Mathematics: CUET Mock Test - 7 - Commerce 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.

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.

Given that, A= 
|A|=(7(5)-5(4))=35-20=15
|A|²-5|A|+1=(15)²-5(15)+1=225-75+1=151.

Given that,

So, by circular determinant property,
Sum of the elements of a row = 0
So, x + 3 + 6 = 2 + x + 7 = 4 + 5 + x = 0
⇒ x = -9

We have, 
So, the value of the  (a – b)(b – c)(c – a)
Now, by circulant determinant,

Multiplying the determinant in row by row,
We get, (a – b)²(b – c)²(c – a)²

The minor of the element 2 can be obtained by deleting the first row and the first column
∴M₁₁=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.

Δ=sin⁡ y 
Δ=sin ⁡y (sin⁡ y-cos⁡ x)-0+sin ⁡y (cos⁡ y-sin ⁡y)
Δ=sin²⁡y-sin ⁡y cos⁡ x+sin⁡ y cos ⁡y-sin^2⁡y=sin ⁡y (cos⁡ y-cos⁡ x)

When a = b or b = c or c = a the determinant reduces to 0
It is not necessary that a = b = c for determinant to be 0
Therefore, the triangle is isosceles.

The given system have non-trivial solution if 
On opening the determinant we get β = sin 2α + cos 2 α
Therefore, -√2 ≤ β ≤ √2
Now, for β = 1,
sin 2α + cos 2 α = 1
⇒(1/√2)sin 2α + (1/√2) cos 2α = (1/√2)
Or, cos(2α – π/4) = 1/√2 = cos(2nπ ± π/4)
⇒ 2α = 2nπ ± π/4 + π/4

The area of triangle formed by collinear points is zero.

Expanding along C₂, we get
1/2 {-2(3-2)+0-k(1-3)}=0
1/2 {-2+2k}=0
∴ k=1

The minor of the element 1 can be obtained by deleting the first row and the first column
∴ M₁₁=8.