Mathematics Test - 1

Hence the option (1) is correct.

Concept:

Range of a Relation:

Let R be a relation from set A to set B. Then, the set of all second components of the ordered pair belonging to relation R forms the range of the relation R

i.e Range (R) = {b : (a, b) ∈ R}.

Calculation:

Given: A = {3, 9, 27, 81}, B = {1, 2, 3, 4, 5, 6} and R is a relation from A to B such that R = {(x, y) : x ∈ A, y ∈ B and y = log₃ x}

As we know that, logx x = 1 for x > 0

∵ R = {(x, y) : x ∈ A, y ∈ B and y = log₃ x}

When x = 3 ∈ A then y = log₃ 3 = 1 ∈ B ⇒ (3, 1) ∈ R

When x = 9 ∈ A then y = log₃ 9 = 2 ∈ B ⇒ (9, 2) ∈ R

When x = 27 ∈ A then y = log₃ 27 = 3 ∈ B ⇒ (27, 3) ∈ R

When x = 81 ∈ A then y = log₃ 81 = 4 ∈ B ⇒ (81, 4) ∈ R

So, the given relation R can be re-written in roaster form as: R = {(3, 1), (9, 2), (27, 3), (81, 4)}

As we know that, Range (R) = {b : (a, b) ∈ R}.

⇒ Range (R) = {1, 2, 3, 4}

Hence, the correct option is 1.

Concept:

Binomial Expansion:

• (a + b)ⁿ = C₀ aⁿ b⁰+ C₁ a^n-1 b¹ + C₂ a^n-2 b² + … + C_r a^n-r b^r + … + C_n-1 a¹ b_n-1 + C_n a⁰ bⁿ, where C₀, C₁, …,C_n are the Binomial Coefficients defined as C_r =  = 

Given equation of hyperbola

3x² - 3y² - 18x + 12y + 2 = 0

3x² - 18x - 3y² + 12y + 2 = 0

3(x² - 6x) - 3(y² - 4y) + 2 = 0

3(x² - 6x) - 3(y² - 4y) + 2 + 27 - 12 = 27 - 12

3(x² - 6x + 9) - 3(y² - 4y + 4) + 2 = 15

3(x - 3)² - 3(y - 2)² = 13

Concept:

• The cube root of unity has three roots, which are 1, ω, ω².
• Here the roots ω and ω² are imaginary roots and one root is a square of the other root. 
• The product of the imaginary roots of the cube root of unity is equal to 1 (ω.ω² = ω³ = 1), and
• The sum of the cube roots of unity is equal to zero.(1 + ω + ω² = 0).

Calculation:

Given, f(x) = (4x3 – 3x2)/6 – 2sin x + (2x – 1)cos x

∴ f '(x) = (2x² – x) – 2cos x + 2cos x – sin x(2x – 1)

= (2x – 1)(x – sin x)

Now, ∀ x ∈ (-∞, 0] ⋃ [1/2, ∞), f '(x) > 0

and, ∀ x ∈ (0, ½), f '(x) < 0

∴ f(x) increases in [1/2, ∞).

The correct answer is Option 1.

CONCEPT:

Symmetric Matrix:

Any real square matrix A = (aij) is said to be symmetric matrix if and only if aij = aji, ∀ i and j or in other words we can say that if A is a real square matrix such that A = At then A is said to be a symmetric matrix.

Calculation:

On comparing 

3 + x = 1 - x

⇒ x = - 1

And, y + 1 = 5 - y

⇒ y = 2

3x + y = 3(-1) + 2

∴ 3x + y = -1

Concept:

Rolle's theorem states that a function f(x) is continuous over the interval [a, b] and differentiable over the interval (a, b) such that f(a) = f(b), then there exists c ϵ (a, b) such that f'(c) = 0.

Calculation:

Given:  f(x) = x³ - 3x, x ∈ [0, √3] 

∴ f'(x) = 3x² - 3 

f'(c) = 3c² - 3 = 0

⇒ 3c²  = 3

⇒ c² = 1

⇒ c = ± 1

∴ c = 1  ∈ (0, √3)

The correct answer is 1.

Given:

The equation x² + 5kx + 16 = 0

Concept Used:

The general quadratic equation ax² + bx + c = 0

Discriminant (D) = b2 – 4ac

If Discriminant (D) < 0, then roots will not be real.

Calculation:

a = 1, b = 5k, c = 16

⇒ (5k)² – 4 × 1 × 16

⇒ 25k² – 64

Now,  25k² – 64 < 0

⇒ (5k – 8)(5k + 8) < 0

⇒ 5k – 8 < 0 or 5k + 8 < 0

⇒ k < 8/5 or k > -8/5

∴ The value of k will be smaller than 8/5 or greater than -8/5.