Mathematics: CUET Mock Test - 2 - Commerce MCQ

Calculations:
In 20 minutes 5/6^th part empty
In 1 minute 5/ 120 part empty
In 9 minutes 5 × 9/120 = 3/8^th Part Empty
Hence, the Correct option is 2.

1. Solve:
2x - 1 ≥ 3
⇒ 2x ≥ 4
⇒ x ≥ 2

2. Solve:
x - 3 > 5
⇒ x > 8

Intersection of both conditions:
x must satisfy both x ≥ 2 and x > 8.
The more restrictive condition is x > 8.

Final Answer: (8, ∞)
Correct option: C

Given inequality:
|3x| ≥ |6 - 3x|

Step 1: Square both sides (since absolute values are non-negative):
(3x)² ≥ (6 - 3x)²
9x² ≥ 36 - 36x + 9x²

Subtract 9x² from both sides:
0 ≥ 36 - 36x
36x ≥ 36
x ≥ 1

Step 2: Also consider the case x ≤ 0:
For x ≤ 0:
|3x| = -3x
|6 - 3x| = 6 - 3x

So, the inequality becomes:
-3x ≥ 6 - 3x
⇒ 0 ≥ 6 (which is false)

Alternative case for x ≤ 0:
|3x| = -3x
|6 - 3x| = -(6 - 3x) = -6 + 3x

So, now check:
-3x ≥ -6 + 3x
⇒ -6x ≥ -6
⇒ x ≤ 1

Since we are already considering x ≤ 0, and this gives x ≤ 1, the inequality holds for x ≤ 0

Final Solution:
x ≤ 0 or x ≥ 1
That is, the solution is: (-∞, 0] ∪ [1, ∞)

Among the options:

Correct Answer: B

Concept used:
An odd order Skew Symmetric matrix having 0 at its diagonal and aij = -aji, so x = u = w = 0

Calculations:
2 = -y ⇒ y = -2
z = -(-1) = 1
v = -6
Hence, the value of x2 + y2 + z2 + u2 + v2 + w² = 0 + 0 + 0 + 1 + 36 + 4 = 41
Hence, the Correct answer is option no 4

Calculations:

y = e^{n x}

.... (1)

Hence, Option 4 is correct

CONCEPT:

The inverse of a matrix: The Inverse of an n × n matrix is given by:
where adj(A) is called an adjoint matrix.
Adjoint Matrix: If Bn× n is a cofactor matrix of matrix An× n then the adjoint matrix of An× n is denoted by adj(A) and is defined as BT. So, adj(A) = BT.

CALCULATION:
Given: A^-1 = x A + yI

Given:

Concept:

Use formula

Calculation:

Hence the option (C) is correct.

Concept Used:-

We know that the value of tan 2θ is,

Also, the value of sin 2θ is,

Explanation:-

Suppose that,

Differentiate these equations with respect to x as,

On putting x=tan θ we get,

Or,

Now, the differential coefficient of u with respect to x,

Hence, the differential coefficient of tan-1() with respect to sin-1() is equal to 1.

Correct option is 4.

Concept:

Some useful formulas are:

sin(sin-1(x)) = x, -1≤ x ≤1

Calculation:

Given expression is sin(tan-1x)

= sin(sin-1)

= , since sin(sin-1x) = x

Given that,  
Separating the variables, we get

log⁡(y - 3) = log⁡(x - 3) + log⁡C₁
log⁡(y - 3) - log⁡(x - 3) = log⁡C₁
 = log⁡C₁
y-3 = C₁(x-3)
This is the general solution for the given differential equation where C₁ is a constant

Given that, 

Separating the variables, we get
2 cos⁡y dy = 3 sin⁡x dx
Integrating both sides, we get
∫ 2 cos⁡y dy = ∫ 3 sin⁡x dx
2 sin⁡y = 3(-cos⁡x) + C
3 cos⁡x + 2 sin⁡y = C

Given that, 
Separating the variables, we get
dy = (3e^x + 2)dx
Integrating both sides, we get
 -----(1)
y = 3e^x + 2x + C which is the general solution of the given differential equation.

We have the differential equation:

dy/dx + 8x = 16x² + 4,

with the condition that y = 1/3 when x = 1.

Rearrange the equation to isolate dy/dx:

dy/dx = 16x² + 4 - 8x = 16x² - 8x + 4.

Integrate both sides with respect to x:

y = ∫(16x² - 8x + 4) dx + C

= (16/3) x³ - 4x² + 4x + C.

Apply the initial condition y(1) = 1/3:

y(1) = (16/3)1³ - 41² + 4*1 + C

mathematica
Copy
 = 16/3 - 4 + 4 + C

 = 16/3 + C
Set this equal to 1/3:

16/3 + C = 1/3

Solve for C:

C = 1/3 - 16/3 = -15/3 = -5.

