Verified Solution • 2025–26 Session moderate

QJEEM25Maa2c5: JEE Main 2025 Mathematics - Class 11 JEEM25Maa2c5 Solution: Let and be the distinct roots of . If m and M are the minimum…

Let αθ\alpha_\theta and βθ\beta_\theta be the distinct roots of 2x2+(cos⁡θ)x−1=0,θ∈(0,2π)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 αθ4+βθ4\alpha_\theta^4+\beta_\theta^4 , then 16(M+m)16(M+m) equals :

Multiple Choice Options

  • A 27
  • B 17
  • C 25
  • D 24

Step-by-Step Verified Solution

amp;(α2+β2)2−2α2β2amp;[(α+β)2−2αβ]2−2(αβ)2amp;[cos⁡2θ4+1]2−2⋅14amp;(cos⁡2θ4+1)2−12amp;M=2516−12=1716amp; m=12,16(M+m)=25\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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