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Questions  

Given A=sin2θ+cos4θ, then for all real θ

a
1≤A≤2
b
34≤A≤1
c
1316≤A≤1
d
34≤A≤1316

detailed solution

Correct option is B

We have,A=sin2⁡θ+cos4⁡θ⇒ A=1−cos⁡2θ2+1+cos⁡2θ22⇒ A=12−12cos⁡2θ+14+12cos⁡2θ+14cos2⁡2θ⇒ A=34+14cos⁡4θ+12=34+18+18cos⁡4θ Now, −1≤cos⁡4θ≤1⇒ −18≤cos⁡4θ8≤18⇒ 34+18−18≤34+141+cos⁡4θ2≤34+18+18⇒ 34≤A≤1

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