Mathematics: CUET Mock Test - 6

Surjective is not a type of relation. It is a type of function. Reflexive, Symmetric and Transitive are type of relations.

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^-1 π = -1 similarly, y = z = -1
Therefore, x + y + z = -1 -1 -1
So, x + y + z = -3.

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.

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(2π + 6 π/7) = sin(π/7)
= sin(3π – 6 π/7) = sin(20π/7)
= sin(-π – 6 π/7) = sin(-15π/7)
= sin(-2π + 6 π/7) = sin(-8π/7)
= sin(-3π – 6 π/7) = sin(-27π/7)
Therefore, sin^-1sin(6 π/7) = sin^-1(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.

Evaluating along R₁, we get
∆ = 5(√3)-(-4)1 = 5√3+4.

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.

Let, θ = sin^-1(-x)
So, -π/2 ≤ θ ≤ π/2
⇒ -x = sinθ
⇒ x = -sinθ
⇒ x = sin(-θ)
Also, -π/2 ≤ -θ ≤ π/2
⇒ -θ = sin^-1(x)
⇒ θ = -sin^-1(x)
So, sin^-1(-x) = -sin^-1(x)

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.

We know that sin(x) = sin(2A * π + x) where A can be positive or negative integer.
If A is -1, then sin(6) = sin(-2π + 6);
If A is 1, then sin(6) = sin(2π + 6);

The given determinant is f(r) =  
Now, _{r = 1}Σⁿ (2r) = 2[(n(n + 1))/2] ……….(1)
= n² + n
_{r = 1}Σⁿ(6r² – 1) = 6[((n + 1)(2n + 1))/6] – n ……….(2)
= n(2n² + 2n + n + 1) – n
= 2n³ + 2n² + n² + n – n
= 2n³ + 3n²
= _{r = 1}Σⁿ(4r³ – 2nr) = n³ (n + 1) ……….(3)
From (1), (2) and (3) we get
_{r = 1}Σⁿ f(x) = 0

The matrix in which the number of columns is greater than the number of rows is called a vertical matrix. There the matrix which follows the condition m>n is a vertical matrix.

Expanding along R₁, we get
∆=-i(i)-(-1)(-1)=-i²-1=-(-1)-1=0.

(2, 3) ∈ R as 2 + 3 = 5, 3 >1, thus satisfying the given condition.
(4, 2) doesn’t belong to R as 4 + 2 ≠ 5.
(2,1) doesn’t belong to R as 2+ 1 ≠ 5.
(5, 0) doesn’tbelong to R as 0 ⊁ 1
 

Let, θ = cos^-1(-x)
So, 0 ≤ θ ≤ π
⇒ -x = cosθ
⇒ x = -cosθ
⇒ x = cos(-θ)
Also, -π ≤ -θ ≤ 0
So, 0 ≤ π -θ ≤ π
⇒ -θ = cos^-1(x)
⇒ θ = -cos^-1(x)
So, cos^-1(x) = π – θ
θ = π – cos^-1(x)
⇒ cos^-1(-x) = π – cos^-1(x)

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 two matrices  and  are equal matrices. Comparing the two matrices, we get a-b=3, c=2, a-b=1, 2c+d=6
Solving the above equations, we get a=2, b=1, c=2, d=2.

∆= 
Expanding along the first row, we get

=1(4-5(2))-1(3-5(-1))-2(6-4(-1))
=(4-10)-(3+5)-2(6+4)
=-6-8-20=-34.

The relation R= {(7, 7), (8, 8), (9, 9)} is reflexive as every element is related to itself i.e. (a, a) ∈ R, for every a∈A. and it is not transitive as it does not satisfy the condition that for a 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 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.

The possible orders in which the matrix with 6 elements can be formed are 2×3, 3×2, 1×6, 6×1. Therefore, the possible orders pairs are (2,3), (3,2), (1,6), (6,1). Thus, (3,1) is not possible.

The matrix is said to be a diagonal matrix if the elements along the diagonal of the matrix are non – zero.
i.e. a_ij=0 for i≠j and a_ij≠0 for i=j.
Therefore, the matrix A=  is a diagonal matrix.

Expanding along the first row, we get

=5(4-6)-4(3-5)+3(18-20)
=5(-2)-4(-2)+3(-2)=-10+8-6=-8.

This is an equivalence relation. A relation R is said to be an equivalence relation when it is reflexive, transitive and symmetric.
Reflexive: We know that a line is always parallel to itself. This implies that I₁ is parallel to I₁ i.e. (I₁, I₂)∈R. Hence, it is a reflexive relation.
Symmetric: Now if a line I₁ || I₂ then the line I₂ || I₁. Therefore, (I₁, I₂)∈R implies that (I₂, I₁)∈R. Hence, it is a symmetric relation.
Transitive: If two lines (I₁, I₃) are parallel to a third line (I₂) then they will be parallel to each other i.e. if (I₁, I₂) ∈R and (I₂, I₃) ∈R implies that (I₁, I₃) ∈R.

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 3×4 matrix as it as 3 rows and 4 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)

Expanding along the first row, we get
∆=8x+1(3x+5)-(2x-2)(x²-1)
=(24x²+43x+5)-(2x³-2x²-2x+2)
=-2x³+26x²+45x+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.

a₃₂ is the element represented in the form a_ij i is the row number and j is the column number. Therefore, the element a₃₂ is the element in the third row (i=3)and second column (j=2) which is 8.

A matrix is called a scalar matrix if the elements along the diagonal of the matrix are equal and are non-zero i.e. a_ij=k for i=j and a_ij=0 for i≠j.
Therefore, the matrix A=  is a scalar matrix.

Given that, A= 

Evaluating along the first row, we get

∆=2(2-24)-5(12-12)+9(48-4)
∆=2(-22)-0+9(44)
∆=-44+9(44)=44(-1+9)=352