If A=\cos ^{2} \theta+\sin ^{4} \theta, then for all values of \theta

\frac{3}{4} \leq A \leq \frac{13}{16}

NIMCET 2009 Question Paper with solutions

Question : 8 of 120

Marks: +1, -0

A=\cos ^{2} θ+\sin ^{4} θ

, then for all values of

θ

Solution:

A=\cos ^{2} θ+\sin ^{4} θ

=\sin ^{4} θ-\sin ^{2} θ+1

=(\sin ^{2} θ-\;{1}/{2})^{2}+\;{3}/{4}

As ,

0 ≤ \sin ^{2} θ ≤ 1

-\;{1}/{2} ≤ \sin ^{2} θ-\;{1}/{2} ≤ \;{1}/{2}

0 ≤(\sin ^{2} θ-\;{1}/{2})^{2} ≤ \;{1}/{4}

\;{3}/{4} ≤(\sin ^{2} θ-\;{1}/{2})^{2}+\;{3}/{4} ≤ 1

⇒ \;\; \;{3}/{4} ≤ A ≤ 1

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