Specific Solution of Equation MCQ Quiz in मल्याळम - Objective Question with Answer for Specific Solution of Equation - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

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 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]

Calculation: 

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

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

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

⇒ \((2 \sin x-1)(\sqrt{3} \sin x+2)=0\)

⇒ \(\sin x=\frac{1}{2}\)

∴ 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 \(\sin 3 x=\cos 2 x\)

\(\implies \cos \left(\dfrac{\pi}{2}-3 x\right)=\cos 2 x\)

\(\implies \dfrac{\pi}{2}-3 x=2 n\pi\pm 2 x,n\in \mathbb{Z}\)

\(\implies \dfrac{\pi}{2}-3 x=2 n\pi+2 x\) or \(\dfrac{\pi}{2}-3 x=2 n\pi-2 x\)

\(\implies x=\dfrac{\pi}{10}-\dfrac{2 n\pi}{5}\) or \(x=\dfrac{\pi}{2}-2 n\pi\)

In the interval \(\left(\dfrac{\pi}{2},\pi\right)\), \(x=\dfrac{9\pi}{10}\)

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 \(\dfrac{1}{\sqrt{2}}\) for two times.

Thus the given equation has two solutions.

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

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

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

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

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

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

Let \(y=\tan 2\theta\)

\(\tan 2\theta = 1 \)

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

\(\implies 2\theta=\dfrac{\pi}{4}+{k\pi},\) where \(k\) is a positive integer

Now,

for \(k=0\Rightarrow \theta=\dfrac{\pi}{4}\) and for \(k=1\Rightarrow \theta=\dfrac{5\pi}{4}\)

\(0\leq \dfrac{\pi}{4} \leq 2\pi\) and \(0\leq \dfrac{5\pi}{4} \leq 2\pi\)

\(\therefore 2\theta=\dfrac{\pi}{4}, \dfrac{5\pi}{4}\)

\(\implies \theta=\dfrac{\pi}{8},\dfrac{5\pi}{8}\)

Hence, the required principal solutions are \(\dfrac{\pi}{8}\) and \(\dfrac{5\pi}{8}\).