Let alpha_theta and beta_theta be the distinct roots of 2x^2+(costheta)x-1=0, thetain(0,2pi). If m and M are the minimum and maximum values of alpha_theta^4+beta_theta^4, then 16(M+m) equals:

Home › MCQs › JEE MAIN Mathematics › Question #978

JEE MAIN Mathematics QUESTION #978

Question 1

Let \(\alpha_\theta\) and \(\beta_\theta\) be the distinct roots of \(2x^2+(\cos\theta)x-1=0,\ \theta\in(0,2\pi)\). If \(m\) and \(M\) are the minimum and maximum values of \(\alpha_\theta^4+\beta_\theta^4\), then \(16(M+m)\) equals:

24
25✔️
17
27

Correct Answer Explanation

By Vieta's: \(\alpha+\beta=-\frac{\cos\theta}{2}\), \(\alpha\beta=-\frac{1}{2}\). Then \(\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta=\frac{\cos^2\theta}{4}+1\). And \(\alpha^4+\beta^4=(\alpha^2+\beta^2)^2-2(\alpha\beta)^2=\left(\frac{\cos^2\theta}{4}+1\right)^2-\frac{1}{2}\). Let \(u=\cos^2\theta\in[0,1]\). Max at \(u=1\): \(M=\frac{25}{16}-\frac{1}{2}=\frac{17}{16}\). Min at \(u=0\): \(m=1-\frac{1}{2}=\frac{1}{2}\). So \(16(M+m)=16\cdot\frac{25}{16}=25\).

More Options