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UPSC NDA Math Model Paper 2

Question : 45 of 120

Marks: +1, -0

\sin^{4} x+\cos^{4} x

Solution:

To Find: maximum value of

\sin^{4} x+\cos^{4} x

Let

f(x)=\sin^{4} x+\cos^{4} x=(\sin^{2} x)^{2}+(\cos^{2} x)^{2}

=(\sin^{2} x)^{2}+(\cos^{2} x)^{2}+2 \sin^{2} x \cos^{2} x

-2 \sin^{2} x \cos^{2} x

=(\sin^{2} x+\cos^{2} x)^{2}-2 \sin^{2} x \cos^{2} x

(\because a^{2}+b^{2}+2 a b=(a+b)^{2})

=1-\frac{1}{2} \times 4 \sin^{2} x \cos^{2} x

(\because \sin^{2} x+\cos^{2} x=1)

=1-\frac{1}{2} \times (2 \sin x \cos x)^{2}

=1-\frac{1}{2} \times \sin^{2} 2 x

(\because 2 \sin x \cos x=\sin 2 x)

As we know

\sin^{2} x

lies between [0,1]

0 \leq \sin^{2} 2 x \leq 1

We need to take the minimum value of sin x as there is a minus sign in front of the sine term.

So we need to take it as zero.

∴ Maximum value of

f(x) = 1 - 0 = 1

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