Mathematics: CUET Mock Test - 7 - CUET MCQ

Calculation:
Given, cable network provider in a small town has 500 subscribers nd he used to collect Rs. 300 per month from each subscriber.
Now, for every increase of Rs. 1, one subscriber will discontinue the service.
⇒ If Rs. x is the monthly increase in subscription amount, then x subscribers will discontinue the service.
⇒ Number of subscribers remaining = 500 - x
∴ If Rs. x is the monthly increase in subscription amount, then the number of subscribers are (500 - x)

Calculation:

Let the increase in monthly charge is x
Total Revenue, R = (500 - x)(300 + x)
⇒ R = 150000 + 200x - x²
For maximum revenue, dR/dx = 0 and d²R/dx² < 0
⇒ 0 + 200 - 2x = 0
⇒ x = 100
Also, d²R/dx² = -2 < 0
⇒ x = 100 is a maxima.
∴ Number of subscribers = 500 - x = 500 - 100 = 400
∴ The number of subscribers which gives the maximum revenue is 400.

Concept:

If the coefficient of dx is the only function of x and coefficient of dy is only a function of y in the given differential equation then we can separate both dx and dy terms and integrate both separately.

Calculation:

Given: ydx - xdy = 0

xdy = ydx

Integrating both sides, we get

Since ln x + ln y = ln (xy) will be:

⇒ y = cx

Solution of the differential equation represents straight line passing through origin.

CONCEPT:

Properties of Scalar Triple Product

• [a b c] = [b c a] = [c a b]
• [a b c] = - [b a c] = - [c b a] = - [a c b]
• [(a + b) c d] = [a c d] + [b c d]
• [λa, b c] = λ [a b c]
• Three non-zero vectors are coplanar if and only if [a b c] = 0

CALCULATION:

Given: are coplanar i.e

⇒ and

⇒

As we know that, [a b c] = [b c a] = [c a b]

⇒

As we know that, [λa, b c] = λ [a b c]

⇒

As we know that, vectors are coplanar if and only if [a b c] = 0

⇒

⇒

Hence, correct option is 3.

We want to find the indefinite integral of |x| with respect to x:

∫ |x| dx

1. Consider x ≥ 0:

   • Here, |x| = x.
   • So, ∫ |x| dx = ∫ x dx = (x²)/2 + C₁

2. Consider x < 0:

   • Here, |x| = -x.
   • So, ∫ |x| dx = ∫ (-x) dx = - (x²)/2 + C₂

To combine these results into a single expression, we note that for an indefinite integral, constants C₁ and C₂ can be absorbed into a single arbitrary constant C. A concise way to write this is:

∫ |x| dx = (x · |x|)/2 + C

• For x ≥ 0, this becomes x·x/2 = x²/2.
• For x < 0, this becomes x·(-x)/2 = -x²/2.
• Combining both the results , we get x|x|/2 + C

Firstly we will calculate the value of k
As we know , 

It is known that the sum of probabilities of a probability distribution of random variables is one.
∴ 0 + k + 2k + 3k + k² + 2k² + (7k² + k) = 1
⇒ 10k² + 9k - 1 = 0
⇒ (10k - 1)(k + 1) = 0
⇒ k = -1, 1/10
k = -1 is not possible as the probability of an event is never negative.
∴ k = 1/10

(ii) P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)
= 0 + k + 2k
= 3k
= 3 × 1/10
= 3/10

Constraints:

2x₁ + 3x₂ ≤ 12 (0, 4)(6, 0)

x₁ ≤ 4 (x₁ = 4)

z(0,0) = 0

z(4,0) = 200

If 2 tan^-1 (cos x) = tan ^-1(2 cosec x),

2tan^-1(cos x) = tan^-1 (2 cosec x)

= tan^-1(2 cosec x) 

= cot x cosec x = cosec x = x = π/4

tan^-1(-2) + tan^-1(-3)

In vector form Distance between two parallel lines  given by :

Given,

x₁ + x₂ ≤ 4; (0, 4)(4, 0)

x₁ + 2x₂ ≥ 6; (0, 3)(6, 0)

only one point:

Z(_0,0) = 0

Z(_0,3) = 18

Z(_2,2) = 20

Z(_4,0) = 16

Simplex method can handle only ≤ type constraints so infeasible solution is not possible.

