Let f(x)=left|begin{array}{ccc}1+sin ^2 x & c | vedclass

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(A) Given $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

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

Vedclass · Answer

(A) Given $f(x)=\left|\begin{array}{ccc}1+\sin ^2 x & \cos ^2 x & \sin 2 x \ \sin ^2 x & 1+\cos ^2 x & \sin 2 x \ \sin ^2 x & \cos ^2 x & 1+\sin 2 x\end{array}ight|$.Applying $C_1 ightarrow C_1+C_2+C_3$,we get:$f(x)=\left|\begin{array}{ccc}1+\sin^2 x+\cos^2 x+\sin 2x & \cos^2 x & \sin 2x \ \sin^2 x+1+\cos^2 x+\sin 2x & 1+\cos^2 x & \sin 2x \ \sin^2 x+\cos^2 x+1+\sin 2x & \cos^2 x & 1+\sin 2x\end{array}ight|$Since $\sin^2 x+\cos^2 x=1$,the first column becomes $2+\sin 2x$:$f(x)=\left|\begin{array}{ccc}2+\sin 2x & \cos^2 x & \sin 2x \ 2+\sin 2x & 1+\cos^2 x & \sin 2x \ 2+\sin 2x & \cos^2 x & 1+\sin 2x\end{array}ight|$Taking $(2+\sin 2x)$ common from $C_1$:$f(x)=(2+\sin 2x)\left|\begin{array}{ccc}1 & \cos^2 x & \sin 2x \ 1 & 1+\cos^2 x & \sin 2x \ 1 & \cos^2 x & 1+\sin 2x\end{array}ight|$Applying $R_2 ightarrow R_2-R_1$ and $R_3 ightarrow R_3-R_1$:$f(x)=(2+\sin 2x)\left|\begin{array}{ccc}1 & \cos^2 x & \sin 2x \ 0 & 1 & 0 \ 0 & 0 & 1\end{array}ight| = (2+\sin 2x)(1) = 2+\sin 2x$.For $x \in \left[\frac{\pi}{6}, \frac{\pi}{3}ight]$,$2x \in \left[\frac{\pi}{3}, \frac{2\pi}{3}ight]$.Thus,$\sin 2x \in \left[\frac{\sqrt{3}}{2}, 1ight]$.So,$f(x) \in \left[2+\frac{\sqrt{3}}{2}, 3ight]$.Therefore,$\alpha = 3$ and $\beta = 2+\frac{\sqrt{3}}{2} = \frac{4+\sqrt{3}}{2}$.Checking option $D$: $\alpha^2+\beta^2 = 3^2 + \left(\frac{4+\sqrt{3}}{2}ight)^2 = 9 + \frac{16+3+8\sqrt{3}}{4} = 9 + \frac{19+8\sqrt{3}}{4} = \frac{36+19+8\sqrt{3}}{4} = \frac{55+8\sqrt{3}}{4}$.Checking option $B$: $\beta^2+2\sqrt{\alpha} = \left(\frac{4+\sqrt{3}}{2}ight)^2 + 2\sqrt{3} = \frac{19+8\sqrt{3}}{4} + 2\sqrt{3} = \frac{19+8\sqrt{3}+8\sqrt{3}}{4} = \frac{19+16\sqrt{3}}{4}$.Re-evaluating: The question asks for the relation. Given the options,let's re-check the calculation. $\beta^2 = \frac{19+8\sqrt{3}}{4}$. Option $B$ is $\frac{19+8\sqrt{3}}{4} + 2\sqrt{3} 
eq \frac{19}{4}$. Actually,$\beta = 2+\frac{\sqrt{3}}{2}$. The correct relation is $\beta^2 - 2\sqrt{3} = \frac{19}{4}$.

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