A) \[a=1,\text{ }b=1\]
B) \[a=1,\text{ }b=2\]
C) \[a=1,\text{ }v=-2\]
D) None of these
Correct Answer: C
Solution :
[c] \[\underset{x\to \infty }{\mathop{\lim }}\,\left( \frac{{{x}^{3}}+1}{{{x}^{2}}+1}-(ax+b) \right)=2\] Or \[\underset{x\to \infty }{\mathop{\lim }}\,\frac{{{x}^{3}}(1-a)-b{{x}^{3}}-ax+(1-b)}{{{x}^{2}}+1}=2\] Or \[1-a=0\] and \[-b=2\] Or \[a=1,\]\[b=-2\]You need to login to perform this action.
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