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Q.

If A=sin2θ+cos4θ , then for all values of θ , where

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a

1≤A≤2

b

34≤A≤1

c

0≤A≤1

d

14≤A≤12

answer is B.

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Detailed Solution

A=sin2θ+cos4θ=sin2θ+cos2θ(1−sin2θ) =sin2θ+cos2θ−sin2θcos2θ=1−14sin22θmax of sin22θ is 1 and min=0⇒1-14 , 1-0⇒34≤A≤1
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If A=sin2θ+cos4θ , then for all values of θ , where