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UPSC CDS 2 2021 Math Paper

Question : 35 of 100

Marks: +1, -0

1+2 \sin^{2} \theta \cos^{2} \theta - \sin^{4} \theta - \cos^{4} \theta

0^{\circ} < \theta < 90^{\circ}

Solution: 👈: Video Solution

Given expression:

1+2 \sin^{2} \theta \cos^{2} \theta - \sin^{4} \theta - \cos^{4} \theta

Where,

0^{\circ} < \theta < 90^{\circ}

\Rightarrow 1+2 \sin^{2} \theta \cos^{2} \theta - \sin^{4} \theta - \cos^{4} \theta

=1-[\sin^{4} \theta+\cos^{4} \theta

-2 \sin^{2} \theta \cos^{2} \theta]

=1-[(\sin^{2} \theta)^{2}+(\cos^{2} \theta)^{2}

-2 \sin^{2} \theta \cdot \cos^{2} \theta]

=1-[(\cos^{2} \theta-\sin^{2} \theta)^{2}]

[ \because a^{2}+b^{2}-2ab=(a-b)^{2}]

=1-[(\cos 2 \theta)^{2}]

[ \because \cos 2x = \cos^{2}x - \sin^{2}x]

=1-\cos^{2} 2 \theta = \sin^{2} 2 \theta

[ \because \sin^{2}A + \cos^{2}A = 1]

For maximum value

\sin^{2} 2 \theta = 1

\Rightarrow \sin 2 \theta = 1 \Rightarrow 2 \theta = 90^{\circ}

\Rightarrow \;\; \theta = 45^{\circ}

\therefore

Maximum value

=1+2 \sin^{2} 45^{\circ} \cos^{2}\;45^{\circ} - \sin^{4} 45^{\circ} - \cos^{4} 45^{\circ}

\;\; =1+2 \cdot \;\frac{1}{2} \cdot \;\frac{1}{2} - \;\frac{1}{4} - \;\frac{1}{4}

\;\; =1+\;\frac{1}{2} - \;\frac{1}{2} = 1

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