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.

Calculation :

7¹ = 7 ,Unit digit is 7

7² = 49 ,Unit digit is 9

7³ = 343 ,Unit digit is 3

7⁴ = 2401 ,Unit digit is 1

17¹²³ 123 when divide by 4 gives remainder 3

Hence, The unit digit of 17¹²³ = 17^{120 + 3 =} 17 ^{4 × 30 + 3} = 17³ and 7³ = 3

Concept used:
The expression "a = b mod n" means that the remainder obtained when b is divided by n is assigned to the variable a. In other words, "mod n" is the modulo operation, which calculates the remainder of the division of b by n.

For example, if we have b = 17 and n = 5, then 17 divided by 5 gives a quotient of 3 with a remainder of 2. Therefore, a = 2, because 17 mod 5 is equal to 2.

Calculation:
(3)³ = 27 ≡ 0 (mod 9) So, b ≡ 0 (mod 9)
(2)⁵ = 32 ≡ 2 (mod 15) So, b ≡ 2 (mod 15)
(4)³ = 64 ≡ 4 (mod 10) So, b ≡ 4 (mod 10)
(5)³ = 125 ≡ 5 (mod 12) So, b ≡ 5 (mod 12)
Hence, option 3 is correct.

Calculation:

Let's denote the original volume of milk in the mixture as M (in liters) and the original volume of water as W (in liters). From the problem, we know that:

1) M/W = 8/x.

2) The total volume of the mixture before adding water is M + W = 33 liters.

3) After adding 3 liters of water, the ratio of milk to water is 2:1, so M / (W + 3) = 2/1.

We can solve these equations to find the values of M, W, and x.

From equation (2) and (3), we get:

M = 2 × (W + 3) (since M/W after adding 3 liters of water is 2)

Substitute M = 33 - W from equation (2) into the above equation, we get:

33 - W = 2 × (W + 3)

33 - W = 2W + 6

33 - 6 = 2W + W

27 = 3W

W = 27 / 3 = 9 liters.

Substitute W = 9 into equation (1), we get:

M / 9 = 8 / x

M = (8/ x) × 9

Since M + W = 33, substitute M and W into the equation, we get:

(8/ x) × 9 + 9 = 33

(72 / x) + 9 = 33

72 / x = 33 - 9 = 24

By cross multiplying, we get:

x = 72 / 24

x = 3.

So, the value of x is 3.

Concept use:

speed of the boat while going upstream is the speed of the boat in still water minus the speed of the current.

Calculation:

When the motorboat goes upstream, it moves against the current, so the effective speed of the boat is reduced. The effective speed of the boat while going upstream is the speed of the boat in still water minus the speed of the current.

In this case, the speed of the boat in still water is 15 km/h, and the speed of the current is 3 km/h. So, the effective speed of the boat while going upstream is:

15 km/h - 3 km/h = 12 km/h

The time it takes to travel a certain distance is the distance divided by the speed. So, the time it takes for the boat to go 36 km upstream is:

36 km / 12 km/h = 3 hours

So, the boat takes 3 hours to go 36 km upstream.

Let A be Hary and B be Ram.

A cover the Distance of 100m in 36 sec and B cover the Distance of 100m in 45sec

Time Difference of A and B is 9 sec

Speed of B = Distance/Time = 100/45 m/s

Distance Covered by B in 9 sec = Speed × Times = 100/45 × 9 = 20 meters

So, A beats B by 20 m

Hence, Hari defeats Ram by 20 m.

Concept:

A function from a set X to a set Y assigns to each element of X exactly one element of Y. The set X is called the domain of the function and the set Y is called the codomain of the function.

Explanation:

Statement 1: The relation f defined by

is a function.

For 0, 1, 2 for the function x³ the values are 0, 1, 8

For 2, 3, 4, 5, 6, 7, 8 for the function 4x the values are

8, 12, 16, 20, 24, 28, 32

So, for x = 2, the function has the same value i.e., 8 (for x3 and 4x).

Hence f(x) is a function.

Statement 2: The relation g defined by

is a function.

For 0, 1, 2, 3, 4 for the function x2 the values are 0, 1, 4, 9, 16.

For 4, 5, 6, 7, 8 for the function 3x the values are 12, 15, 18, 21, 24.

So, for x = 4 the function has different values i.e., 16 (for x2) and

12 (for 3x).

Hence g(x) is a not function.

∴ Correct answer is option (1)

Concept:

n(A × B) = Number of elements in A × Number of elements in B

Calculation:

Given that

A = Finite set, B = Finite set

A number of elements in the Cartesian product (n(A × B)):

n(A × B) = Number of elements in A × Number of elements in B

Therefore, n(A × B) = n (A) × n (B)

Concept:

Codomain - A codomain is the group of possible values that the dependent variable can take.

Range - The range is all the elements from set B that have the corresponding pre-image in set A.

Calculation:

Given,

For f(x) to be defined, 3x + 5 ≠ 0

⇒ x =

∴ A =

Now, y = f(x) is onto

⇒ y =

⇒ 3xy + 5y = 2x + 3

⇒ 3xy - 2x = 3 - 5y

⇒ x(3y - 2) = 3 - 5y

⇒ x =

For x to be defined for

Since, for onto functions co-domain = range

∴ B =

∴

Concept:

F of G of x is a composite function made of two functions f(x) and g(x).

It is denoted by f(g(x)) or (f ∘ g)(x) and it means that x = g(x) should be substituted in f(x).

It is an operation that combines two functions to form another new function.

For finding f(g(x)), we have to first find g(x) and then take g(x) as input of f(x) and simplify.

Formula Used:

(a + b)² = a² + b² + 2ab

Calculation:

We have,

⇒ f(x) = x²

⇒ g(x) = x + 3

⇒ f(g(x))

⇒ f(x + 3)

⇒ (x + 3)²

⇒ x² + 9 + 6x

∴ Then the value of F(g) is x² + 6x + 9.

Calculation:
log32 ⋅ log43 ⋅ log54 ⋅ log65 ⋅ log76 ⋅ log87
⇒ (log32 ⋅ log43) (log54 ⋅ log65) (log76 ⋅ log87)
[logbM × logab = logaM]
⇒ log₄2 .log₆4. log₈6
⇒ (log₄2 .log₆4) log₈6
⇒ log62 ⋅ log₈6
⇒ log₈2
⇒ 1/log₂8
⇒ 1/log₂2³ = 1/3log₂2 =
∴ log32 ⋅ log43 ⋅ log54 ⋅ log65 ⋅ log76 ⋅ log87 = 1/3

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.