Specific Solution of Equation MCQ Quiz - Objective Question with Answer for Specific Solution of Equation - Download Free PDF

Concept:

• Range of Sine Function: The output of the sine function is always between −1 and 1, i.e., sin(θ) ∈ [−1, 1] for all real θ.
• Quadratic Function: A quadratic function of the form ax² + bx + c represents a parabola. If a > 0, the parabola opens upward, and its minimum value occurs at x = −b / 2a.
• Key Idea: To find how many solutions exist for sin(expression) = quadratic, we determine how many values of x make the quadratic expression lie within [−1, 1].

Calculation:

Given,

Let f(x) = x² − 4x + 6

Minimum of f(x) occurs at:

x = 4 / 2 = 2

⇒ f(2) = (2)² − 4×2 + 6 = 4 − 8 + 6 = 2

Since the parabola opens upward, the range of f(x) is [2, ∞)

But, sin(θ) ∈ [−1, 1]

⇒ The equation has solutions only if x² − 4x + 6 ∈ [−1, 1]

But f(x) ≥ 2 for all x, and 2 > 1

⇒ No value of x satisfies f(x) ∈ [−1, 1]

∴ Number of real solutions is zero.

Calculation: 

5θ = 2kπ or 10θ = 2kπ

⇒ 

m = 5, n = 5

∴ m.n = 25 

Hence, the correct answer is 25. 

We know that, .

So, from , we get that .

So, we have to find the solutions of the equation , where .

From this data, we get that or .

That is, or .

Given equation is ,

The number of common solutions in the given interval are .

Given :

......

Also, ..... [Given]

Now, we know that

So, the possible values for are

But, from we get the values for as and .

Principal solutions for are and .

Concept:

- 1 ≤ sin x ≤ 1

Calculation:

Given: 9 sin 7x + 20 = 0

⇒ sin 7x = - 20/9

As we know that, for any argument θ, - 1 ≤ sin θ ≤ 1

Hence, there is no solution for this equation.

Concept:

The minimum and maximum value of a sin x + b cos x 

- ≤ a sin x + b cos x ≤ 

Calculation:

As we know, - ≤ a sin x + b cos x ≤ 

⇒ - ≤  2sin x + 0 cos x ≤ 

⇒ -2 ≤ 2sin x ≤ 2

⇒ -2 ≤ 2k + 1 ≤ 2

⇒ -3 ≤ 2k ≤ 1

⇒ -1.5 ≤ k ≤ 0.5

k = 0, -1 

Hence, Option 3 is correct.

CONCEPT:

The general solution of the equation tan x = tan α is given by: x = nπ + α, where  and n ∈ Z

Note: The solutions of a trigonometric equation for which 0 ≤ x

CALCULATION:

Given: 

As we know that, 

⇒ 

As we know that, if tan x = tan α then x = nπ + α, where  and n ∈ Z

⇒ x = nπ + (5π/6) where n ∈ Z.

As we know that, if the solutions of a trigonometric equation for which 0 ≤ x

So, the principal solutions of the given equation are x = 5π/6 and 11π/6

Hence, the correct option is 3.

Concept:

If sin θ = sin α then θ = nπ + (- 1)ⁿ α, α ∈ [-π/2, π/2], n ∈ Z.

Calculation:

Given: 4 sin x + 2√3 = 0 such that x ∈ [0, 2π]

⇒ 4 sinx = - 2√3

⇒ sin x = - √3/2

As we know that, sin x is negative in 3^rd and 4^th quadrant.

∵ x ∈ [0, 2π]

⇒ x = 4π/3, 5π/3

Hence, x = 4π/3, 5π/3 are the solution of the given equation.

Concept:

Trigonometry:

• tan  = 1.

Algebra:

• (a ± b)² = a² ± 2ab + b².

Calculation:

1 + tan2 x - 2 tan x = 0

⇒ 1² - 2 (1) (tan x) + (tan x)² = 0

⇒ (1 - tan x)² = 0

⇒ 1 - tan x = 0

⇒ tan x = 1

⇒ x = .

CONCEPT:

The general solution of the equation cos x = cos α is given by x = 2nπ ± α where α ∈ [0, π] and n ∈ Z

Note: The solutions of a trigonometric equation for which 0 ≤ x

CALCULATION:

Given: sec x = 2

⇒ cos x = 1/2

As we know that, cos (π/3) = 1/2

⇒ cos x = cos π/3

As we know that, if cos x = cos α then x = 2nπ ± α where α ∈ [0, π] and n ∈ Z

⇒ x = 2nπ ± (π/3), where n ∈ Z

As we know that, if the solutions of a trigonometric equation for which 0 ≤ x

So, the principal solutions of the given equation are x = π/3 and 5π/3

Hence, the correct option is 4.

Formula used:

sin^-1 x = 

x = sin ()

sin () = 1

Calculation:

sin-1 x + sin-1 y + sin-1 z =  

This is possible only when

sin-1 x = sin-1 y = sin-1 z =  

⇒ sin-1 x = 

⇒ x = 1    ----(i)

⇒ sin-1 y = 

⇒ y = 1      ----(ii)

⇒ sin-1 z = 

⇒ z = 1      ----(iii)

Now, we have to find the value of 

x1000 + y1001 + z1002

Now, from (i), (ii), and (iii), we get

⇒ (1)1000 + (1)1001 + (1)1002

⇒ 1 + 1 + 1

⇒ 3

∴ The value of x1000 + y1001 + z1002 is 3.

Concept:

• sin (A ± B) = sin A cos B ± sin B cos A.
• sin 2A + sin 2B = 2 sin (A + B) cos (A - B).
• cos (2nπ + θ) = cos θ.

Calculation:

sin x + sin 5x = sin 3x

Using sin 2A + sin 2B = 2 sin (A + B) cos (A - B),  we get:

2 sin 3x cos 2x = sin 3x

⇒ sin 3x (2 cos 2x - 1) = 0

⇒ sin 3x = 0 OR 2 cos 2x - 1 = 0

CASE 1: sin 3x = 0 = sin nπ, n ∈ Z.

⇒ 

⇒ 

CASE 2: 2 cos 2x - 1 = 0

⇒ , n ∈ Z.

⇒ 

⇒ 

Therefore, there are 6 possible values of x in the interval [0, π].

Additional Information

Trigonometric Ratios for Allied Angles: 

• sin (-θ) = -sin θ.
• cos (-θ) = cos θ.
• sin (2nπ + θ) = sin θ.
• cos (2nπ + θ) = cos θ.
• sin (nπ + θ) = (-1)ⁿ sin θ.
• cos (nπ + θ) = (-1)ⁿ cos θ.
•  = (-1)ⁿ cos θ.
•  = (-1)ⁿ (-sin θ).

Concept:

The solution to the quadratic equation ax2 + bx + c = 0 is given by: .

• sin (-θ) = -sin θ.
• sin θ = sin (2nπ + θ).

Calculation:

sin2 x - sin x - 2 = 0

Using the formula for the roots of a quadratic equation: 

⇒ 

⇒  OR 

⇒ sin x = 2 OR sin x = -1.

∵ -1 ≤ sin x ≤ 1, ∴ sin x = 2 is not possible.

sin x = -1 = , n ∈ Z.

For n = 0, x = .

For n = 1, x = .

For n = 2, x = .

The only value of x on the interval 0 ≤ x .

Additional Information

• Period of sin θ is 2π. ⇒ sin θ = sin (2nπ + θ).
• Period of cos θ is 2π. ⇒ cos θ = cos (2nπ + θ).
• Period of tan θ is π. ⇒ tan θ = tan (nπ + θ).