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UPSC NDA I 2015 Mathematics Solved Paper

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For the next two (2) items that follow :

Given that

\frac{d}{dx}\left(\frac{1+x^2+x^4}{1+x+x^2}\right)=Ax+B

.

[NDA I 2015]

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Question : 92 of 120

Marks: +1, -0

What is the value of B?

$-1$
1
2
4

Solution:

\frac{d}{dx}\left(\frac{1+x^2+x^4}{1+x+x^2}\right)

=\frac{d}{dx}\left(\frac{1+x+x^2-x+x^4}{1+x+x^2}\right)

=\frac{d}{dx}\left(1+\frac{x(x^3-1)}{1+x+x^2}\right)

=\frac{d}{dx}\left(1+\frac{x(x-1)(1+x+x^2)}{1+x+x^2}\right)

=\frac{d}{dx}\left(1+x(x-1)\right)

=2x-1

Comparing with the expression in the question, we get A = 2 and B = -1

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