A) \[a=1\] and \[b=2\]
B) \[a=1,b\in R\]
C) \[a\in R,b=2\]
D) \[a\in R,b\in R\]
Correct Answer: B
Solution :
[b] We know that \[\underset{x\to \infty }{\mathop{\lim }}\,{{\left( 1+x \right)}^{\frac{1}{x}}}=e\] We have \[\underset{x\to \infty }{\mathop{\lim }}\,{{\left( 1+\frac{a}{x}+\frac{b}{{{x}^{2}}} \right)}^{2x}}={{e}^{2}}\] \[\Rightarrow \underset{x\to \infty }{\mathop{\lim }}\,{{\left[ {{\left( 1+\frac{a}{x}+\frac{b}{{{x}^{2}}} \right)}^{\left( \frac{1}{\frac{a}{x}+\frac{b}{{{x}^{2}}}} \right)}} \right]}^{2x\left( \frac{a}{x}+\frac{b}{{{x}^{2}}} \right)}}={{e}^{2}}\] \[\Rightarrow {{e}^{\underset{x\,\to \,\infty }{\mathop{\lim }}\,2}}^{\left[ a+\frac{b}{x} \right]}={{e}^{2}}\Rightarrow {{e}^{2a}}={{e}^{2}}\] \[\Rightarrow a=1\] and \[b\in R\]
You need to login to perform this action.
You will be redirected in
3 sec
