Mathematics: CUET Mock Test - 1 - CUET MCQ

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

A binary operation ‘*’ defined on a set A is said to be commutative only if a * b = b *a, ∀ a, b ∈ A.
If (a * b) * c = a * (b * c), then the operation is said to associative ∀ a, b∈ A.
If (b ο c) * a = (b * a) ο (c * a), then the operation is said to be distributive ∀ a, b, c ∈ A.

tan⁻¹ √3 = π/3, sec⁻¹2 = π/3, cos⁻¹1 = 0
 tan⁻¹√3 + sec⁻¹2 – cos−¹1 = π/3 + π/3
= 2π/3

Let sin^-1⁡x = y
⇒ x = sin⁡y
⇒ x = √1 - cos²y
⇒ x² = 1 - cos²y
⇒ cos²y = 1 - x²
∴ y = cos^-1⁡ √1 - x² = sin^-1⁡x

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.

Given that, g(x) = 3 x 2 + 7 and f(x) = √x
∴ gοf(x) = g(f(x)) = g(√x) = 3(√x)2 + 7 = 3x + 7.
Hence, gοf(x) = 3x + 7.

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.

Let  = y
sec y = 2/√3
sec⁡ y = secπ/6
⇒ y = π/6

Let sec^-1⁡x = y
⇒ x = sec⁡y
⇒ x = √ 1 + tan²y
⇒ x² - 1 = tan²y
∴ y = tan^-1√x² - 1 = sec^-1⁡x

Given that, A =   and B =
Then A + B = 
= 

Given that f(x) = (5 + x⁴)^1/4
∴ fοf(x) = f(f(x)) = (5 + {(5 + x⁴)^1/4}⁴)^1/4
= (5 + (5 + x⁴))^1/4 = (10+x⁴)^1/4

The binary operation is defined by a * b = a - b + ab².
∴ 4 * 5 = 4 - 5 + 4(5²) = -1 + 100 = 99.

[-1, 1] is the domain for sin^-1⁡x.
The domain for cot^-1⁡x is (-∞,∞).
The domain for tan^-1⁡⁡x is (-∞,∞).
The domain for sec^-1⁡⁡x is (-∞,-1] ∪ [1,∞).

Given that,
A + B =  and A = 
⇒ B = (A + B) - A = 
B = 

A function is invertible if and only if it is bijective i.e. the function is both injective and surjective. If a function f: A → B is bijective, then there exists a function g: B → A such that f(x) = y ⇔ g(y) = x, then g is called the inverse of the function.

The function f = {(7,7),(8,8),(9,9)} is invertible as it is both one – one and onto. The function is one – one as every element in the domain has a distinct image in the co – domain. The function is onto because every element in the codomain M = {7,8,9} has a pre – image in the domain.

(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

Given that, f(x) = 3x - 5, g(y) = 6y² and h(z) = tan⁡z,
Then, ho(gof) = hο(g(f(x)) = h(6(3x-5)²) = tan⁡(6(3x - 5)²)
∴ ho(gof) = tan⁡(6(3x - 5)²)

The binary operation ‘*’ is both commutative and associative for a * b = a + b.
The operation is commutative on a * b = a + b because a + b = b + a.
The operation is associative on a * b = a + b because (a + b) + c = a + (b + c).

Given that, a * b = 4ab.
Then, (a * b) * a = (4ab) * a
= 4(4ab)(a) = 16a² b.

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.

It is given that a * b=3a^b + 5.
Then, 8 * 3 = 3(8³) + 5 = 3(512) + 5 = 1536 + 5 = 1541.

Transitive is not a type of binary operation. It is a type of relation. Distributive, associative, commutative are different types of binary operations.

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.

Matrix multiplication is never commutative i.e. AB ≠ BA. Therefore, the condition AB = BA is incorrect.

 (AB)’ = (BA)’is incorrect. We know that matrix multiplication is not commutative i.e. AB ≠ BA. Hence, its transpose will also not be commutative.
(AB)’=B’A’

To find the transpose of the matrix of the given matrix, interchange the rows with columns and columns with rows.
Hence, we get A’ = 

According to the reverse law of transposes the transpose of the product is the product of the transposes taken in the reverse order i.e. (AB)’ = B’ A’.

Given that, A = 
⇒ A’ = 
i.e.A=A’. Hence, it is a symmetric matrix.

A matrix is said to be skew-symmetric if it is equal to the negative of its transpose i.e. A = -A’.