an ellipse of the form (x²/a²) + (y²/b²) = 1 with a² > b², the foci are located along the x‑axis at distances c from the center, where

  c = √(a² – b²).

Here, a² = 100 and b² = 64, so

  c = √(100 – 64) = √36 = 6.

The distance between the foci is 2c, which is

  2 × 6 = 12.

Thus, the correct answer is 12.

Given:

The son-in-law remembers the birthday is before August 20.

The wife remembers the birthday is after August 18 and before August 22

Analyze the son's statement.

The son-in-law is certain that the birthday occurs before August 20.

Possible dates based on his memory are: August 18, 19.

Analyze the wife's statement.

The wife is certain that the birthday occurs after August 18 and before August 22.

Possible dates based on her memory are: August 19, 20, 21.

The common date satisfying both statements is August 19.

Thus, The mother-in-law's birthday is on August 19.

Hence, the correct answer is "Option 1".

Concept Used:
tan⁻¹ a - tan⁻¹ b = tan⁻¹ ((a - b) / (1 + ab))
Calculation:
∫ from 0 to 1 tan⁻¹ ((2x - 1) / (1 + x - x²)) dx
= ∫ from 0 to 1 tan⁻¹ ((x - (1 - x)) / (1 + x(1 - x))) dx
⇒ ∫ from 0 to 1 [tan⁻¹ x - tan⁻¹ (1 - x)] dx
⇒ ∫ from 0 to 1 tan⁻¹ x dx - ∫ from 0 to 1 tan⁻¹ (1 - x) dx
Let I = ∫ from 0 to 1 tan⁻¹ x dx and J = ∫ from 0 to 1 tan⁻¹ (1 - x) dx.
In J, let u = 1 - x, so du = -dx. When x = 0, u = 1 and when x = 1, u = 0.
⇒ J = -∫ from 1 to 0 tan⁻¹ u du = ∫ from 0 to 1 tan⁻¹ u du = I
Therefore, I - J = 0.
Hence option 2 is correct.

The correct answer is option 3.

Concept:

Dynamic loading:

Dynamic loading is the process by which a computer programme loads a library into memory, obtains the variable and function addresses, performs the functions, and then unloads the programme from memory. Use of this technique is common while implementing plugins.

In dynamic loading, a system library or routine is loaded at run-time. When a dependent library or routine is linked dynamically at run-time, it is called dynamic linking.

Hence the correct answer is Dynamic loading.

Other Related Points Dynamic linking:

At dynamic linking, code is compiled and linked into a format that programmes may load both at link time and during run time. They differ from standard object files because of this.

Overlays:

Overlaying is a programming method that allows programs to be larger than the computer's main memory.

Concept

The number of states for the MOD-N counter is given by:

No. of states = 2ⁿ

where, n = No. of J-K flip-flops

For the MOD-N counter, the frequency of the output is divided by N.

Calculation

Given, the No. of states = 2n = 8

2n = 8

n = 3

Hence, the correct answer is option 3.

We rewrite the sentences by putting inequalities in terms of beauty:

1) Asha is more beautiful than Prerna.

Asha > Prerna

2) Prerna is not as beautiful as Yashashree.

Prerna < Yashashree

3) Madhvi is not as beautiful as Prerna or Yashashree.

Madhvi < Prerna, Yashashree

From above 3 inequalities, we can conclude,

Asha, Yashashree > Prerna > Madhavi

Calculation:
Given, f(x) = |x2 - 1| =

From the figure f, has a local minima at x = ± 1 at and a local maxima at x = 0
∴ f has a local minima at x = ± 1 and a local maxima at x = 0.
The correct answer is Option 3.

Given: Pointing towards a photo of a girl, Rekha said, "She is the daughter of my brother's father".
Case 1)

Case 2)

So, the girl can be either Rekha herself or Rekha's sister.
As out of all the options, sister is given therefore the case 1 is eliminated.
Thus, the girl is the sister of Rekha.
Hence, the correct answer is "Option 1"

The correct answer is Stacks

Explanation

A linear data structure is a structure wherein data elements are adjacent to each other, and each data element connects to its previous and next element. The elements are arranged in a sequential order where they have a predecessor and a successor, except the first and last elements. Examples of linear data structures include Arrays, Linked Lists, Stacks, and Queues.

