Q21 / 60 · JEE Advanced 2012 (Paper 1) · Mathematics
If $\lim _{x \rightarrow \infty}\left(\frac{x^{2}+x+1}{x+1}-a x-b\right)=4$, then
- A$a=1, b=4$
- Correct answerB$a=1, b=-4$
- C$a=2, b=-3$
- D$a=2, b=3$
Solution
Given: $\lim _{x \rightarrow \infty}\left(\frac{x^{2}+x+1}{x+1}-a x-b\right)=4$
$\begin{array}{l}
\Rightarrow \lim _{x \rightarrow \infty} \frac{x^{2}+x+1-a x^{2}-a x-b x-b}{x+1}=4 \\
\Rightarrow \lim _{x \rightarrow \infty} \frac{(1-a) x^{2}+(1-a-b) x+(1-b)}{x+1}=4
\end{array}$
For this limit to be finite $1-a=0 \Rightarrow a=1$ then given limit reduces to
$\lim _{x \rightarrow \infty} \frac{-b x+(1-b)}{x+1}=4 \Rightarrow \lim _{x \rightarrow \infty} \frac{-b+\frac{(1-b)}{x}}{1+\frac{1}{x}}=4$
$\Rightarrow-b=4$ or $b=-4, \therefore a=1, b=-4$
Hence, the correct answer is option: 'B'.