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

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 = $\frac{1+\mathrm{t}\mathrm{a}{\mathrm{n}}^2\theta }{1-\mathrm{t}\mathrm{a}{\mathrm{n}}^2\theta }$

Calculator:

From statement 2

sinθ = sinα

⇒ sinθ - sinα = 0

$\Rightarrow 2.cos\frac{\theta +\alpha }{2}.sin\frac{\theta -\alpha }{2}=0$

So, either 

$cos\frac{\theta +\alpha }{2}=0$ or $sin\frac{\theta -\alpha }{2}=0$

Now from $cos\frac{\theta +\alpha }{2}=0$

$\frac{\theta +\alpha }{2}=(2k+1)\frac{\pi }{2},kϵinteger$

⇒ θ = (2k + 1)π - α        ----(i)

Now from $\frac{\theta -\alpha }{2}=k\pi ,kϵinteger$

⇒ 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

⇒ $\frac{1-\cos 2\theta }{2}=\frac{1-\cos 2\alpha }{2}$

⇒ cos 2θ = cos 2α

⇒ 2θ = 2nπ ± 2α 

$\theta =\mathrm{n}\pi \pm \alpha$, n ∈ Z

Similarly for cos2 θ = cos2 α 

⇒ $\frac{\cos 2\theta +1}{2}=\frac{\cos 2\alpha +1}{2}$

⇒ cos 2θ + 1 = cos 2α + 1

⇒ cos 2θ = cos 2α 

⇒ $\theta =\mathrm{n}\pi \pm \alpha$, n ∈ Z

tan2 θ = tan2 α

Componendo and dividendo

$\frac{1+\mathrm{t}\mathrm{a}{\mathrm{n}}^2\theta }{1-\mathrm{t}\mathrm{a}{\mathrm{n}}^2\theta }=\frac{1+\mathrm{t}\mathrm{a}{\mathrm{n}}^2\alpha }{1-\mathrm{t}\mathrm{a}{\mathrm{n}}^2\alpha }$

⇒ cos 2θ = cos 2α 

So, $\theta =\mathrm{n}\pi \pm \alpha$, 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,

$\tan \theta \sin \theta \times {\displaystyle \frac{1}{6}}={\cos }^2\theta$

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

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

We get $\cos \theta ={\displaystyle \frac{1}{2}}$ by observation, since its value has to lie between $0$ and $1$.

Since $\cos \theta ={\displaystyle \frac{1}{2}}$, the general solution becomes $2n\pi \pm {\displaystyle \frac{\pi }{3}}$

Concept:

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

α ∈ $[\frac{-\pi }{2},\frac{\pi }{2}]$ 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 $(\theta -\frac{\pi }{4})$ = $-\frac{1}{\sqrt{2}}$

⇒ cos$(\theta -\frac{\pi }{4})$ = cos$\left(\frac{3\pi }{4}\right)$

⇒ θ - $\frac{\pi }{4}$ = 2kπ ± $\frac{3\pi }{4}$

⇒ θ = 2k π + π or θ = 2kπ - $\frac{\pi }{2}$

⇒ θ = π, $\frac{3\pi }{2}$ 

Case - 2 : If n = 0

cos θ  + sin θ = 0

⇒ cos$(\theta -\frac{\pi }{4})$ = 0

⇒ θ - $\frac{\pi }{4}$ = 2kπ ± $\frac{\pi }{2}$

⇒ θ = 2kπ + $\frac{3\pi }{4}$ or θ = 2kπ - $\frac{\pi }{4}$

⇒ θ = $\frac{3\pi }{4}$, $\frac{7\pi }{4}$

Case - 3 : If n = 1

cosθ + sinθ = $\frac{1}{\sqrt{2}}$

⇒ cos$(\theta -\frac{\pi }{4})$ = cos$\left(\frac{\pi }{4}\right)$

⇒ θ - $\frac{\pi }{4}$ = 2kπ ± $\frac{\pi }{4}$

⇒ θ = 2kπ + $\frac{\pi }{2}$ or θ = 2kπ 

⇒ θ = $\frac{\pi }{2}$, 0

∴ θ = $\{0,\frac{\pi }{2},\pi ,\frac{3\pi }{2},\frac{3\pi }{4},\frac{7\pi }{4}\}$

So, sin$(\theta +\frac{\pi }{4})$ = $\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}$

Now, $\sum _{\theta \in S}{\sin }^2$ $(\theta +\frac{\pi }{4})$ = $\frac{1}{2}$ + $\frac{1}{2}$ + $\frac{1}{2}$ + $\frac{1}{2}$ = 2

∴ The value of ${\displaystyle \sum _{\theta \in S}{\sin }^2(\theta +\frac{\pi }{4})}$ is 2.

Given:

 3 sin2 x - 7 sin x + 2 = 0

Concept:

Use concept of general value of sin x

$x=\mathrm{n}\pi +(-1{)}^{\mathrm{n}}\theta$

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 

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

Hence the option (3) is correct.

Given:

 3 sin2 x - 7 sin x + 2 = 0

Concept:

Use concept of general value of sin x

$x=\mathrm{n}\pi +(-1{)}^{\mathrm{n}}\theta$

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 

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

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