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UPSC NDA I 2015 Mathematics Solved Paper

For the next two (2) items that follow :

Given that

\lim\limits_{x\to\infty}\left(\frac{2+x^2}{1+x}-Ax-B\right)=3

[NDA I 2015]

Question : 93 of 120

Marks: +1, -0

What is the value of A ?

[NDA I 2015]

Validate

Solution:

Consider the following transformation

x=\frac{1}{n}

x\rightarrow\infty,\ n\rightarrow0

f(x)=\frac{2+\frac{1}{n^2}}{1+\frac{1}{n}}-\frac{A}{n}-B

=\frac{2n^2+1}{n^2+n}-\frac{A}{n}-B

=\frac{2n^2+1-A(n+1)-B(n^2+n)}{n^2+n}

=\frac{n^2(2-B)-n(A+B)+1-A}{n^2+n}

Now as ‘n’ tends to 0 the numerator and denominator, both should yield 0 to obtain an indefinite form. Hence for this to happen, the value of A has to 1.

f(x)=\frac{n^2(2-B)-n(1+B)}{n^2+n}

\lim\limits_{n\to 0}f(x)

=\lim\limits_{n\to 0}\left[\frac{n^2(2-B)-n(1+B)}{n^2+n}\right]

=\lim\limits_{n\to 0}\left[\frac{2n(2-B)-(1+B)}{2n+1}\right]

(Using L Hopital's rule )

=\frac{-(1+B)}{1}

=3

B+1=-3

B=-4

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