Let $S=\left\{\sin ^{2} 2 heta:\left(\sin ^{4} heta+\cos ^{4} heta ight) x^{2}+(\sin 2 heta

Question

Let $S=\left\{\sin ^{2} 2 	heta:\left(\sin ^{4} 	heta+\cos ^{4} 	hetaight) x^{2}+(\sin 2 	heta) x+\left(\sin ^{6} 	hetaight.ight. \left.+\cos ^{6} 	hetaight)=0$ has real roots}. If $\alpha$ and $\beta$ be the smallest and largest elements of the set $S$, respectively, then $3\left((\alpha-2)^{2}+(\beta-1)^{2}ight)$ equals $\_\_\_\_$ .\

Professor PC Thomas Classes · Answer

(4): For real roots $D \geq 0$ $$ \begin{gathered} \Rightarrow \quad(\sin 2 	heta)^{2}-4\left(\sin ^{4} 	heta+\cos ^{4} 	hetaight) \ \left(\sin ^{6} 	heta+\cos ^{6} 	hetaight) \geq 0 \ \Rightarrow \quad \sin ^{2} 2 	heta \geq 4\left(\sin ^{4} 	heta+\cos ^{4} 	hetaight) \ \quad\left(\sin ^{6} 	heta+\cos ^{6} 	hetaight) \end{gathered} $$ Put $\sin ^{2} 2 	heta=t$ $$ \begin{aligned} & \Rightarrow \quad t \geq 4\left(1-\frac{t}{2}ight)\left(1-\frac{3 t}{4}ight) \ & \Rightarrow \quad 2 t \geq(2-t)(4-3 t) \ & \Rightarrow \quad 3 t^{2}-12 t+8 \leq 0 \Rightarrow t^{2}-4 t+\frac{8}{3} \leq 0 \ & \Rightarrow \quad(t-2)^{2}+\frac{8}{3}-4 \leq 0 \Rightarrow(t-2)^{2} \leq \frac{4}{3} \ & \Rightarrow \quad-\frac{2}{\sqrt{3}} \leq t-2 \leq \frac{2}{\sqrt{3}} \ & \Rightarrow \quad 2-\frac{2}{\sqrt{3}} \leq t \leq 2+\frac{2}{\sqrt{3}} \end{aligned} $$ $\because \quad t \in[0,1] \Rightarrow 2-\frac{2}{\sqrt{3}} \leq t \leq 1$\ Since, $\alpha$ and $\beta$ be the smallest and largest elements of set S $$ 	herefore \quad \alpha=2-\frac{2}{\sqrt{3}}, \beta=1 $$ Hence, $3\left[(\alpha-2)^{2}+(\beta-1)^{2}ight]$ $$ \begin{aligned} & =3\left[\left(2-\frac{2}{\sqrt{3}}-2ight)^{2}+(1-1)^{2}ight] \ & =3\left[\frac{4}{3}+0ight]=4 \end{aligned} $$