A) \[a=1,\text{ }b=-2\]
B) \[a=-1,\text{ }b=2\]
C) \[a=-1,\text{ }b=-2\]
D) None of these
Correct Answer: A
Solution :
\[\underset{x\to \infty }{\mathop{\lim }}\,\left[ \frac{{{x}^{3}}+1}{{{x}^{2}}+1}-(ax+b) \right]=2\] \[\Rightarrow \]\[=\underset{x\to \infty }{\mathop{\lim }}\,\frac{{{x}^{3}}(1-a)-b{{x}^{2}}-ax+(1-b)}{{{x}^{2}}+1}=2\] \[\Rightarrow \]\[1-a=0\]and\[-b=2\Rightarrow a=1,b=-2\]
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