The area of the triangle with the given point as vertices is,


Now, by performing the row operation R₂ = R₂ – R₁ and R₃ = R₃ – R₂

Now, breaking the determinant we get,
= 1/2 (2m + 2 – 2m – 4)
= -1
Thus, it is independent of m.

The area of the triangle (0,2), (0, 0), (3, 0) with vertices is given by

Expanding along R₃, we get
Δ= (1/2)  {0-0+3(2-0)}
Δ=3 sq.units.

• (A) Chain Rule → (I) A method used to differentiate composite functions, applying derivatives to outer and inner functions.
• (B) Derivative of log(x) → (II) The derivative of the natural logarithmic function, log(x), is 1/x.
• (C) Rolle's Theorem → (IV) If a function is continuous on a closed interval and differentiable on an open interval, there exists at least one point where the derivative is zero.
• (D) Second-Order Derivative → (III) The second derivative provides information on the concavity or curvature of the function.

Thus, the correct answer is (1) (A) - (I), (B) - (II), (C) - (IV), (D) - (III).

• (A) ∫0 to 2 (x + 5) dx: The integral of (x + 5) from 0 to 2 gives 12 (which is (II)).
• (B) ∫0 to 3 (2x + 1) dx: The integral of (2x + 1) from 0 to 3 gives 14 (which is (I)).
• (C) ∫0 to 1 (x² + x) dx: The integral of (x² + x) from 0 to 1 gives 5/3 (which is (III)).
• (D) ∫0 to 1 (x³ + 1) dx: The integral of (x³ + 1) from 0 to 1 gives 2 (which is (IV)).

Given that,

We can observe that both the matrices above are circular matrices.

So, by circular determinant property,
Sum of the elements of a row = 0
So, x + 3 + 6 = 2 + x + 7 == 0
⇒ x = -9

The element 3 is in the second row (i=2) and first column(j=1).
∴ M₂₁=5 (obtained by deleting R₂ and C₁ in Δ)
A₂₁=(-1)¹⁺² M₂₁=-1×5 =-5.

There are 2 curves.

The black curve is the graph of y = cotx
The red curve is the graph for y = cot^-1x
This curve does not pass through the origin but approaches to infinity in the direction of x axis only.
The part of the curve that lies in the (x, y) coordinate gradually meets to the x-axis.
This graph lies above +x axis and –x axis.

The elements following the condition i=j will have the same row number and column number. The elements are a₁₁, a₂₂, a₃₃ which in the matrix A are 2, 9 and 7 respectively.

P (A⋃B) = P (A) + P (B) – P (A⋂B)

= 4/6
= 2/3
We have, P(A).P(B) = P(A⋂B), for independent events.
P(A).P(B) = (2/3) × (1/2) = 1/3
This is equal to P(A⋂B).
Thus events A and B are independent events.
[Note that, for mutually exclusive events, P (A⋂B) = 0. Also, for mutually exhaustive events, P (A⋃B) = 1. Both of these conditions are not true here.]

Given:
A, B, and C are three mutually exclusive and exhaustive events such that if P(B) = 3/2 P(A) and P(C) = 1/2 P(B).

Concept:
If A, B, and C are three mutually exclusive and exhaustive events then,
P (A U B U C) = P(A) + P(B) + P(C) = 1

Explanation:
According to the question,
A, B, and C are three mutually exclusive and exhaustive events such that if P(B) = 3/2 P(A) and P(C) = 1/2 P(B).

Also,
P (A U B U C) = 1

Hence, option 4 is correct.

• (A) Cross Product of Two Vectors → (III) Produces a vector perpendicular to both given vectors, calculated using:  A × B = |A| |B| sin(θ) n
• (B) Dot Product of Two Vectors → (IV) Produces a scalar value measuring alignment between two vectors, given by:  A · B = |A| |B| cos(θ)
• (C) Scalar Projection → (I) A scalar value representing the component of one vector along another, given by: (A · B) / |B|
• (D) Vector Projection → (II) A vector obtained by projecting one vector onto another, calculated as: ((A · B) / |B|²) B

A relation R on a non empty set A is said to be transitive if fx Ry and yRz ⇒ x Rz, for all x ∈ R.