Let $\alpha$ and $\beta$ be two real roots of the $(k + 1) \tan^2 x - \sqrt{2} \cdot \lambda \tan x = (1 - k)$, where $k (\neq -1)$ and $\lambda$ are real numbers. If $\tan^2 (\alpha + \beta) = 50$, then a value of $\lambda$ is: Options: (A) 10 (B) 5 (C) 7 (D) 12

### Analyzing the Setup The given equation is $(k + 1) \tan^2 x - \sqrt{2} \lambda \tan x = (1 - k)$. To reveal the underlying structure, we rearrange all terms to one side: $$(k + 1) \tan^2 x - \sqrt{2} \lambda \tan x + (k - 1) = 0$$ By substituting $t = \tan x$, we transform this into a standard quadratic equation in $t$: $$(k + 1)t^2 - \sqrt{2} \lambda t + (k - 1) = 0$$ Here, the coefficients are $a = (k + 1)$, $b = -\sqrt{2} \lambda$, and $c = (k - 1)$. ### The Power of Vieta Let the roots of this quadratic be $t_1 = \tan \alpha$ and $t_2 = \tan \beta$. According to Vieta's formulas, the sum and product of the roots are given by: $$\tan \alpha + \tan \beta = -\frac{b}{a} = \frac{\sqrt{2} \lambda}{k + 1}$$ $$\tan \alpha \tan \beta = \frac{c}{a} = \frac{k - 1}{k + 1}$$ These expressions serve as the fundamental bridge between the roots $\alpha, \beta$ and the parameters $\lambda, k$. ### The Bridge The problem provides the condition $\tan^2(\alpha + \beta) = 50$. We utilize the compound angle formula for tangent: $$\tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta}$$ Substituting our Vieta expressions into this formula, we obtain: $$\tan(\alpha + \beta) = \frac{\frac{\sqrt{2} \lambda}{k + 1}}{1 - \frac{k - 1}{k + 1}}$$ ### The Elegant Collapse We simplify the denominator by finding a common denominator: $$1 - \frac{k - 1}{k + 1} = \frac{(k + 1) - (k - 1)}{k + 1} = \frac{2}{k + 1}$$ Substituting this back into the expression for $\tan(\alpha + \beta)$, the $(k + 1)$ terms cancel out: $$\tan(\alpha + \beta) = \frac{\frac{\sqrt{2} \lambda}{k + 1}}{\frac{2}{k + 1}} = \frac{\sqrt{2} \lambda}{2} = \frac{\lambda}{\sqrt{2}}$$ ### Final Calculation Given the condition $\tan^2(\alpha + \beta) = 50$, we substitute our simplified result: $$\left( \frac{\lambda}{\sqrt{2}} \right)^2 = 50$$ $$\frac{\lambda^2}{2} = 50$$ $$\lambda^2 = 100 \Rightarrow \lambda = \pm 10$$ The final result is **$\lambda = 10$** (considering the positive magnitude). You have successfully navigated the complexity to reach the elegant solution.

What if the denominator $1 - \tan \alpha \tan \beta$ becomes zero?

That is a very sharp observation! If $1 - \tan \alpha \tan \beta = 0$, then $\tan(\alpha + \beta)$ would be undefined, meaning $\alpha + \beta$ would be an odd multiple of $\pi/2$. However, the problem gives us a specific value for $\tan^2(\alpha + \beta)$, which implies that $\tan(\alpha + \beta)$ is defined. Therefore, we can safely proceed with the formula.

JEE Main 2020 - 7 Jan (Morning)

LEVELJEE Main

Animated Solution for Mathematics - Quadratic Equations: Let $α$ and $β$ be two real roots of the $(k + 1) tan^{2} x - 2 \cdot λ tan x = (1 - k)$ , where $k (\neq = - 1)$ and $λ$ are real numbers. If $tan^{2} (α + β) = 50$ , then a value of $λ$ is:

Q: Why did we get $\lambda = \pm 10$ but only one option?

