Mathematics: CUET Mock Test - 6 - CUET MCQ

Concept:

Conditional probability:

It gives the probability of happening of any event if the other has already occurred.
 Probability o fgetting the event E₁ when E₂ is already occured.

Calculation:
Given:
P (E₁) = Probability of stopping at the gas station and ask for tyre checked = 0.12
P (E₂) = Probability of stopping at the gas station and ask for oil checked = 0.29
P (E₁∩ E₂) = Probability of both checked = 0.07
 = Probability of person who has his tyre checked will also have oil checked

Given -

Given differential equation is

Concept -

Order : The order of a differential equation is the highest power of the derivative present in the equation.

Explanation -

The highest derivative in the given differential equation is second order. Hence the order = 2

Hence option (ii) is correct.

The number of elements for a matrix with the order m×n is equal to mn, where m is the number of rows and n is the number of columns in the matrix.

A square matrix is a matrix in which the number of rows(m) is equal to the number of columns(n). Therefore, the matrix which follows the condition m = n is a square matrix.

Expanding along R₁, we get
∆=2(-1)-5(-1)=-2+5
= 3.

• (A) Expected Value of Discrete Distribution: The expected value of a discrete distribution is calculated by Σ(x * P(x)) (which is (I)).
• (B) Variance of Discrete Distribution: The variance of a discrete distribution is calculated by Σ((x - μ)² * P(x)) (which is (II)).
• (C) Standard Deviation of Discrete Distribution: The standard deviation is the square root of the variance (which is (III)).
• (D) Probability of an Event in Discrete Distribution: This is simply the probability associated with each outcome (which is (IV)).
   

y = tan^-1 x
=> tan y = tan(tan^-1 x)
x = tan y
sec² y dy/dx = 1
=> 1/(cos²y) dy/dx = 1
dy/dx = cos²y
Differentiate it with respect to x
d²y/dx² = 2cos y(-sin y)(dy/dx)..................(1)
Put the value of dy/dx in eq(1)
=> d²y/dx² = 2cos y(-sin y)(cos² y)
=> d²y/dx² = 2cos³ y siny

x^a y^b = (x−y)^a+b
taking log both the sides.
log(x^a y^b)=log(x−y)^(a+b)
alogx+blogy=(a+b)log(x−y)
differentiating both sides w.r.t 'x'.
(a/x+b/y)dy/dx = (a+b)/(x−y)[1−dy/dx]
dy/dx[b/y+(a+b)/(x−y)] = (a+b)/(x−y) − a/x
dy/dx[(bx−by+ay+by)/y(x−y)] = (ax+bx−ax+ay)/x(x−y)
dy/dx[(bx+ay)/y]=(bx+ay)/x
dy/dx = y/x

√2 - 1 + 2√3 - 2√2 + 6 - 3√3
5 - √2 - √3

CONCEPT:

• If , and , then .
• If vectors are coplanar then

​CALCULATION:

Given: The vectors are coplanar.

As we know that, if vectors are coplanar then

⇒

⇒ 2(0 - 4p) -(-1)(-15 - 0) + 1(12 - 0) = 0

⇒ -8p - 15 + 12 = 0

⇒ -8p - 3 =0

⇒ -8p = 3

⇒ p = -3/8

Hence, correct option is 3.

Determinant of the matrix A= is not possible as it is a rectangular matrix and not a square matrix. Determinants can be calculated only if the matrix is a square matrix.

sin^-1sin(6 π/7)
Now, sin(6 π/7) = sin(π – 6 π/7)
= sin(π/7)
Therefore, sin^-1sin(6 π/7) = sin^-1sin(π/7) = π/7

a₁₁=1+1=2, a₁₂=1+2=3, a₂₁=2+1=3, a₂₂=2+2=4
∴ 

The matrix in which number of rows is smaller than the number of columns is called is called a horizontal matrix. In the given matrix A=  m = 3 and n = 2 i.e.
3<2. Hence, it is a horizontal matrix.

A relation in a set A is said to be symmetric if (a₁, a₂)∈R implies that (a₁, a₂)∈R,for every a₁, a₂∈R.
Hence, for the given set A={1, 2, 3}, R={(1, 2), (2, 1)} is symmetric. It is not reflexive since every element is not related to itself and neither transitive as it does not satisfy the condition that for a given relation R in a set A if (a₁, a2)∈R and (a₂, a₃)∈R implies that (a₁, a₃)∈ R for every a₁, a₂, a₃∈R.

