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

Question

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

Aicharya · Accepted Answer

Given equation is $\begin{aligned} & (\mathrm{k}+1) 	an ^2 x-\sqrt{2} \lambda 	an x=(1-\mathrm{k}) \quad \ldots (i) \ & \Rightarrow(\mathrm{k}+1) 	an ^2 x-\sqrt{2} \lambda 	an x+(\mathrm{k}-1)=0 \quad \ldots (ii) \end{aligned}$ $\alpha$ and $\beta$ are two real roots. $\begin{aligned} 	herefore \quad & 	an \alpha+	an \beta=\frac{\sqrt{2} \lambda}{k+1} \ & 	an \alpha \cdot 	an \beta=\frac{k-1}{k+1} \end{aligned}$ Now, $	an (\alpha+\beta)=\frac{	an \alpha+	an \beta}{1-	an \alpha 	an \beta}$ $\begin{aligned} & =\frac{\frac{\sqrt{2} \lambda}{\mathrm{k}+1}}{1-\frac{(\mathrm{k}-1)}{\mathrm{k}+1}} \ & =\frac{\sqrt{2} \lambda}{2} \end{aligned}$ $\begin{aligned} & \Rightarrow 	an (\alpha+\beta)=\frac{\lambda}{\sqrt{2}} \ & \Rightarrow 	an ^2(\alpha+\beta)=\frac{\lambda^2}{2} \ & \Rightarrow 50=\frac{\lambda^2}{2} \ & \Rightarrow \lambda^2=100 \ & \Rightarrow \lambda=10\end{aligned}$

Let $\alpha$ and $\beta$ be two real roots of the equation $(\mathrm{k}+1) \tan ^2 x-\sqrt{2} \lambda \tan…

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