General Solution of Equation MCQ Quiz in தமிழ் - Objective Question with Answer for General Solution of Equation - இலவச PDF ஐப் பதிவிறக்கவும்

Calculation

Given

sin-1 α + sin-1 β + sin-1 γ = π

Let 

sin-1 α = A,  sin-1 β = B, sin-1 γ = C

⇒ A + B + C =  π ⇒ B + C = π - A  ⇒ A = π - (B + C)

(α + β + γ) (α  - γ + β) = 3αβ 

⇒ (α + β)² - γ² = 3αβ 

⇒ α² + β² - γ2 = αβ 

⇒ sin²A + sin²B - sin²C = sinA sinB

⇒ sin2A + sin2(B + C) sin2(B - C)  = sinA sinB

⇒ sin2A + sin(A) sin(B - C)  = sinA sinB

⇒ sinA(sin(A)  + sin(B - C))  = sinA sinB

⇒ sinA(sin(B + C)  + sin(B - C))  = sinA sinB

⇒ sinA(2sin(B)cos(C) - sinB)  = 0

⇒ sinAsinB(2cos(C) - 1 ) = 0

⇒ (2cos(C) - 1 ) = 0 {∵ α, β, γ ≠ 0)

⇒ cosC =  \(\frac{1}{2}\)

⇒ sinC =\(\frac{\sqrt{3}}{2}\)

⇒ γ = \(\frac{\sqrt{3}}{2}\) 

Hence option 1 is correct

Concept:

cos 2θ = 1 - 2 sin² θ 

cos 2θ = 2 cos2θ - 1

cos 2θ = \(\rm \frac{1 + tan^{2}θ}{1 - tan^{2}θ}\)

Calculator:

From statement 2

sinθ = sinα

⇒ sinθ - sinα = 0

\(⇒ 2. cos \frac {θ + α}{2}. sin \frac {θ - α}{2} = 0\)

So, either 

\(cos \frac {θ + α}{2} = 0\) or \(sin \frac {θ - α}{2} = 0\)

Now from \(cos \frac {θ + α}{2} = 0\)

\( \frac {θ + α}{2} = (2k + 1)\frac {π}{2}, \ k \ ϵ \ integer\)

⇒ θ = (2k + 1)π - α        ----(i)

Now from \( \frac {θ - α}{2} = kπ, \ k \ ϵ \ integer\)

⇒ θ = 2kπ + α          ----(ii) 

Now, On combining equation (i) and (ii), we get 

⇒ θ = nπ + (-1)^α, where α ϵ integer

So, The general solution of sinθ = sinα is θ = nπ + (-1)α

Hence, Statement 2 is not correct.

Now from statement 1

sin2 θ = sin2 α

⇒ \(\rm \frac{1 - \cos2θ}{2} = \frac{1 - \cos2α}{2}\)

⇒ cos 2θ = cos 2α

⇒ 2θ = 2nπ ± 2α 

\(\rm θ = nπ ± α \), n ∈ Z

Similarly for cos2 θ = cos2 α 

⇒ \(\rm \frac{\cos2θ + 1}{2} = \frac{\cos2α + 1}{2}\)

⇒ cos 2θ + 1 = cos 2α + 1

⇒ cos 2θ = cos 2α 

⇒ \(\rm θ = nπ ± α \), n ∈ Z

tan2 θ = tan2 α

Componendo and dividendo

\(\rm \frac{1 + tan^{2}θ}{1 - tan^{2}θ} = \frac{1 + tan^{2}α}{1 - tan^{2}α} \)

⇒ cos 2θ = cos 2α 

So, \(\rm θ = nπ ± α \), n ∈ Z

So, the statement 1 is correct.

∴ Only option (i) is correct.

Calculation:

Given, x3 + 2x2 + 5x + 2cosx = 0

⇒ x3 + 2x2 + 5x = - 2cosx 

From the graph, we can observe that for x ∈ [0, 2π] the graphs do not intersect.

