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UPSEE 2018 Solved Paper

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Question : 103 of 150

Marks: +1, -0

If

f_{n}(x)=\frac{1}{n}(\cos^{n} x+\sin^{n} x)

n=1,2,3\dots

f_{4}(x)-f_{6}(x)

is equal to

10
$\frac{1}{12}$
$\frac{1}{10}$
12

Solution:

f_{n}(x)=\frac{1}{n}(\cos^{n} x+\sin^{n} x)

,

where

f_{n}(x)=\frac{1}{n}(\cos^{n} x+\sin^{n} x)

,

Now,

f_{4}(x)-f_{6}(x)=\frac{1}{4}(\cos^{4} x+\sin^{4} x)

-\frac{1}{6}(\cos^{6} x+\sin^{6} x)

=\frac{1}{4}[(\cos^{2} x+\sin^{2} x)^{2}-2\cos^{2} x\sin^{2} x]

-\frac{1}{6}[(\cos^{2} x)^{3}+(\sin^{2} x)^{3}]

=\frac{1}{4}[1-2\sin^{2}\cos^{2}x]-\frac{1}{6}[(\cos^{2} x+\sin^{2} x)^{3}

-3\sin^{2} x\cos^{2} x(\sin^{2} x+\cos^{2} x)]

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

-\frac{1}{6}[1-3\sin^{2} x\cos^{2} x]

=\frac{1}{4}-\frac{1}{2}\sin^{2} x\cos^{2} x-\frac{1}{6}

+\frac{1}{2}\sin^{2} x\cos^{2} x

=\frac{1}{4}-\frac{1}{6}=\frac{1}{12}

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