The number of rows (m) and the number of columns (n) in the given matrix A=   is 2. Therefore, the order of the matrix is 2×2(m×n).

The given matrix A =  is of the order 3×1. The matrix has only one column (n=1). Hence, it is a column matrix.

Expanding along R₁, we get
∆=-sinθ(sinθ)-(-1)1=-sin²⁡θ+1=cos^2⁡θ.

For the above given set S = {3, 4, 6}, R = {(3, 4), (4, 6), (3, 6)} is transitive as (3, 4)∈R and (4, 6) ∈R and (3,6) also belongs to R . It is not a reflexive relation as it does not satisfy the condition (a, a) ∈ R, for every a ∈ A for a relation R in the set A.

Solving the given system of equation by Cramer’s rule, we get,
x = D₁/D, y = D₂/D, z = D₃/D where,


Now, performing, C₃ = C₃ – C₂ and C₂ = C₂ – C₁ we get,

As two columns have identical values, so,
D = 0
Similarly,

Now, performing, C₁ = C₁ – C₃

Now, performing, C₃ = C₃ – C₂

As two columns have identical values, so,
D₁ = 0

Now, performing,

Now, performing, C₂ = C₂ – C₃ and C₃ = C₃ – C₁

As two columns have identical values, so,
D₂ = 0


Now, performing, C₂ = C₂ – C₂ and C₃ = C₃ – C₂

As two columns have identical values, so,
D₃ = 0
Since, D = D₁ = D₂ = D₃ = 0, thus, it has infinitely many solutions.

The following graph represents 2 equations.

The pink curve is the graph of y = sinx
The blue curve is the graph for y = sin^-1x
This curve passes through the origin and approaches to infinity in both positive and negative axes.

There are 2 curves.

The green curve is the graph of y = cosx
The red curve is the graph for y = cos^-1x
This curve origin from some point before π/3 and approaches to infinity in both positive y axis by intersecting at a point near 1.5 in y axis.

The matrix A=  is a 4×3 matrix as it as 4 rows and 3 columns.

Given that the vertices are (1,2), (3,5), (k,0)
Therefore, the area of the triangle with vertices (1,2), (3,5), (k,0) is given by

Expanding along R₃, we get
(1/2){k(2-5)-0+1(5-6)}=(1/2){-3k-1}=(7/2)
⇒ -(1/2)(3k+1)=7/2
3k=-8
k = -(8/3)

R= {(4, 5), (5, 4), (4, 4)} is symmetric since (4, 5) and (5, 4) are converse of each other thus satisfying the condition for a symmetric relation and it is transitive as (4, 5)∈R and (5, 4)∈R implies that (4, 4) ∈R. It is not reflexive as every element in the set I is not related to itself.

There are 2 curves.

The blue curve is the graph of y = tanx
The red curve is the graph for y = tan^-1x
This curve passes through the origin and approaches to infinity in the direction of x axis only.
This graph lies below –x axis and above +x axis.

dy/dx = (x² + 3y²)/2xy………….(1)
Let y = vx
dy/dx = v + xdv/dx
Substitute the value of y and dy/dx in (1)
v + x dv/dx = (1+3v²)/2v
x dv/dx = (1+3v²)/2v - v
x dv/dx = (1 + 3v² - 2v²)/2
x dv/dx = (1+ v²)/2v
2v/(1+v²) dv = dx/x…………(2)
Integrating both the sides
∫2v/(1+v²) dv = ∫dx/x
Put t = 1 + v²
dt = 2vdv
∫dt/t  = ∫dx/x
=> log|t| = log|x| + log|c|
=> log|t/x| = log|c|
t/x = +- c
(1+v²)/x = +-c
(1 + (y²)/(x²))/x = +-c
x² + y² = Cx³……….(3)
y(1) = 3
1 + 9 = c(1)³
c = 10
From eq(3), we get x²+ y² = 10x³

Correct Answer: a) The solution may contain alternative optimal solutions

To determine if the linear programming problem has alternative optimal solutions, we check if any constraint is parallel to the objective function, which would result in multiple optimal solutions along the boundary. The constraint 4A + 6B ≤ 30 can be simplified to 2A + 3B ≤ 15, which has the same slope as the objective function Z = 2A + 3B. This means the objective function is parallel to this constraint, so the linear programming problem may have alternative (multiple) optimal solutions. Therefore, option a) The solution may contain alternative optimal solutions is correct.