∴ The number of solutions of the equation x3 + 2x2 + 5x + 2cosx = 0 in [0, 2π] is zero.

The correct answer is Option 4.

Concept:

If cos x = cos α, then x = 2nπ ± α, n = 0, ± 1, ± 2, … 

Calculation:

Given, 3sin²x + 10cos x - 6 = 0.

⇒ 3(1 - cos²x) + 10cos x - 6 = 0

⇒ 3 - 3cos²x + 10cos x - 6 = 0

⇒ 3cos²x - 10cos x + 3 = 0

⇒ 3cos²x - 9cos x - cos x + 3 = 0

⇒ 3cos x (cos x - 3) - (cos x - 3) = 0

⇒ (cos x - 3)(3cos x - 1) = 0

⇒ cos x = 3 or 3cos x = 1

But cos x = 3 cannot be possible as - 1 ≤ cos x ≤ 1 for all x ∈ R.

∴ 3cos x = 1

⇒ cos x = \(\frac{1}{3}\)

⇒ cos x = cos(cos ^-1(\(\frac{1}{3}\)))

⇒ x = 2nπ ± cos–1(1/3)

∴ The equation is satisfied for x = x = 2nπ ± cos–1(1/3).

The correct answer is Option 3.

\(ac = b^2\)

Similarly,

\(\tan{\theta} \sin{\theta} \times \dfrac{1}{6} = \cos^2{\theta}\)

\(\Rightarrow \sin^2{\theta} = 6\cos^3{\theta}\)

\(\Rightarrow 6\cos^3{\theta} + \cos^2{\theta} - 1 = 0\)

We get \(\cos{\theta} = \dfrac{1}{2}\) by observation, since its value has to lie between \(0\) and \(1\).

Since \(\cos{\theta} = \dfrac{1}{2}\), the general solution becomes \(2n\pi \pm \dfrac{\pi}{3}\)

Concept:

(1) General solution of tanθ = tanα is θ = nπ + α; 

α ∈ \(\left[\frac{-π}{2}, \frac{π}{2}\right]\) n ∈ I

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

Calculation:

Given: S = {θ ∈[0, 2 λ); tan (π cos θ) + tan (π sin θ) = 0}

⇒ tan (π cos θ) = – tan (π sin θ)

⇒ tan (π cos θ) = tan (– π sin θ)

∴ π cos θ = nπ – π sin θ; n ∈ I

⇒ π cos θ + π sin θ = nπ

⇒ cos θ + sin θ = n

Since, \(-\sqrt{2}\) ≤ cos θ + sin θ ≤ \(\sqrt{2}\)

∴ n = –1, 0, 1

Case 1 : If n = –1

cos θ + sin θ = -1

⇒ cos \(\left(θ-\frac{π}{4}\right)\) = \(\rm -\frac{1}{\sqrt2}\)

⇒ cos\(\left(θ-\frac{π}{4}\right)\) = cos\(\left(\frac{3 π}{4}\right)\)

⇒ θ - \(\frac{π}{4}\) = 2kπ ± \(\frac{3 π}{4}\)

⇒ θ = 2k π + π or θ = 2kπ - \(\frac{π}{2}\)

⇒ θ = π, \(\frac{3 π}{2}\) 

Case - 2 : If n = 0

cos θ  + sin θ = 0

⇒ cos\(\left(θ-\frac{π}{4}\right)\) = 0

⇒ θ - \(\frac{π}{4}\) = 2kπ ± \(\frac{π}{2}\)

⇒ θ = 2kπ + \(\frac{3 π}{4}\) or θ = 2kπ - \(\frac{π}{4}\)

⇒ θ = \(\frac{3 π}{4}\), \(\frac{7 π}{4}\)

Case - 3 : If n = 1

cosθ + sinθ = \(\frac{1}{\sqrt{2}}\)

⇒ cos\(\left(θ-\frac{π}{4}\right)\) = cos\(\left(\frac{π}{4}\right)\)

⇒ θ - \(\frac{π}{4}\) = 2kπ ± \(\frac{π}{4}\)

