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(B) The given equation is $(k+1) 	an^{2} x - (\sqrt{2} \lambda) 	an x + (k-1) = 0$.Let $t = 	an x$. Then $(k+1) t^{2} - (\sqrt{2} \lambda) t + (k-1

Vedclass · Accepted Answer

$10$

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(B) The given equation is $(k+1) 	an^{2} x - (\sqrt{2} \lambda) 	an x + (k-1) = 0$.Let $t = 	an x$. Then $(k+1) t^{2} - (\sqrt{2} \lambda) t + (k-1) = 0$.Since $\alpha$ and $\beta$ are roots,$	an \alpha$ and $	an \beta$ are roots of this quadratic equation.Sum of roots: $	an \alpha + 	an \beta = \frac{\sqrt{2} \lambda}{k+1}$.Product of roots: $	an \alpha 	an \beta = \frac{k-1}{k+1}$.Using the formula $	an(\alpha+\beta) = \frac{	an \alpha + 	an \beta}{1 - 	an \alpha 	an \beta}$,we get:$	an(\alpha+\beta) = \frac{\frac{\sqrt{2} \lambda}{k+1}}{1 - \frac{k-1}{k+1}} = \frac{\sqrt{2} \lambda}{k+1 - (k-1)} = \frac{\sqrt{2} \lambda}{2} = \frac{\lambda}{\sqrt{2}}$.Given $	an^{2}(\alpha+\beta) = 50$,so $\left(\frac{\lambda}{\sqrt{2}}ight)^{2} = 50$.$\frac{\lambda^{2}}{2} = 50 \implies \lambda^{2} = 100 \implies \lambda = \pm 10$.Thus,a possible value of $\lambda$ is $10$.

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

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