Let x,y,z be real numbers with x≥y≥z≥π12 such that x+y+z=π2 and let P=cosx·siny·cosz then

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$Let x, y, z be real numbers with x \geq y \geq z \geq \frac{π}{12} such that x + y + z = \frac{π}{2} and let$ $P = \cos x \cdot \sin y \cdot \cos z then$

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$Minimum value of P is \frac{1}{4}$

$Maximum value of P is \frac{2 + \sqrt{3}}{4}$

$Maximum value of P is \frac{2 + \sqrt{3}}{8}$

$Minimum value of P is \frac{1}{8}$

answer is A, D.

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Detailed Solution

$(A, D)$

$\begin{array}{l} P = \frac{1}{2} \cos x [\sin (y + z) + \sin (y - z)] \geq \frac{1}{2} \cos x \cdot \sin (y + z) = \frac{1}{2} \cos^{2} x \\ But x = \frac{π}{2} - (y + z) \leq \frac{π}{2} - 2 \cdot \frac{π}{12} = \frac{π}{3} ∴ P \geq \frac{1}{8} (\sin c e y \geq \frac{π}{12}, a n d z \geq \frac{π}{12} \Rightarrow y + z \geq \frac{2 π}{12}) \\ Again, P = \frac{1}{2} \cos z [(\sin (x + y) - \sin (x - y))] \\ P \leq \frac{1}{2} \cos^{2} z = \frac{1 + \cos 2 z}{4} \Rightarrow P \leq \frac{2 + \sqrt{3}}{8} (\sin c e z \geq \frac{π}{12} \Rightarrow 2 z \geq \frac{π}{6} \Rightarrow \cos 2 z \leq \cos \frac{π}{6} = \frac{\sqrt{3}}{2}) \end{array}$