⇒ θ = 2kπ + \(\frac{π}{2}\) or θ = 2kπ 

⇒ θ = \(\frac{π}{2}\), 0

∴ θ = \(\left\{0, \frac{π}{2}, π, \frac{3 π}{2}, \frac{3 π}{4}, \frac{7 π}{4}\right\}\)

So, sin\(\left(θ+\frac{π}{4}\right)\) = \(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\)

Now, \(\sum_{θ ∈ S} \sin ^2\) \(\left(θ+\frac{π}{4}\right)\) = \(\frac{1}{2}\) + \(\frac{1}{2}\) + \(\frac{1}{2}\) + \(\frac{1}{2}\) = 2

∴ The value of \(\displaystyle \sum_{θ \in S} \sin ^2\left(θ+\frac{\pi}{4}\right)\) is 2.

Given:

 3 sin2 x - 7 sin x + 2 = 0

Concept:

Use concept of general value of sin x

\(x=\rm n\pi+(-1)^n \theta\)

Calculation:

 3 sin2 x - 7 sin x + 2 = 0

Put sin x = y

⇒3y² - 7y + 2 = 0

⇒3y2 - 6y - y + 2 = 0

⇒3y(y - 2) - (y - 2) = 0

⇒(y - 2)(3y - 1) = 0

⇒y = 2 , 1/3

we know that range of sin x is [-1,1] then 2 is not in range than 

⇒ sin x = 1/3

⇒ x = sin^-1(1/3)

Hence the general value 

\(\rm x={n\pi}+(-1)^n\sin^{-1}\frac{1}{3}\)

Hence the option (3) is correct.

Given:

 3 sin2 x - 7 sin x + 2 = 0

Concept:

Use concept of general value of sin x

\(x=\rm n\pi+(-1)^n \theta\)

Calculation:

 3 sin2 x - 7 sin x + 2 = 0

Put sin x = y

⇒3y² - 7y + 2 = 0

⇒3y2 - 6y - y + 2 = 0

⇒3y(y - 2) - (y - 2) = 0

⇒(y - 2)(3y - 1) = 0

⇒y = 2 , 1/3

we know that range of sin x is [-1,1] then 2 is not in range than 

⇒ sin x = 1/3

⇒ x = sin^-1(1/3)

Hence the general value 

\(\rm x={n\pi}+(-1)^n\sin^{-1}\frac{1}{3}\)

Hence the option (3) is correct.

Concept:

The table below shows the sign of trigonometric functions in different quadrants:

Calculation:

Given: α is the root of the equation 25 cos² θ + 5 cos θ - 12 = 0

∵ α is the root of the equation 25 cos² θ + 5 cos θ - 12 = 0

⇒ 25 cos² α + 5 cos α – 12 = 0

Let us suppose 5 cos α = x. So, the given equation 25 cos² θ + 5 cos θ - 12 = 0 can be written as:

⇒ x² + x – 12 = 0

⇒ x² + 4x – 3x – 12 = 0

⇒ (x + 4) (x - 3) = 0

⇒ x = - 4 or x = 3

Now, by substituting the value of 5 cos α = x in the above equation, we get

⇒ cos α = - 4 / 5 or cos α = 3 / 5

\(\because \alpha \in \left( {\pi ,\frac{{3\pi }}{2}} \right) \Rightarrow \cos \alpha = \; - ve\)

Hence, cos α = - 4 / 5

\( \Rightarrow \sin \alpha = \sqrt {1 - {{\left( {-\frac{4}{5}} \right)}^2}} = \; \pm \frac{3}{5}\)

\(\because \alpha \in \;\left( {\pi ,\frac{{3\pi }}{2}} \right) \Rightarrow \sin \alpha = \; - ve\)

Hence, sin α = - 3 / 5

\(\Rightarrow \tan \alpha = \frac{{\sin \alpha }}{{\cos \alpha }} = \;\frac{3}{4}\)

Explanation:

tan θ = tan α

∴ θ = nπ + α