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UPSC NDA I 2015 Mathematics Paper

Question Numbers: 91-92

Given that:

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

Question : 91 of 120

Marks: +1, -0

What is the value of A?

Solution:

Concept:

We differentiate the given rational expression after simplifying it by factoring the numerator.

Explanation:

First, factor the numerator. Note that

1 + x^{2} + x^{4} = x^{4} + x^{2} + 1

This can be expressed as

(x^{2} + x + 1)(x^{2} - x + 1)

Then the given expression becomes:

\frac{1+x^{2}+x^{4}}{1+x+x^{2}} = \frac{(x^{2}+x+1)(x^{2}-x+1)}{x^{2}+x+1} = x^{2} - x + 1

Now differentiate:

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

Comparing with the right side

Ax + B

, we get

A = 2

and

B = -1

Answer:

A = 2

, so the correct option is C.

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