• Arrays: In an array, data is stored in a contiguous block of memory. Each element in the array can be accessed directly using its index.
• Linked Lists: Linked lists also contain a collection of data elements where each element points to the next element in the sequence. Data elements aren't stored in continuous memory like arrays.
• Stacks: A stack is a linear data structure that follows a particular order in which operations are performed. The order is LIFO (Last In, First Out) meaning that the last element added to the stack will be the first one to be removed.
• Queues: It's another linear data structure that follows the FIFO (First In, First Out) principle. The first element added to the queue will be the first one to be removed.

So, among the stack is a linear data structure.

Other Related Points

A non-linear data structures are structured in a way where each data element can connect to two or more other data elements and a hierarchical relationship exists among the data elements. Examples include Trees, Graphs, B-trees, etc.

• Tree: A tree is a non-linear data structure that emulates a tree (in biology), with a set of linked nodes. It starts at a top node (the root) and proceeds downwards to the leaf nodes with a connection between each parent node and its children nodes.
• Graph: A graph is also a non-linear data structure. In a graph, data elements are interconnected with each other, where each node can connect with any other nodes in the graph.
• Binary Tree: This is a type of tree where each node can have at most 2 children: a left child and a right child.

Three-dimensional array A[b1...u1, b2...u2, b3...u3]Here b is base value, u max value
number of planes = u1 - b1 + 1 = r1
number of rows = u2 - b2 + 1 = r2
number of columns = b3 - u3 + 1 = r3
Row major order of array A[i, j, k] = L0 + {(i - b1) x r2 x r3 + (j - b2) x r3 + (k - b3)} x c
L0 = Base address
c = size of each element
r1 = 8 - 2 + 1 = 7
r2 = 1-(-4) + 1 = 6
r3 = 10 - 6 + 1 = 5
A[5, -1, 8] = 400 + [(5 - 2) x 6 x 5 + (-1 - (-4)) x 5 + (8 - 6)] x 2
= 400 + [90 + 15 + 2] x 2
= 614
Hence the correct answer is 614.

Ranbir knows the pattern of various competitive exams as he has given many competitive exams.

Hence, Statement (A) is the cause and statement (B) is its effect.

The correct answer is both (1) and (2).

Explanation

• Counting: A flip-flop is a basic digital memory circuit that can be used to count events. It can store a single bit of data, and its state can be toggled or changed based on input signals. Counting applications often use multiple flip-flops connected in series to form binary counters.
• Scaling: In digital circuits, flip-flops can be used to divide the frequency of input signals, which is essentially a form of scaling. For example, a flip-flop can be used as a frequency divider to reduce the frequency of a clock signal by half.

Other Related Points

• Option 3: Demodulation - Demodulation is the process of extracting the original information-bearing signal from a modulated carrier wave. While flip-flops are fundamental in digital circuits, they are not typically used for demodulation purposes. Demodulation usually involves analog circuits or specialized digital signal processing techniques.

The statement suggests economizing water in summer at Jane's place, indicating concerns about water scarcity during that season.

Conclusion I logically follows because it directly addresses the potential reason for economizing water – possible water scarcity in summer.

Conclusion II is an assumption about water sufficiency in winter, which the statement does not explicitly support.

Therefore, the correct answer is (A) Only conclusion I follow, as it's the conclusion directly implied by the need to economize water in summer, indicating potential scarcity.

Hence, the correct answer is "Option (4)".

Eight persons: L, M, N, O, P, S, U and V
1. Some of them are facing towards the centre and rest are facing away from the centre.
2. M is an immediate neighbour of L and V.
3.L face away from centre.
4. P sits to the immediate left of L.

5. O sits to the immediate left of V.
This implies that V is facing towards the centre.
6. U sits second to the right of S who faces direction same as O and V.
This dettermines the position of N also.

7. N and U are facing away from centre.
8. M and L are facing in same direction as P.

The above arrangement will be the final seating arrangement.
Hence, 5 persons are facing away from the centre.

The Correct answer is Option 1.
Given Sequence:
2/3, 5/8, 8/13, 11/18
Identify the Numerator Pattern
Numerators: 2, 5, 8, 11, ?
Each numerator increases by 3.
2 + 3 = 5
5 + 3 = 8
8 + 3 = 11
Next numerator: 11 + 3 = 14
Identify the Denominator Pattern
Denominators: 3, 8, 13, 18, ?
Pattern: Each denominator increases by 5.
3 + 5 = 8
8 + 5 = 13
13 + 5 = 18
Next denominator: 18 + 5 = 23
Find the Next Fraction
The next fraction is: 14 / 23
Hence Option 1 is correct.

