Question

If P=4sinx+cos2x, then which of the following is/are true ?

A
Maximum value of P is 5
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B
Minimum value of P is 4
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C
Maximum value of P occurs when sinx=0
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D
Minimum value of P occurs when sinx=1
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Solution

The correct options are B Minimum value of P is −4 D Minimum value of P occurs when sinx=−1P=4sinx+cos2x⇒P=4sinx+(1−sin2x) Assuming sinx=t⇒t∈[−1,1] P=4t+1−t2⇒f(t)=−(t−2)2+5, t∈[−1,1] x coordinate of vertex is 2 and 2∉[−1,1] f(1)=4 f(−1)=−4 ⇒ The mimimum value of P is −4 at sinx=−1 ⇒ The maximum value of P is 4 at sinx=1 Alternate Solution: P=4sinx+cos2x⇒P=4sinx+(1−sin2x)⇒P=−(sin2x−4sinx+4)+5 ⇒P=5−(sinx−2)2 −1≤sinx≤1⇒−3≤sinx−2≤−1⇒1≤(sinx−2)2≤9⇒−9≤−(sinx−2)2≤−1⇒−4≤5−(sinx−2)2≤4⇒−4≤P≤4 P=4⇒5−(sinx−2)2=4⇒sinx−2=±1⇒sinx=1 (∵sinx≠3) P=−4⇒5−(sinx−2)2=−4⇒sinx−2=±3⇒sinx=−1 (∵sinx≠5) The mimimum value of P is −4 when sinx=−1. The maximum value of P is 4 when sinx=1.

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