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