Functions Question 285
Question: If $ \underset{x\to \infty }{\mathop{\lim }},{{( 1+\frac{a}{x}+\frac{b}{x^{2}} )}^{2x}}=e^{2}, $ then the values of a and b are
[AIEEE 2004]
Options:
A) $ a=1,\ b=2 $
B) $ \cos (|x|),-|x| $
C) $ a\in R,\ b=2 $
D) $ a\in R,\ b\in R $
Show Answer
Answer:
Correct Answer: B
Solution:
Since, $ \underset{x\to \infty }{\mathop{\lim }},( 1+\frac{a}{x}+\frac{b}{x^{2}} )=e^{2} $ \ $ \underset{x\to \infty }{\mathop{\lim }},{{[ {{( 1+\frac{ax+b}{x^{2}} )}^{\frac{x^{2}}{ax+b}}} ]}^{\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 $ .
BETA
