Functions Question 620
Question: If $ {e^{f(x)}}=\frac{10+x}{10-x},\ x\in (-10,\ 10) $ and $ f(x)=kf( \frac{200x}{100+x^{2}} ) $ , then $ k= $
[EAMCET 2003]
Options:
A) 0.5
B) 0.6
C) 0.7
D) 0.8
Show Answer
Answer:
Correct Answer: A
Solution:
$ {e^{f(x)}}=\frac{10+x}{10-x},,x\in (-10,,10) $
Þ $ f(x)=\log ( \frac{10+x}{10-x} ) $
Þ $ f( \frac{200x}{100+x^{2}} )=\log [ \frac{10+\frac{200x}{100+x^{2}}}{10-\frac{200x}{100+x^{2}}} ]=\log {{[ \frac{10(10+x)}{10(10-x)} ]}^{2}} $ $ =2\log ( \frac{10+x}{10-x} )=2f(x) $ \ $ f(x)=\frac{1}{2}f( \frac{200x}{100+x^{2}} )\Rightarrow k=\frac{1}{2}=0.5. $
BETA