Excellent catch! When we solve $\lambda^2 = 100$, we get both positive and negative roots. In many JEE problems, you might find that only one of the values is provided in the options. Always check the options provided; if both were present, the question would likely specify a condition like $\lambda > 0$.

Select Answer:

Visualized Solution

Identifying the Quadratic Structure

Given equation: $(k + 1) tan^{2} x - 2 λ tan x = (1 - k)$
Rearranging into standard quadratic form $a t^{2} + b t + c = 0$ :
$(k + 1) tan^{2} x - 2 λ tan x + (k - 1) = 0$

Defining the Roots

Let $t = tan x$ . The roots of the original equation are $α$ and $β$ .
So, the roots of the quadratic in $t$ are $t_{1} = tan α$ and $t_{2} = tan β$ .
Comparing with $a t^{2} + b t + c = 0$ :
$a = (k + 1)$ , $b = - 2 λ$ , $c = (k - 1)$

Calculating Sum of Roots

Sum of roots: $tan α + tan β = - \frac{b}{a}$
Substituting values: $tan α + tan β = - \frac{( - 2 λ )}{k + 1}$
$tan α + tan β = \frac{2 λ}{k + 1}$

Calculating Product of Roots

Product of roots: $tan α tan β = \frac{c}{a}$
Substituting values: $tan α tan β = \frac{k - 1}{k + 1}$

Applying the Compound Angle Formula

We need to connect our results to $tan (α + β)$ .
Using the identity: $tan (α + β) = \frac{t a n α + t a n β}{1 - t a n α t a n β}$

Substituting the Values

Substitute the sum and product into the formula:
$tan (α + β) = \frac{\frac{2 λ}{k + 1}}{1 - \frac{k - 1}{k + 1}}$

Simplifying the Denominator

Focus on the denominator: $1 - \frac{k - 1}{k + 1}$
Take the common denominator: $\frac{( k + 1 ) - ( k - 1 )}{k + 1}$
Simplify the numerator: $\frac{k + 1 - k + 1}{k + 1} = \frac{2}{k + 1}$

Final Expression for $tan (α + β)$

Substitute the simplified denominator back:
$tan (α + β) = \frac{\frac{2 λ}{k + 1}}{\frac{2}{k + 1}}$
The $(k + 1)$ terms cancel out: $tan (α + β) = \frac{2 λ}{2}$
Simplify further: $tan (α + β) = \frac{λ}{2}$

Squaring and Equating to $50$

Given condition: $tan^{2} (α + β) = 50$
Substitute our result: $(\frac{λ}{2})^{2} = 50$
Square the terms: $\frac{λ ^{2}}{2} = 50$

Solving for $λ$

Multiply by $2$ : $λ^{2} = 50 \times 2$
$λ^{2} = 100$
Take the square root: $λ = \pm 10$
From the given options, $λ = 10$ .

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The Sigma Insight: Relation Between Roots and Coefficients

Analyzing the Setup

The given equation is

(k + 1) tan^{2} x - 2 λ tan x = (1 - k)

. To reveal the underlying structure, we rearrange all terms to one side:

(k + 1) tan^{2} x - 2 λ tan x + (k - 1) = 0

By substituting

t = tan x

, we transform this into a standard quadratic equation in

t

(k + 1) t^{2} - 2 λ t + (k - 1) = 0

Here, the coefficients are

a = (k + 1)

b = - 2 λ

, and

c = (k - 1)

