Let f(x)=1+sin2⁡xcos2⁡xsin⁡2xsin2⁡x1+cos2⁡xsin⁡2xsin2⁡xcos2⁡x1+sin2⁡x,x∈π6,π3. If α and β respectively are the maximum and the minimum values of f, then

C1→C1+C2+C3f(x)=2+sin⁡2xcos2⁡xsin⁡2x2+sin⁡2x1+cos2⁡xsin⁡2x2+sin⁡2xcos2⁡x1+sin⁡2xf(x)=(2+sin⁡2x)1cos2⁡xsin⁡2x11+cos2⁡xsin⁡2x1cos2⁡x1+sin⁡2xR2→R2−R1R3→R3−R1f(x)=2+sin⁡2x)1cos2⁡xsin⁡2x010001=(2+sin⁡2x)(1)=2+sin⁡2xf'(x) = 2cos2x = 0⇒x=π4f(π/6)=2+32=βf(π/3)=2+32f(π/4)=3=α β2−2α=194

Let f(x)=1+sin2⁡xcos2⁡xsin⁡2xsin2⁡x1+cos2⁡xsin⁡2xsin2⁡xcos2⁡x1+sin2⁡x,x∈π6,π3. If α and β respectively are the maximum and the minimum values of f, then

C1→C1+C2+C3f(x)=2+sin⁡2xcos2⁡xsin⁡2x2+sin⁡2x1+cos2⁡xsin⁡2x2+sin⁡2xcos2⁡x1+sin⁡2xf(x)=(2+sin⁡2x)1cos2⁡xsin⁡2x11+cos2⁡xsin⁡2x1cos2⁡x1+sin⁡2xR2→R2−R1R3→R3−R1f(x)=2+sin⁡2x)1cos2⁡xsin⁡2x010001=(2+sin⁡2x)(1)=2+sin⁡2xf'(x) = 2cos2x = 0⇒x=π4f(π/6)=2+32=βf(π/3)=2+32f(π/4)=3=α β2−2α=194

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

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$β^{2} + 2 \sqrt{α} = \frac{19}{4}$

$α^{2} - β^{2} = 4 \sqrt{3}$

$β^{2} - 2 \sqrt{α} = \frac{19}{4}$

$α^{2} + β^{2} = \frac{9}{2}$

answer is D.

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Detailed Solution

$\begin{array}{l} C_{1} \to C_{1} + C_{2} + C_{3} \\ f (x) = |\begin{array}{lcc} 2 + \sin 2 x & \cos^{2} x & \sin 2 x \\ 2 + \sin 2 x & 1 + \cos^{2} x & \sin 2 x \\ 2 + \sin 2 x & \cos^{2} x & 1 + \sin 2 x \end{array}| \\ f (x) = (2 + \sin 2 x) |\begin{array}{ccc} 1 & \cos^{2} x & \sin 2 x \\ 1 & 1 + \cos^{2} x & \sin 2 x \\ 1 & \cos^{2} x & 1 + \sin 2 x \end{array}| \end{array}$
$\begin{array}{l} R_{2} \to R_{2} - R_{1} \\ R_{3} \to R_{3} - R_{1} \\ f (x) = 2 + \sin 2 x) |\begin{array}{ccc} 1 & \cos^{2} x & \sin 2 x \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}| \\ = (2 + \sin 2 x) (1) = 2 + \sin 2 x \end{array}$

f'(x) = 2cos2x = 0
$\begin{array}{l} \Rightarrow x = \frac{π}{4} \\ f (π / 6) = 2 + \frac{\sqrt{3}}{2} = β \\ f (π / 3) = 2 + \frac{\sqrt{3}}{2} \\ f (π / 4) = 3 = α β^{2} - 2 \sqrt{α} = \frac{19}{4} \end{array}$