Q116 / 200 · Manipal (MET) 2012 · Mathematics

If $\lim _{x \rightarrow \infty}\left(\frac{x^3+1}{x^2+1}-(a x+b)\right)=2$, then

  • A
    $a=1, b=1$
  • B
    $a=1, b=2$
  • Correct answer
    C
    $a=1, b=-2$
  • D
    None of these

Solution

$\begin{aligned} & \lim _{x \rightarrow \infty}\left(\frac{x^3+1}{x^2+1}-(a x+b)\right)=2 \\ \Rightarrow \quad & \lim _{x \rightarrow \infty} \frac{x^3(1-a)-b x^2-a x+(1-b)}{x^2+1}=2 \\ \Rightarrow \quad & \lim _{x \rightarrow \infty} \frac{x(1-a)-b-\frac{a}{x}+\frac{1-b}{x^2}}{1+\frac{1}{x^2}}=2 \\ \Rightarrow \quad & 1-a=0 \text { and }-b=2 \\ \Rightarrow \quad & a=1 \text { and } b=-2\end{aligned}$

Hence, the correct answer is option: 'C'.

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