JEE MAIN - Mathematics (2025 - 22nd January Evening Shift - No. 1)
Let $\alpha_\theta$ and $\beta_\theta$ be the distinct roots of $2 x^2+(\cos \theta) x-1=0, \theta \in(0,2 \pi)$. If m and M are the minimum and the maximum values of $\alpha_\theta^4+\beta_\theta^4$, then $16(M+m)$ equals :
27
17
25
24
Explanation
$$\begin{aligned}
& \left(\alpha^2+\beta^2\right)^2-2 \alpha^2 \beta^2 \\
& {\left[(\alpha+\beta)^2-2 \alpha \beta\right]^2-2(\alpha \beta)^2} \\
& {\left[\frac{\cos ^2 \theta}{4}+1\right]^2-2 \cdot \frac{1}{4}} \\
& \left(\frac{\cos ^2 \theta}{4}+1\right)^2-\frac{1}{2} \\
& \mathrm{M}=\frac{25}{16}-\frac{1}{2}=\frac{17}{16} \\
& \mathrm{~m}=\frac{1}{2}, 16(\mathrm{M}+\mathrm{m})=25
\end{aligned}$$
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