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Question

# Let x,y,z be real numbers with x≥y≥z≥π12 such that x+y+z=π2. If p=cosxsinycosz, then

A
the maximum value of p is 2+38
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B
the minimum value of p is 18
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C
the maximum value of p is attained when x=y=5π24 and z=π12
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D
the minimum value of p is attained when x=y=z=π6
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Solution

## The correct options are A the maximum value of p is 2+√38 B the minimum value of p is 18 C the maximum value of p is attained when x=y=5π24 and z=π12 Since, π2>x≥y≥z≥π12 ⇒sin(y−z)≥0, sin(x−y)≥0 p=cosxsinycosz ⇒p=12cosx[sin(y+z)+sin(y−z)] ⇒p≥12cosxsin(y+z) [∵sin(y−z)≥0] ⇒p≥12cos2x x+y+z=π2 x=π2−(y+z)≤π2−2×π12=π3 ⇒x≤π3 Hence, the minimum value of p is 12cos2π3=18, which is obtained when x=π3 and sin(y−z)=0 i.e., p is minimum when x=π3 and y=z=π12 as x+y+z=π2 On the other hand, we also have p=12cosz[sin(x+y)−sin(x−y)] ⇒p≤12coszsin(x+y) [∵sin(x−y)≥0] ⇒p≤12cos2z ⇒p≤14(1+cos2z) ⇒p≤14(1+cosπ6)=2+√38 This maximum value is obtained whenx=y=5π24 and z=π12

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