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