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JEE Main 20244 AprilEvening Shift
Exam
JEE Main
Year
2024
Date
4 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Trigonometric Equations

JEE Main 2024 · 4 April · Evening Shift · Mathematics · Trigonometric Equations

Let S=\ ^2 2 θ:( ^4 θ+ ^4 θ) x^2+( 2 θ) x+( ^6 θ+ ^6 θ)=0. has real...

Let S={sin⁡22θ:(sin⁡4θ+cos⁡4θ)x2+(sin⁡2θ)x+(sin⁡6θ+cos⁡6θ)=0 has  real  roots }S=\left\{\sin ^2 2 \theta:\left(\sin ^4 \theta+\cos ^4 \theta\right) x^2+(\sin 2 \theta) x+\left(\sin ^6 \theta+\cos ^6 \theta\right)=0\right.\text{ has }\text{ real }\text{ roots } \}. If α\alpha and β\beta be the smallest and largest elements of the set SS, respectively, then 3((α−2)2+(β−1)2)3\left((\alpha-2)^2+(\beta-1)^2\right) equals _________

Correct answer

0

Step-by-step solution

D=(sin⁡2θ)2−4(1−sin⁡22θ2)(1−34sin⁡22θ)=(sin⁡2θ)2−4(1−54sin⁡22θ+38sin⁡42θ)D=−32sin⁡42θ+6sin⁡22θ−4>03sin⁡42θ−12sin⁡22θ+8<0sin⁡22θ=12±122−12.86=12±436=6±233sin⁡22θ=2±23, but sin⁡22θ∈[0,1]∴α=2−23,β=1→(α−2)2=43,(β−1)2=03(α−2)2+(β−1)2=4\begin{aligned} & D=(\sin 2 \theta)^2-4\left(1-\frac{\sin ^2 2 \theta}{2}\right)\left(1-\frac{3}{4} \sin ^2 2 \theta\right) \\ & =(\sin 2 \theta)^2-4\left(1-\frac{5}{4} \sin ^2 2 \theta+\frac{3}{8} \sin ^4 2 \theta\right) \\ & D=-\frac{3}{2} \sin ^4 2 \theta+6 \sin ^2 2 \theta-4>0 \\ & 3 \sin ^4 2 \theta-12 \sin ^2 2 \theta+8 < 0 \\ & \sin ^2 2 \theta=\frac{12 \pm \sqrt{12^2-12.8}}{6}=\frac{12 \pm 4 \sqrt{3}}{6}=\frac{6 \pm 2 \sqrt{3}}{3} \\ & \sin ^2 2 \theta=2 \pm \frac{2}{\sqrt{3}}, \text{ but } \sin ^2 2 \theta \in[0,1] \\ & \therefore \alpha=2-\frac{2}{\sqrt{3}}, \beta=1 \rightarrow(\alpha-2)^2=\frac{4}{3},(\beta-1)^2=0 \\ & 3(\alpha-2)^2+(\beta-1)^2=4\end{aligned}
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