A) \[a=1,\ b=2\]
B) \[\cos (|x|)\,-|x|\]
C) \[a\in R,\ b=2\]
D) \[a\in R,\ b\in R\]
Correct Answer: B
Solution :
Since, \[\underset{x\to \infty }{\mathop{\lim }}\,\left( 1+\frac{a}{x}+\frac{b}{{{x}^{2}}} \right)={{e}^{2}}\] \ \[\underset{x\to \infty }{\mathop{\lim }}\,{{\left[ {{\left( 1+\frac{ax+b}{{{x}^{2}}} \right)}^{\frac{{{x}^{2}}}{ax+b}}} \right]}^{\frac{2(ax+b)}{x}}}={{e}^{2}}\] Þ \[\underset{x\to \infty }{\mathop{\lim }}\,{{e}^{\frac{2(ax+b)}{x}}}={{e}^{2}}\Rightarrow \underset{x\to \infty }{\mathop{\lim }}\,\frac{2(ax+b)}{x}=2\]Þ \[2a=2\Rightarrow a=1\] Thus \[a=1\] and \[b\in R\].
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