A) \[a=1\]and \[b=1\]
B) \[a=1\]and \[b=-1\]
C) \[a=1\]and \[b=-2\]
D) \[a=1\]and \[b=2\]
Correct Answer: C
Solution :
\[\underset{x\to \infty }{\mathop{\lim }}\,\left[ \frac{{{x}^{3}}+1}{{{x}^{2}}+1}-\frac{(ax+b)}{1} \right]=2\] \[\underset{x\to \infty }{\mathop{\lim }}\,\frac{(1-a)-b{{x}^{2}}-ax+(1-b)}{{{x}^{2}}+1}=2\] or \[\underset{x\to \infty }{\mathop{\lim }}\,[{{x}^{3}}(1-a)-b{{x}^{2}}-ax+(a-b)]=2\] Comparing the coefficients of both sides, we get \[1-a=0,\] \[-b=2\] \[a=1,\] \[b=-2\]
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