Let If and be the smallest and largest elements of the set respectively, then equals

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Solution

Q.
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 _________ . [2024]

Ans.
(4)

For real roots $D \geq 0$

$\Rightarrow {(\sin 2 θ)}^{2} - 4 (\sin^{4} θ + \cos^{4} θ) (\sin^{6} θ + \cos^{6} θ) \geq 0$

$\Rightarrow \sin^{2} 2 θ \geq 4 (\sin^{4} θ + \cos^{4} θ) (\sin^{6} θ + \cos^{6} θ)$

Put $\sin^{2} 2 θ = t$

$\Rightarrow t \geq 4 (1 - \frac{t}{2}) (1 - \frac{3 t}{4}) \Rightarrow 2 t \geq (2 - t) (4 - 3 t)$

$\Rightarrow 3 t^{2} - 12 t + 8 \leq 0$ $\Rightarrow t^{2} - 4 t + \frac{8}{3} \leq 0$

$\Rightarrow {(t - 2)}^{2} + \frac{8}{3} - 4 \leq 0$ $\Rightarrow {(t - 2)}^{2} \leq \frac{4}{3}$

$\Rightarrow - \frac{2}{\sqrt{3}} \leq t - 2 \leq \frac{2}{\sqrt{3}}$ $\Rightarrow 2 - \frac{2}{\sqrt{3}} \leq t \leq 2 + \frac{2}{\sqrt{3}}$

$∵$ $t \in [0, 1]$ $\Rightarrow 2 - \frac{2}{\sqrt{3}} \leq t \leq 1$

Since, $α$ and $β$ be the smallest and largest elements of set $S$

$∴$ $α = 2 - \frac{2}{\sqrt{3}}, β = 1$

$Hence, 3 [{(α - 2)}^{2} + {(β - 1)}^{2}] = 3 [{(2 - \frac{2}{\sqrt{3}} - 2)}^{2} + {(1 - 1)}^{2}]$ $= 3 [\frac{4}{3} + 0] = 4 .$

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