Let S= \ sin 2 2 θ: (sin 4 θ+cos 4 θ ) x 2 +(sin 2 θ) x + . (sin 6 θ+cos 6 θ )=0 has real roots \ . If α and β be the smallest and largest elements of the set S, respectively, then 3 ((α-2) 2 +(β-1) 2 ) equals.....

Mathematics · Trigonometry Ratios and Identities

JEE Main 2024 — 4 April, Shift 2 — Question 21

Let $S=\left\{\sin ^{2} 2 \theta:\left(\sin ^{4} \theta+\cos ^{4} \theta\right) x^{2}+(\sin 2 \theta) \mathrm{x}+\right.$ $\left(\sin ^{6} \theta+\cos ^{6} \theta\right)=0$ has real roots $\}$ .

If $\alpha$ and $\beta$ be the smallest and largest elements of the set S, respectively, then $3\left((\alpha-2)^{2}+(\beta-1)^{2}\right)$ equals.....

Answer: 4

Numerical answer — enter this value.

Step-by-step solution

$\quad \mathrm{D}=(\sin 2 \theta)^{2}-4\left(1-\frac{\sin ^{2} 2 \theta}{2}\right)\left(1-\frac{3}{4} \sin ^{2} 2 \theta\right)$

$=(\sin 2 \theta)^{2}-4\left(1-\frac{5}{4} \sin ^{2} 2 \theta+\frac{3}{8} \sin ^{4} 2 \theta\right)$

$\mathrm{D}=-\frac{3}{2} \sin ^{4} 2 \theta+6 \sin ^{2} 2 \theta-4>0$

$3 \sin ^{4} 2 \theta-12 \sin ^{2} 2 \theta+8<0$

$\sin ^{2} 2 \theta=\frac{12 \pm \sqrt{12^{2}-12.8}}{6}=\frac{12 \pm 4 \sqrt{3}}{6}=\frac{6 \pm 2 \sqrt{3}}{3}$

$\sin ^{2} 2 \theta=2 \pm \frac{2}{\sqrt{3}}$ , but $\sin ^{2} 2 \theta \in[0,1]$

$\therefore \alpha=2-\frac{2}{\sqrt{3}}, \beta=1 \rightarrow(\alpha-2)^{2}=\frac{4}{3},(\beta-1)^{2}=0$

$3(\alpha-2)^{2}+(\beta-1)^{2}=4$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 4 April, Shift 2
Subject: Mathematics
Chapter: Trigonometry Ratios and Identities
Topic: Trigonometric Ratios of Multiples and Submultiples of angles