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  • Let <img alt="f(x)=\left|\begin{array}{ccc}1+\sin ^2 x & \cos ^2 x & \sin 2 x \\ \sin ^2 x & 1+\cos ^2 x & \sin 2 x \\ \sin ^2 x & \cos ^2 x & 1+\sin 2 x\end{array}\right|,

Let f(x)=\left|\begin{array}{ccc}1+\sin ^2 x & \cos ^2 x & \sin 2 x \\ \sin ^2 x & 1+\cos ^2 x & \sin 2 x \\ \sin ^2 x & \cos ^2 x & 1+\sin 2 x\end{array}\right|, x \in\left[\frac{\pi}{6}, \frac{\pi}{3}\right]. If \alpha and \beta respectively are the maximum and the minimum values of f, then

Option: 1

\alpha^2+\beta^2=\frac{9}{2}


Option: 2

\beta^2-2 \sqrt{\alpha}=\frac{19}{4}


Option: 3

\alpha^2-\beta^2=4 \sqrt{3}


Option: 4

\beta^2+2 \sqrt{\alpha}=\frac{19}{4}


Answers (1)

best_answer

\begin{aligned} & f(\mathrm{x})=\left|\begin{array}{ccc} 1+\sin ^2 x & \cos ^2 x & \sin 2 x \\ \sin ^2 \mathrm{x} & 1+\cos ^2 x & \sin 2 x \\ \sin ^2 x & \cos ^2 x & 1+\sin x \end{array}\right| \\ & \mathrm{R}_2 \rightarrow \mathrm{R}_2-\mathrm{R}_1 \\ & \mathrm{R}_3 \rightarrow \mathrm{R}_3-\mathrm{R}_1 \\ & =\left|\begin{array}{ccc} 1+\sin ^2 x & \cos ^2 x & \sin 2 x \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{array}\right| \\ & =\left(1+\sin ^2 \mathrm{x}\right)-\cos ^2 \mathrm{x}(-1)+\sin 2 \mathrm{x} \\ & f(\mathrm{x})=2+\sin 2 \mathrm{x} \\ & 2 x \in\left[\frac{\pi}{3}, \frac{2 \pi}{3}\right] \Rightarrow \frac{\sqrt{3}}{2} \leq \sin 2 x \leq 1 \\ & \alpha=2+1=3 \\ & \beta=2+\frac{\sqrt{3}}{2} \\ & \beta^2-2 \sqrt{\alpha}=\left(2+\frac{\sqrt{3}}{2}\right)^2-2 \sqrt{3} \\ & =4+\frac{3}{4}+2 \sqrt{3}-2 \sqrt{3} \\ & =\frac{19}{4} \end{aligned}

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Gautam harsolia

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