Key Concepts and Formulas

• Quadratic Equations and Vieta's Formulas: For a quadratic equation $ax^2 + bx + c = 0$, if $x_1$ and $x_2$ are the roots, then the sum of roots is $x_1 + x_2 = -b/a$, and the product of roots is $x_1 x_2 = c/a$.
• Tangent Addition Formula: $\tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta}$.

Step 1: Rewrite the Equation as a Quadratic in tan x

We are given the equation $(k + 1)\tan^2 x - \sqrt{2} \lambda \tan x = (1 - k)$. To use Vieta's formulas, we need to rewrite this as a standard quadratic equation in the form $at^2 + bt + c = 0$, where $t = \tan x$. Rearranging the terms, we get:

$(k + 1)\tan^2 x - \sqrt{2} \lambda \tan x - (1 - k) = 0$

Thus, $a = k + 1$, $b = -\sqrt{2} \lambda$, and $c = -(1 - k) = k - 1$. The condition $k \ne -1$ ensures $a \ne 0$, so it's a valid quadratic equation.

Step 2: Apply Vieta's Formulas

We are given that $\alpha$ and $\beta$ are roots of the given equation. This means $\tan \alpha$ and $\tan \beta$ are the roots of the quadratic equation in $\tan x$ that we derived in Step 1. We can now apply Vieta's formulas to find the sum and product of these roots.

1. Sum of roots:

   $\tan \alpha + \tan \beta = -\frac{b}{a} = -\frac{-\sqrt{2} \lambda}{k + 1} = \frac{\sqrt{2} \lambda}{k + 1}$

   This gives us the sum of the tangent values of the roots.

2. Product of roots:

   $\tan \alpha \cdot \tan \beta = \frac{c}{a} = \frac{k - 1}{k + 1}$

   This gives us the product of the tangent values of the roots.

Step 3: Use the Tangent Addition Formula

We are given $\tan^2(\alpha + \beta) = 50$. We need to relate this to the sum and product of $\tan \alpha$ and $\tan \beta$, which we found in Step 2. We use the tangent addition formula:

$\tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta}$

Now, substitute the expressions we found in Step 2 using Vieta's formulas:

$\tan(\alpha + \beta) = \frac{\frac{\sqrt{2} \lambda}{k + 1}}{1 - \frac{k - 1}{k + 1}}$

Simplify the denominator:

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

Now substitute this back into the expression for $\tan(\alpha + \beta)$:

$\tan(\alpha + \beta) = \frac{\frac{\sqrt{2} \lambda}{k + 1}}{\frac{2}{k + 1}} = \frac{\sqrt{2} \lambda}{k + 1} \cdot \frac{k + 1}{2} = \frac{\sqrt{2} \lambda}{2} = \frac{\lambda}{\sqrt{2}}$

This simplifies the problem considerably.

Step 4: Solve for λ

We are given that $\tan^2(\alpha + \beta) = 50$. Substitute the expression we found for $\tan(\alpha + \beta)$ in Step 3:

$\left(\frac{\lambda}{\sqrt{2}}\right)^2 = 50$

Squaring the left side, we get:

$\frac{\lambda^2}{2} = 50$

Multiply both sides by 2:

$\lambda^2 = 100$

Taking the square root of both sides:

$\lambda = \pm \sqrt{100} = \pm 10$

The question asks for a value of $\lambda$, so either $10$ or $-10$ is acceptable.

Step 5: Check the options

Looking at the options, we want to see if $10$ or $-10$ can be expressed as one of the options. The correct option is $\lambda = 5\sqrt{2}$ (option A). However, this is incorrect, since we found $\lambda = \pm 10$.

The question states the answer is A. There must be an error in the question statement. If $\tan^2(\alpha + \beta) = 50$, then $\tan(\alpha + \beta) = \pm \sqrt{50} = \pm 5\sqrt{2}$. If $\tan(\alpha + \beta) = \pm 5\sqrt{2}$, then $\frac{\lambda}{\sqrt{2}} = \pm 5\sqrt{2}$, so $\lambda = \pm 5\sqrt{2} \cdot \sqrt{2} = \pm 10$.

If the problem stated $\tan(\alpha + \beta) = 5\sqrt{2}$, then $\frac{\lambda}{\sqrt{2}} = 5\sqrt{2}$, so $\lambda = 5\sqrt{2} \cdot \sqrt{2} = 10$. This is not option A. The question still seems to be incorrect.

If the problem stated that $\tan^2(\alpha + \beta) = 50$, and we wanted to find an approximate value of $\lambda$, then we would have $\frac{\lambda}{\sqrt{2}} = \pm \sqrt{50} = \pm 5\sqrt{2}$, which gives $\lambda = \pm 10$.

However, if the problem stated that $\tan(\alpha + \beta) = \frac{5\sqrt{2}}{2}$, then $\frac{\lambda}{\sqrt{2}} = \frac{5\sqrt{2}}{\sqrt{2} \sqrt{2}} = \frac{5\sqrt{2}}{2}$ so $\lambda = \frac{5\sqrt{2}}{\sqrt{2} \sqrt{2}} \cdot \sqrt{2} = \frac{5\sqrt{2}}{\sqrt{2}}$.

The question has an error. Let's assume that $\tan(\alpha + \beta) = 5\sqrt{2}$. Then $\frac{\lambda}{\sqrt{2}} = 5\sqrt{2}$ so $\lambda = 5\sqrt{2} \cdot \sqrt{2} = 10$.

Common Mistakes & Tips

• Sign Errors: Be extremely careful with the signs when applying Vieta's formulas.
• Algebraic Manipulation: Double-check all algebraic manipulations, especially when simplifying fractions.
• Square Roots: Remember to consider both positive and negative roots when taking the square root.

Summary

By rewriting the given equation as a quadratic in $\tan x$, applying Vieta's formulas, and using the tangent addition formula, we derived a relationship between $\tan(\alpha + \beta)$ and $\lambda$. Using the given condition $\tan^2(\alpha + \beta) = 50$, we solved for $\lambda$ and found $\lambda = \pm 10$. However, if $\tan(\alpha + \beta) = 5\sqrt{2}$, then $\lambda = 10$.

The final answer is \boxed{10}, which corresponds to option (B).