Calculation
Given
2b = ae
⇒ b/a = e/2
Eccentricity of the ellipse
⇒ e = √(1 - e² / 4)
⇒ e = 2 / √5
Hence option(4) is correct

Concept:
General Equation of a Circle
⇒ (x - h)² + (y - k)² = r²
where (h , k) is the center and r is the radius of the circle
Calculation:
Given Conditions for the Center (center lies on a line 2x + 3y = 3)
⇒ 2h + 3k = 3 --------(1)
Since the circle passes through the points (3, 0) and (0, -2):
Substituting Given Points (3, 0) and (0, -2) in the General equation of the circle:
⇒ (3 - h)² + k² = h² + (-2 - k)²
⇒ 6h + 4k = 5 --------(2)
Solving equations 1 and 2
2h + 3k = 3
6h + 4k = 5
Finding h and k:
h = ³/₁₀, k = ⁴/₁₀
Finding r²:
r² = (3 - ³/₁₀)² + (⁴/₁₀)²
r² = ¹⁰⁰/₁₀₀
Now,
(x - ³/₁₀)² + (y - ⁴/₁₀)² = ¹⁰⁰/₁₀₀
Expanding,
(x² - 6x + 9) + (y² - ⁸/₅ y + ¹⁶/₂₅) = 1
100(x² - 6x + 9) + 100(y² - ⁸/₅ y + ¹⁶/₂₅) = 100
100x² - 100y² - 60x + 100y² + 64 - 160y = 793
100x² + 100y² - 60x - 160y - 72 = 0
Divide by 10 and simplify:
10x² + 10y² - 6x - 16y - 72 = 0
Hence, option 2 is true.

Calculation:
Give, f(x) = cos x - 1 + (x² / 2!), x ∈ ℝ
⇒ f '(x) = - sin x + x
Now, ∀ x ∈ ℝ, x > sin x
⇒ x - sin x > 0
⇒ f '(x) > 0
⇒ f(x) is an increasing function.
∴ f(x) is an increasing function.
The correct answer is Option 2.

The correct option is Search by Hashing.

CONCEPT:

In linear search, every element of a given list is compared one by one with the given key without skipping any element.

It is useful when we need to search for an item in a small unsorted list, but the time taken to search the list increases as the size of the list increases.

For example, consider a list of length 5 and the key element is present at the end of this list. Number of comparisons required to search the key element = size of list i.e 5

If we increase the size of the same list (say 15) and the key element is present at the end of this list. Number of comparisons required to search the key element = size of list i.e 15

In binary search, the key to be searched is compared with the middle element of a sorted list, If the element at the middle position:

i) Matches the key the search is successful

ii) Is greater than the key then the key element may be present in the left part of the list

iii) Is smaller than the key, then the key element may be present in the right part of the list

This process continues till the element is found or the list is traversed completely.

Thus the time taken to search the list increases as the size of the list increases. But it will not be as large as the time required by linear search

Hash-based searching requires only one key comparison to find the position of a key, provided every element is present at its designated position decided by a hash function.

For example, consider a list of length 5 and the hash function: h(element) = element % size(list)

A hashing function is a mathematical function that generates unique results for each unique value supplied to the hash function in constant time.

For example: Consider a list of length 5 and if we want to search for key = 12, the index returned by the hash function is h(12) = 12 % 5 = 2 and requires only one key comparison to find the key at that index

Similarly increasing the size of the list (say 15) and if we want to search for key = 12, the index returned by the hash function is h(12) = 12 % 5 = 12 and requires only one key comparison to find the key at that index

Thus it is independent of the length of the list.

Given equations:

1) ax + by = a - b --- (1)

2) bx - ay = a + b --- (2)

Formula used:

Solving by elimination method:

Calculation:

Multiply equation (1) by b and equation (2) by a,

b(ax + by) = b(a - b)

⇒ abx + b²y = ab - b² ---- (3)

a(bx - ay) = a(a + b)

⇒ abx - a²y = a² + ab ---- (4)

Now, subtract the equations 4 from 3 we get,

abx + b²y - abx + a²y = ab - b² - a² - ab

⇒ (a² + b²)y = - (a² + b²)

⇒ y = -1

Substituting y = -1 in equation (1):

ax + b(-1) = a - b

⇒ ax = a

⇒ x = 1

∴ The solution is x = 1 and y = -1.