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Question

If A=sin8 θ+cos14 θ, then for all real values of θ

A
A1
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B
0 < A1
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C
1 < 2A3
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D
None of these
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Solution

The correct option is B 0 < A1
Given that A=sin8 θ+cos14 θ,
we know that (sin2 θ)4sin2 θ and (cos2 θ)7cos2 θ
Adding, we that sin8 θ+cos14 θ1
or 0 < sin8 θ+cos14 θ1. Hence 0 < A1.

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