The Power of Vieta

Let the roots of this quadratic be

t_{1} = tan α

and

t_{2} = tan β

. According to Vieta's formulas, the sum and product of the roots are given by:

tan α + tan β = - \frac{b}{a} = \frac{2 λ}{k + 1}

tan α tan β = \frac{c}{a} = \frac{k - 1}{k + 1}

These expressions serve as the fundamental bridge between the roots

α, β

and the parameters

λ, k

The Bridge

The problem provides the condition

tan^{2} (α + β) = 50

. We utilize the compound angle formula for tangent:

tan (α + β) = \frac{tan α + tan β}{1 - tan α tan β}

Substituting our Vieta expressions into this formula, we obtain:

tan (α + β) = \frac{\frac{2 λ}{k + 1}}{1 - \frac{k - 1}{k + 1}}

The Elegant Collapse

We simplify the denominator by finding a common denominator:

1 - \frac{k - 1}{k + 1} = \frac{( k + 1 ) - ( k - 1 )}{k + 1} = \frac{2}{k + 1}

Substituting this back into the expression for

tan (α + β)

, the

(k + 1)

terms cancel out:

tan (α + β) = \frac{\frac{2 λ}{k + 1}}{\frac{2}{k + 1}} = \frac{2 λ}{2} = \frac{λ}{2}

Final Calculation

Given the condition

tan^{2} (α + β) = 50

, we substitute our simplified result:

(\frac{λ}{2})^{2} = 50

\frac{λ ^{2}}{2} = 50

λ^{2} = 100 \Rightarrow λ = \pm 10

The final result is $λ = 10$ (considering the positive magnitude). You have successfully navigated the complexity to reach the elegant solution.

Common Traps & FAQs

Wait, why can we treat

tan x

as a variable

t

Great question! In trigonometry, when you see an equation involving

tan x

and its square, it is often a quadratic equation in terms of

tan x

. By substituting

t = tan x

, we simplify the problem into a standard algebraic form

a t^{2} + b t + c = 0

, which allows us to use powerful tools like Vieta's formulas to relate the roots

α

and

β

to the coefficients of the equation.

What if the denominator

1 - tan α tan β

becomes zero?

That is a very sharp observation! If

1 - tan α tan β = 0

, then

tan (α + β)

would be undefined, meaning

α + β

would be an odd multiple of

π /2

. However, the problem gives us a specific value for

tan^{2} (α + β)

, which implies that

tan (α + β)

is defined. Therefore, we can safely proceed with the formula.

Why did we get

λ = \pm 10

but only one option?

Excellent catch! When we solve

λ^{2} = 100

, we get both positive and negative roots. In many JEE problems, you might find that only one of the values is provided in the options. Always check the options provided; if both were present, the question would likely specify a condition like

λ > 0

Animated Solution for Mathematics - Quadratic Equations: Let $α$ and $β$ be two real roots of the $(k + 1) tan^{2} x - 2 \cdot λ tan x = (1 - k)$ , where $k (\neq = - 1)$ and $λ$ are real numbers. If $tan^{2} (α + β) = 50$ , then a value of $λ$ is:

Select Answer:

Visualized Solution

Identifying the Quadratic Structure

Defining the Roots

Calculating Sum of Roots

Calculating Product of Roots

Applying the Compound Angle Formula

Substituting the Values

Simplifying the Denominator

Final Expression for $tan (α + β)$

Squaring and Equating to $50$

Solving for $λ$

The Sigma Insight: Relation Between Roots and Coefficients

Analyzing the Setup

The Power of Vieta

The Bridge

The Elegant Collapse

Final Calculation

Similar Questions

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Select Answer:

Visualized Solution

Identifying the Quadratic Structure

Defining the Roots

Calculating Sum of Roots

Calculating Product of Roots

Applying the Compound Angle Formula

Substituting the Values

Simplifying the Denominator

Final Expression for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } tan(α+β)

Squaring and Equating to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 50

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λ

The Sigma Insight: Relation Between Roots and Coefficients

Analyzing the Setup

The Power of Vieta

The Bridge

The Elegant Collapse

Final Calculation

Similar Questions

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Final Expression for $tan (α + β)$

Squaring and Equating to $50$

Solving for $λ$

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