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,

\(\sin (\frac{\pi x }{3\sqrt{2}}) = x^2-4x+6 \)   

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: 

\(\cos 2 \theta \cdot \cos \frac{\theta}{2}=\cos 3 \theta \cdot \cos \frac{9 \theta}{2} \)

\(\Rightarrow 2 \cos 2 \theta \cdot \cos \frac{\theta}{2}=2 \cos \frac{9 \theta}{2} \cdot \cos 3 \theta \)

\(\Rightarrow \cos \frac{5 \theta}{2}+\cos \frac{3 \theta}{2}=\cos \frac{15 \theta}{2}+\cos \frac{3 \theta}{2} \)

\(\Rightarrow \cos \frac{15 \theta}{2}=\cos \frac{5 \theta}{2} \)

\(\Rightarrow \frac{15 \theta}{2}=2 \mathrm{k} \pi \pm \frac{5 \theta}{2} \)

5θ = 2kπ or 10θ = 2kπ

⇒ \(\theta=\frac{2 \mathrm{k} \pi}{5} \)

\(\therefore \theta=\left\{-\pi, \frac{-4 \pi}{5}, \frac{-3 \pi}{5}, \frac{-2 \pi}{5}, \frac{-\pi}{5}, 0, \frac{\pi}{5}, \frac{2 \pi}{5}, \frac{3 \pi}{5}, \frac{4 \pi}{5}, \pi\right\} \)

m = 5, n = 5

∴ m.n = 25 

Hence, the correct answer is 25. 

So, from \(\text{sin }x \text{ cos }x = \frac{1}{4}\), we get that \(\text{ sin }(2x) = \frac{1}{2}\).

So, we have to find the solutions of the equation \(\text{sin} (2x) = \frac{1}{2}\), where \(x \in (0, \frac{\pi}{2})\).

From this data, we get that \(2x = \frac{\pi}{6}\) or \(\frac{5\pi}{6}\).

That is, \(x = \frac{\pi}{12}\) or \(\frac{5\pi}{12}\).

\(\Longrightarrow \sin x + \sin 5x = -\sin 3x\)

\(\Longrightarrow 2 \sin 3x \cos 2x = -\sin 3x\)

\(\Longrightarrow \sin 3x \left( 2 \cos 2x + 1 \right) = 0\)

\(\Longrightarrow \sin 3x = 0 \quad \text{or} \quad \cos 2x = \dfrac { -1 }{ 2 }\)

\(\Longrightarrow x = \dfrac { 2\pi }{ 3 } , \pi , \dfrac { 4\pi }{ 3 } \quad \text{or} \quad x = \dfrac { 2\pi }{ 3 } , \dfrac { 4\pi }{ 3 }\)

The number of common solutions in the given interval are \(3\).

\(\implies 2\pi<2x<4\pi\) ...... \((i)\)

Also, \(\sin 2x+\cos 2x=0\) ..... \((ii)\) [Given]

\(\implies \sin 2x=-\cos 2x\)

\(\implies \dfrac{\sin 2x}{\cos 2x}=-1\)

\(\implies \tan 2x=-1\)

\(\implies 2x=\tan^{-1}(-1)\)

Now, we know that

\(\tan \dfrac{3\pi}{4}=\tan \dfrac{7\pi}{4}=\tan \dfrac{11\pi}{4}=...... =\tan \dfrac{(4n+3)\pi}{4}=-1\)

So, the possible values for \(2x\) are \(\dfrac{3\pi}{4}, \dfrac{7\pi}{4}, \dfrac{11\pi}{4},.....,\dfrac{(4n+3)\pi}{4}\)

But, from \((i)\) we get the values for \(2x\) as \(\dfrac{11\pi}{4}\) and \(\dfrac{15 \pi}{4}\).

\(\therefore 2x=\dfrac{11\pi}{4}, \dfrac{15\pi}{4}\)

\(\implies x=\dfrac{11\pi}{8}, \dfrac{15\pi}{8}\)

\(\therefore\) Principal solutions for \((ii)\) are \(\dfrac{11\pi}{8}\) and \(\dfrac{15\pi}{8}\).

Concept:

- 1 ≤ sin x ≤ 1

Calculation:

Given: 9 sin 7x + 20 = 0

⇒ sin 7x = - 20/9 < - 1

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

Concept:

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

-\(\rm \sqrt{a^{2} + b^{2}}\) ≤ a sin x + b cos x ≤ \(\rm \sqrt{a^{2} + b^{2}}\)

Calculation:

As we know, -\(\rm \sqrt{a^{2} + b^{2}}\) ≤ a sin x + b cos x ≤ \(\rm \sqrt{a^{2} + b^{2}}\)

⇒ -\(\rm \sqrt{2^{2} + 0^{2}}\) ≤  2sin x + 0 cos x ≤ \(\rm \sqrt{2^{2} + 0^{2}}\)

⇒ -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 \(\alpha ∈ \left( { - \frac{π }{2},\frac{π }{2}} \right)\) and n ∈ Z

Note: The solutions of a trigonometric equation for which 0 ≤ x < 2π are called principal solution.

