Specific Solution of Equation MCQ Quiz in తెలుగు - Objective Question with Answer for Specific Solution of Equation - ముఫ్త్ [PDF] డౌన్‌లోడ్ కరెన్

CONCEPT:

The general solution of the equation tan x = tan α is given by: x = nπ + α, where $\alpha \in \left(-\frac{\pi }{2},\frac{\pi }{2}\right)$ 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, $\tan \frac{5\pi }{6}=-\frac{1}{\sqrt{3}}$

⇒ $\tan x=\tan \frac{5\pi }{6}$

As we know that, if tan x = tan α then x = nπ + α, where $\alpha \in \left(-\frac{\pi }{2},\frac{\pi }{2}\right)$ 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 tan θ = tan α then θ = nπ + α, α ∈ (-π/2, π/2), n ∈ Z.

Calculation:

Given: tan x = 1/√3 such that x ∈ [0, π /2]

As we know that, tan (π/6) = 1/√3

⇒ tan x = 1/√3 = tan (π/6)

As we know that, If tan θ = tan α then θ = nπ + α, α ∈ (-π/2, π/2), n ∈ Z.

⇒ x = nπ + π/6, where n ∈ Z.

For n = 0, x = π/6 ∈ [0, π /2]

So, x = π/6 is a solution of the given equation.

For n ≥ 1, x = nπ + π/6 ∉ [0, π /2]

For n < 1, x = nπ + π/6 ∉ [0, π /2]

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

⇒ $\theta =\frac{2\mathrm{k}\pi }{5}$

$\therefore \theta =\{-\pi ,\frac{-4\pi }{5},\frac{-3\pi }{5},\frac{-2\pi }{5},\frac{-\pi }{5},0,\frac{\pi }{5},\frac{2\pi }{5},\frac{3\pi }{5},\frac{4\pi }{5},\pi \}$

m = 5, n = 5

∴ m.n = 25 

Hence, the correct answer is 25. 

Calculation: 

$(4-\sqrt{3})\sin x-2\sqrt{3}{\cos }^{2}x=\frac{-4}{1+\sqrt{3}},x\in [-2\pi ,\frac{5\pi }{2}]$

⇒ $(4-\sqrt{3})\sin x-2\sqrt{3}(1-{\sin }^{2}x)=2(1-\sqrt{3})$

⇒ $2\sqrt{3}{\sin }^{2}x+4\sin x-\sqrt{3}\sin x-2=0$

⇒ $(2\sin x-1)(\sqrt{3}\sin x+2)=0$

⇒ $\sin x=\frac{1}{2}$

∴ Number of solution = 5  

Hence, the correct answer is Option 5. 

Calculation:

Given: sin 2x + 4 cos x = 2 + sin x

⇒ 2 sin x cos x + 4 cos x = 2 + sin x

⇒ 2 sin x cos x − sin x + 4 cos x − 2 = 0

Group terms: sin x(2 cos x − 1) + 4 cos x − 2 = 0

Let f(x) = sin x(2 cos x − 1) + 4 cos x − 2

Solve f(x) = 0 in [−π, 4π]

Use graphical methods or trial substitution over each period

Check critical intervals:

• In each period of 2π, this equation yields 1 solution.
• Total interval length = 4π − (−π) = 5π
• Hence, it spans 2 full periods of 2π + 1π = 2 solutions + additional interval
• Final solution count = 4

∴ Total number of solutions in [−π, 4π] is 4

Given $\sin 3x=\cos 2x$

$⟹\cos ({\displaystyle \frac{\pi }{2}}-3x)=\cos 2x$

$⟹{\displaystyle \frac{\pi }{2}}-3x=2n\pi \pm 2x,n\in \mathbb{Z}$

$⟹{\displaystyle \frac{\pi }{2}}-3x=2n\pi +2x$ or ${\displaystyle \frac{\pi }{2}}-3x=2n\pi -2x$

$⟹x={\displaystyle \frac{\pi }{10}}-{\displaystyle \frac{2n\pi }{5}}$ or $x={\displaystyle \frac{\pi }{2}}-2n\pi$

In the interval $({\displaystyle \frac{\pi }{2}},\pi )$, $x={\displaystyle \frac{9\pi }{10}}$

So number of solutions is $1$

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 $\frac{5\pi }{12}$.

We know that between $[0,\pi ]$ sine is positive and takes value ${\displaystyle \frac{1}{\sqrt{2}}}$ for two times.

Thus the given equation has two solutions.