A) \[a=0,\,\,b=0\]
B) \[a=1,\,\,b=-1\]
C) \[a=-1,\,\,b=1\]
D) \[a=2,\,\,b=-1\]
Correct Answer: B
Solution :
\[\underset{x\to \infty }{\mathop{\lim }}\,\left( \frac{{{x}^{2}}+1}{x+1}-ax-b \right)=0\] \[\Rightarrow \underset{x\to \infty }{\mathop{\lim }}\,\left[ \frac{{{x}^{2}}(1-a)-(a+b)x-b+1}{x+1} \right]=0\] \[\Rightarrow \,1-a=0\]and\[a+b=0\] \[\Rightarrow a=1,\,b=-1\]
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