Multiple choice

Let $\alpha$ and $\beta$ be two real roots of the equation $(k+1)\tan^2x-\sqrt{2}\lambda\tan x=(1-k)$, where $k(\neq\,\,-1)$ and $\lambda$ are real numbers. If $tan^2(\alpha+\beta)=50$, then a value of $\lambda$ is:

  1. $10\sqrt{2}$
  2. $5\sqrt{2}$
  3. $10$
  4. $5$
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C Correct answer
Explanation

Equation: (k+1)tan^2x - sqrt(2)lambda tan x + (k-1) = 0. Let tan alpha, tan beta be roots. tan alpha + tan beta = sqrt(2)lambda / (k+1). tan alpha * tan beta = (k-1)/(k+1). tan(alpha+beta) = (tan alpha + tan beta) / (1 - tan alpha * tan beta). tan^2(alpha+beta) = 50. Solving leads to lambda = 10.

AI explanation

Let y = tan(x), transforming the given equation into the quadratic (k+1)y^2 - sqrt(2)*lambda*y + (k-1) = 0. Using the sum of roots formula, tan(a) + tan(b) = sqrt(2)*lambda / (k+1). The formula for the tangent of a sum gives tan(a+b) = [sqrt(2)*lambda / (k+1)] / [1 - (-k+1)/(k+1)], which simplifies to lambda / sqrt(2). Setting [lambda / sqrt(2)]^2 = 50 yields lambda^2 = 100, so a valid value for lambda is 10.

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