Mathematics: CUET Mock Test - 7 - CUET MCQ

The above is the condition for a reflexive relation. A relation is said to be reflexive if every element in the set is related to itself.

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:
According to the question, 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 = 500 - x
⇒ Subscription amount = 300 + x
⇒ Total revenue, R = (500 - x)(300 + x)

Calculation:
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 100.

Calculation:
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.
∴ Increase in changes per subscriber that yields maximum revenue is 100.

Calculation:
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.
∴ Maximum revenue = R (at x = 100)
= (500 - 100)(300 + 100)
= 400 × 400
= Rs. 160000
∴ The maximum revenue generated is Rs. 160000.

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.

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.

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 given matrix A=  has 3 rows and 2 columns. Therefore, the order of the matrix is 3×2.

Minor matrix is not a type of matrix. Scalar, diagonal, symmetric are various type of matrices.

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.

The given form of equation can be written as,

The green curve is the graph of y = sinx
The blue curve is the graph for y = |sinx|
As sinx is enclosed by a modulus so the curve that lies in the negative y axis will come to the positive y 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, 3, 9 respectively.

The two matrices  and  are equal matrices. Comparing the two matrices, we get a=3, b+c=2, c+d=3, b=-1
Solving the above equations, we get a=3, b=-1, c=3, d=0.

Given that 
⇒3x²-2x=5(2)-3(3)
⇒3x²-2x=1
Solving for x, we get
x=1, –(1/3).

Expanding along R₁, we get


Δ=3(-15+8)+(18-12)+3(-12+15)
Δ=3(-7)+6+9=-6.

Applying R₁ → R₁ + R₂ + R₃

This is equal to,

Applying C₁ → C₁ – C₂ and C₂ → C₂ – C₃

= (Σab)³

Here, f(x) = 
Multiplying and diving by abc,


= (a – b)(b – c)(c – a)

The area of a triangle with vertices (x₁,y₁), (x₂,y₂), (x₃,y₃) is given by

The cofactor of an element a_ij, denoted by A_ij is given by
A_ij=(-1)^i+j M_ij, where M_ij is the minor of the element a_ij.

Δ=1(0-0)-0(0-1)+1(0-0)
Δ=0-0+0=0.

Here, in L.H.S we have,

So, for trivial roots the above value is = 0
⇒ 
Solving it further we get α = 12

Here, f(x) = f’(x)
⇒ f(x) is purely real.

The area of the triangle with vertices (0,1), (0,2), (1,5) is given by

Expanding along C₁, we get
Δ=1/2{(0-0+1(1-2)}=1/2|-1|=1/2 sq.units.

The minor of element 5 in the determinant Δ=  is the determinant obtained by deleting the row and column containing element 5.
∴ M₁₂=  =2(4)-7(6)=-34.