CALCULATION:

Given: \(\tan \ x = -\frac{1}{\sqrt 3} \)

As we know that, \(\tan \ \frac{5π}{6} = -\frac{1}{\sqrt 3} \)

⇒ \(\tan \ x = \tan \ \frac{5π}{6}\)

As we know that, if tan x = tan α then x = nπ + α, where \(\alpha ∈ \left( { - \frac{π }{2},\frac{π }{2}} \right)\) 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 < 2π are called principal solution.

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

Concept:

Trigonometry:

• tan \(\rm\frac{\pi}{4}\) = 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 = \(\rm\frac{\pi}{4}\).

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 < 2π are called principal solution.

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 < 2π are called principal solution.

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 = \(\rm \frac{\pi}{2}\)

x = sin (\(\rm \frac{\pi}{2}\))

sin (\(\rm \frac{\pi}{2}\)) = 1

Calculation:

sin-1 x + sin-1 y + sin-1 z = \(\frac{3\pi}{2}\) 

This is possible only when

sin-1 x = sin-1 y = sin-1 z = \(\rm \frac{\pi}{2}\) 

⇒ sin-1 x = \(\rm \frac{\pi}{2}\)

⇒ x = 1    ----(i)

⇒ sin-1 y = \(\rm \frac{\pi}{2}\)

⇒ y = 1      ----(ii)

⇒ sin-1 z = \(\rm \frac{\pi}{2}\)

⇒ 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.

⇒ \(\rm x=\frac{n\pi}{3}\)

⇒ \(\rm x=...,0, \frac{\pi}{3},\frac{2\pi}{3},\pi,...\)

CASE 2: 2 cos 2x - 1 = 0

⇒ \(\rm \cos 2x=\frac12=\cos\left(2n\pi\pm \frac{\pi}{3}\right)\), n ∈ Z.

⇒ \(\rm x=n\pi\pm \frac{\pi}{6}=(6n\pm1) \frac{\pi}{6}\)

⇒ \(\rm x=..., \frac{\pi}{6},\frac{5\pi}{6},...\)

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.
• \(\rm \sin \left [(2n+1)\frac{π}{2}+\theta \right ]\) = (-1)ⁿ cos θ.
• \(\rm \cos \left [(2n+1)\frac{π}{2}+\theta \right ]\) = (-1)ⁿ (-sin θ).

If the equation involves a variable 0 ≤ x < 2π, then the solutions are called principal solutions.

tan^-1 (1) = \((\rm n\pi - \frac{3\pi}{4}) \)

Calculations:

Given tan 2θ = 1

⇒ 2θ = tan -1 (1)

⇒ 2θ = \(\rm \dfrac{\pi}{4},\dfrac{5\pi}{4},\dfrac{9\pi}{4},\dfrac{13\pi}{4},\dfrac{17\pi}{4} ...\) 

⇒ θ = \(\rm \dfrac{\pi}{8},\dfrac{5\pi}{8},\dfrac{9\pi}{8},\dfrac{13\pi}{8},\dfrac{17\pi}{8} ...\) 

for principal solution 0 ≤ θ < 2π

Hence,θ = \(\rm \dfrac{\pi}{8},\dfrac{5\pi}{8},\dfrac{9\pi}{8},\dfrac{13\pi}{8}\)  

The number of principal solutions of tan 2 θ = 1 is four

Concept:

The solution to the quadratic equation ax2 + bx + c = 0 is given by: \(\rm x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\).

• 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: 

⇒ \(\rm \sin x = \frac{-(-1) \pm \sqrt{(-1)^2-4(1)(-2)}}{2(1)}\)

⇒ \(\rm \sin x = \dfrac{1 +\sqrt{9}}{2}\) OR \(\rm \sin x = \dfrac{1 -\sqrt{9}}{2}\)

⇒ sin x = 2 OR sin x = -1.

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

sin x = -1 = \(\rm -\sin\dfrac{\pi}{2}=\sin\left(-\dfrac{\pi}{2}\right)=\sin\left(2n\pi-\dfrac{\pi}{2}\right)\), n ∈ Z.

For n = 0, x = \(-\dfrac{\pi}{2}\).

For n = 1, x = \(\dfrac{3\pi}{2}\).

For n = 2, x = \(\dfrac{7\pi}{2}\).

The only value of x on the interval 0 ≤ x < 2π is x = \(\dfrac{3\pi}{2}\).

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).