Let alpha and beta be two real roots of the e | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

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

Vedclass · Accepted Answer

$10$

Vedclass · Answer

(B) The given equation is $(k+1) 	an^2 x - (\sqrt{2} \lambda) 	an x + (k-1) = 0$.Let $t = 	an x$. The roots of the quadratic equation $(k+1)t^2 - (\sqrt{2} \lambda)t + (k-1) = 0$ are $	an \alpha$ and $	an \beta$.From the properties of quadratic equations,the sum of roots is $	an \alpha + 	an \beta = \frac{\sqrt{2} \lambda}{k+1}$ and the product of roots is $	an \alpha 	an \beta = \frac{k-1}{k+1}$.Using the tangent addition formula,$	an(\alpha + \beta) = \frac{	an \alpha + 	an \beta}{1 - 	an \alpha 	an \beta}$.Substituting the values,$	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$,we have $\left(\frac{\lambda}{\sqrt{2}}ight)^2 = 50$.$\frac{\lambda^2}{2} = 50 \Rightarrow \lambda^2 = 100 \Rightarrow \lambda = \pm 10$.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 :

Similar Questions

If $\sin^2 \theta = \frac{x^2 + y^2 + 1}{2x}$,then $x$ must be

If $\tan x = \frac{2b}{a - c}$ $(a \ne c)$,$y = a \cos^2 x + 2b \sin x \cos x + c \sin^2 x$ and $z = a \sin^2 x - 2b \sin x \cos x + c \cos^2 x$,then:

The maximum value of $f(x) = \sin x + \cos x$ is

If $\pi < \alpha < \frac{3\pi}{2}$,then $\sqrt{\frac{1 - \cos \alpha}{1 + \cos \alpha}} + \sqrt{\frac{1 + \cos \alpha}{1 - \cos \alpha}} = $

The value of $\tan^{-1} \left( \frac{\sin 2 - 1}{\cos 2} \right)$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module