Mathematics: CUET Mock Test - 2 - CUET MCQ

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.

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

Split the integral

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

Second term: 

Combine:

So, Option (c) is the correct answer.

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

The Poisson distribution, Statement 1 is correct because it accurately reflects the nature of Poisson events, which can occur multiple times within a given timeframe. Statement 2 is incorrect as the defining characteristic of a Poisson process is that events occur independently; thus, the occurrence of one event does not affect the probability of another event occurring. Therefore, the correct answer is that only Statement 1 is correct.

We assume that the particle moves with uniform acceleration 2f m/sec².
Let, x m be the distance of the particle from a fixed point on the straight line at time t seconds.
Let, v be the velocity of the particle at time t seconds, then,
So, dv/dt = 2f
Or ∫dv = ∫2f dt
Or v = 2ft + b   ……….(1)
Or dx/dt = 2ft + b
Or ∫dx = 2f∫tdt + ∫b dt
Or x = ft² + bt + a   ……….(2)
Where, a and b are constants of integration.
Given, x = 21, when t = 2; x = 43, when t = 4 and x = 91, when t = 7.
Putting these values in (2) we get,
4f + 2b + a = 21   ……….(3)
16f + 4b + a = 43   ……….(4)
49f + 7b + a = 91  ……….(5)
Solving (3), (4) and (5) we get,
a = 7, b = 5 and f = 1
Therefore, from (2) we get,
x = t² + 5t + 7
Putting t = 3, f = 1 and b = 5 in (1),
We get, the velocity of the particle in 3 seconds,
= [v]_{t = 3} = (2*1*3 + 5)m/sec = 11m/sec.

We have, t = ax² + bx + c  ……….(1)
Differentiating both sides of (1) with respect to x we get,
dt/dx = d(ax² + bx + c)/dx = 2ax + b
Thus, v = velocity of the particle at time t
= dx/dt = 1/(dt/dx) = 1/(2ax + b) = (2ax + b)^-1  ……….(2)
Initially, when t = 0 and v = u, let x = x₀; hence, from (1) we get,
ax₀² + bx₀ + c = 0
Or ax₀² + bx₀ = -c   ……….(3)
And from (2) we get, u = 1/(2ax₀ + b)
Thus, 1/v² – 1/u² = (2ax + b)² – (2ax₀ + b)²
= 4a²x² + 4abx – 4a²x₀² – 4abx₀
= 4a²x² + 4abx – 4a(ax₀² – bx₀)
= 4a²x² + 4abx – 4a(-c)   [using (3)]
= 4a(ax² + bx + c)
Or 1/v² – 1/u² = 4at

Assume that the velocity of the particle at time t second is vcm/sec.
Then, v = dx/dt = 4t³/12 – 6t²/3 + 6t/2 + 1
So, v = dx/dt = t³/3 – 2t²/ + 3t + 1
Thus, dv/dt = t² – 4t + 3
And d²v/dt² = 2t – 4
For maximum and minimum value of v we have,
dv/dt = 0
Or t² – 4t + 3 = 0
Or (t – 1)(t – 3) = 0
Thus, t – 1 = 0 i.e., t = 1 Or t – 3 = 0 i.e., t = 3
Now, [d²v/dt²]_{t = 3} = 2*3 – 4 = 2 > 0
Thus, v is minimum at t = 3.
Putting t = 3 in (1) we get,
3³/3 – 2(3)²+ 3(3) + 1
= 1 cm/sec.

We have, x²/a + y²/b = 1 ……….(1)
and
x²/c + y²/d = 1 ……….(2)
Let, us assume curves (1) and (2) intersect at (x₁, y₁). Then
x₁²/a + y₁²/b = 1 ……….(3)
and
x₁²/c + y₁²/d = 1 ……….(4)
Differentiating both side of (1) and (2) with respect to x we get,
2x/a + 2y/b(dy/dx) = 0
Or dy/dx = -xb/ya
Let, m₁ and m₂ be the slopes of the tangents to the curves (1) and (2) respectively at the point (x₁, y₁); then,
m₁ = [dy/dx]_{(x1, y1)} = -(bx₁/ay₁) and m₂ = [dy/dx]_{(x1, y1)} = -(dx₁/cy₁)
By question as the curves (1) and (2) intersects at right angle, so, m₁m₂ = -1
Or -(bx₁/ay₁)*-(dx₁/cy₁) = -1
Or bdx₁² = -acy₁² ……….(5)
Now, (3) – (4) gives,
bdx₁²(c – a) = acy₁²(d – b) ……….(6)
Dividing (6) by (5) we get,
c – a = d – b
Or a – b = c – d.

The above is a condition for a symmetric relation.
For example, a relation R on set A = {1,2,3,4} is given by R={(a,b):a+b=3, a>0, b>0}
1+2 = 3, 1>0 and 2>0 which implies (1,2) ∈ R.
Similarly, 2+1 = 3, 2>0 and 1>0 which implies (2,1)∈R. Therefore both (1, 2) and (2, 1) are converse of each other and is a part of the relation. Hence, they are symmetric.

Poisson distribution shows the number of times an event is likely to occur within a specified time. The Poisson distribution probability formula is P(x; μ) = (e^-μ) (μ^x) / x!

The equation is cos^-1 x + cos^-1 y + cos^-1 z = 3π
This means cos^-1 x = π, cos^-1 y = π and cos^-1 z = π
This will be only possible when it is in maxima.
As, cos^-1 x = π so, x = cos π = -1 similarly, y = z = -1
Therefore, x + y + z = -1 -1 -1
So, x + y + z = -3.

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

• (A) Random Variable X: A random variable is a variable whose value is determined by the outcome of a random event. So it matches with (I).
• (B) Continuous Random Variable: A continuous random variable can take any value within a given interval, matching with (II).
• (C) Discrete Random Variable: A discrete random variable takes countable outcomes, which fits with (III).
• (D) Probability Mass Function (PMF): A PMF assigns probabilities to each value of a discrete random variable, so it matches with (IV).

• (A) Unit Vector → (I) A vector with a magnitude of 1, often used to specify direction.
• (B) Position Vector → (II) A vector representing a point’s position in space with respect to the origin.
• (C) Zero Vector → (III) A vector with zero magnitude and no particular direction.
• (D) Collinear Vectors → (IV) Vectors that are parallel or lie along the same straight line.

Thus, the correct answer is A (A) - (I), (B) - (II), (C) - (III), (D) - (IV).

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,

We have :

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.

Here, 0 ≤ x- 3 ≤ 7 - x  
⇒0 ≤ x - 3 and x - 3 ≤ 7 - x
By solvation, we will get 3 ≤ x ≤ 5
So x = 3,4,5 find the values of ^7-x P_{x - 3} by substituting the values of x
at x = 3 ⁴P₀ = 1
at x = 4 ³P₁ = 3 
at x = 5 ²P₂ = 2

Apply C₂ → C₂ + C_3,