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UPSC NDA II 2014 Mathematics Paper

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For the next three (03) items that follow:

Given that

16\sin^5 x = p\sin 5x + q\sin 3x + r\sin x

[NDA II 2014]

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Question : 45 of 120

Marks: +1, -0

What is the value of r

5
8
10
-10

Solution:

Give that

16\sin^5 x = p\sin 5x + q\sin 3x + r\sin x

16\sin^5 x = p(16\sin^5 x + 5\sin 3x - 10\sin x) + q\sin 3x + r\sin x

\Rightarrow 16\sin^5 x = p(16\sin^5 x + 5(3\sin x - 4\sin^3 x) - 10\sin x) + q(3\sin x - 4\sin^3) + r\sin x

\Rightarrow 16\sin^5 x = p(16\sin^5 x + 15\sin x - 20\sin^3 x) + q(3\sin x - 4\sin^3 x) + r\sin x

\Rightarrow p = 1

and

-20p - 4q = 0 \Rightarrow q = -5

5p + 3q + r = 0

\Rightarrow 5 - 15 + r = 0

\Rightarrow r = 10

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