A) \[a=1,\,\,b=2\]
B) \[a=2,\,\,b=1\]
C) \[a=1,\,\,b\in R\]
D) None of these
Correct Answer: C
Solution :
Since,\[\underset{x\to \infty }{\mathop{\lim }}\,{{\left( 1+\frac{a}{x}+\frac{b}{{{x}^{2}}} \right)}^{2x}}-{{e}^{2}}\] \[\Rightarrow \] \[\underset{x\to \infty }{\mathop{\lim }}\,{{\left( 1+\frac{ax+b}{{{x}^{2}}} \right)}^{2x}}-{{e}^{2}}\] ... (i) \[\Rightarrow \] \[\underset{x\to \infty }{\mathop{\lim }}\,=\frac{ax+b}{{{x}^{2}}}\]must be equal to zero. \[\Rightarrow \] \[\underset{x\to \infty }{\mathop{\lim }}\,\frac{ax+b}{{{x}^{2}}}=0\] \[\Rightarrow \] \[\underset{x\to \infty }{\mathop{\lim }}\,\left( \frac{a}{x}+\frac{b}{{{x}^{2}}} \right)=0\] \[\Rightarrow \] \[a\]and\[b\ne 0\] \[\therefore \]From Eq. (i) \[{{e}^{\underset{x\to \infty }{\mathop{\lim }}\,\frac{2(ax+b)}{x}}}={{e}^{2}}\] \[\Rightarrow \] \[{{e}^{2a}}={{e}^{2}}\] \[\Rightarrow \] \[a=1\]and\[b\in R\]
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