Write the particular solution:

y = (16/3) x³ - 4x² + 4x - 5.

You can verify by differentiating y:

dy/dx = 16x² - 8x + 4,

so

dy/dx + 8x = (16x² - 8x + 4) + 8x = 16x² + 4,

Step 1: Separate Variables

We are given:
dy/dx = (9y * log x) / (5x * log y)

Now rearrange the terms to separate x and y:

(log y)/y dy = (9 log x)/(5x) dx

Step 2: Integrate Both Sides

Left side:
∫ (log y)/y dy
Let u = log y → du = (1/y) dy
So, the integral becomes: ∫ u du = (u²)/2 = (log y)² / 2

Right side:
∫ (9 log x)/(5x) dx
Take constant 9/5 out: (9/5) ∫ (log x)/x dx
Let v = log x → dv = (1/x) dx
So, the integral becomes: ∫ v dv = (v²)/2 = (log x)² / 2
Thus the right side = (9/5) * (log x)² / 2

Step 3: Combine Results

So we get:
(log y)² / 2 = (9/5) * (log x)² / 2 + C

Multiply both sides by 2 to simplify:
(log y)² = (9/5) * (log x)² + K (where K = 2C)

Step 4: Analyze the Solution

We want a particular form from the options.
If K = 0, then:
(log y)² = (9/5) * (log x)²

This can be rearranged to:
(log y)² - (9/5)(log x)² = 0

Final Answer: Option B

Step 1: Expand the integrand

Step 2: Integrate term-wise

Step 3: Simplify

So, Option (c) is the correct answer.

Split the integral

First term (substitution):
Let u = x³
⇒ du = 3x² dx

Second term: 

Combine:

So, Option (c) is the correct answer.

We have :

On comparing the given equations with: 
, we get: 

The distance of a point whose position vector is  from the plane  given by :

check for interval (7/3, ∞ ) the whole would be +ve
check for interval (-∞, 3/2 ) the whole would be +ve

(7x-5)/(8x+3) > 4
(7x-5)/(8x+3) - 4 > 0
7x - 5 - 4 ( 8x + 3 ) / 8x + 3 > 0
- 25 x - 17 / 8x + 3 > 0
Now furthermore solving for general range:
x ∈ ( -17/ 25, - 3/8)

Since,
AM ≥ GM
⇒ (a + b) / 2 ≥ √ab
⇒ (a + b) / 2 ≥ √4 (∴ ab = 4, given)
⇒ a + b ≥ 4

(x - 1)(x² - 5x + 7) < (x - 1)

Bring all terms to one side: (x - 1)(x² - 5x + 7) - (x - 1) < 0

Factor out the common term (x - 1): (x - 1)(x² - 5x + 6) < 0

Factor the quadratic: (x - 1)(x - 2)(x - 3) < 0

Critical points are x = 1, 2, 3. Test intervals:

For x < 1: signs (−)(−)(−) → product < 0, so (−∞, 1) satisfies.

For 1 < x < 2: signs (+)(−)(−) → product > 0, not included.

For 2 < x < 3: signs (+)(+)(−) → product < 0, so (2, 3) satisfies.

For x > 3: signs (+)(+)(+) → product > 0, not included.

Therefore the solution set is (−∞, 1) ∪ (2, 3), option C.

Since the particle is moving in a straight line under a retardation kv³, hence, we have,
dv/dt = -kv³   ……….(1)
Or dv/v³ = -k dt
Or ∫v^-3 dv = -k∫dt
Or v^-3+1/(-3 + 1) = -kt – c [c = constant of integration]
Or 1/2v² = kt + c   ……….(2)
Given, u = v when, t = 0; hence, from (2) we get,
1/2u² = c
Thus, putting c = 1/2u² in (2) we get,
1/2v² = kt + 1/2u²
Or 1/v² = 1/u² + 2kt

Let v be the velocity of the particle at time t seconds after start (that is at a distance s from O). Then,
v = ds/dt = d[(t – 1)²(t – 2)²]/dt
Or v = (t – 2)(3t – 4)
Clearly, v = 0, when (t – 2)(3t – 4) = 0
That is, when t = 2
Or 3t – 4 = 0 i.e., t = 4/3
Now, s = (t – 1)(t – 2)²
Therefore, when t = 4/3, then s = (4/3 – 1)(4/3 – 2)² = 4/27
And when t = 2, then s =(2 – 1)(2 – 2)² = 0
Therefore, the velocity of the particle is zero, when its distance from O is 4/27 units and when it is at O.