Functions Question 436
Question: Let $ f(x)=\frac{\alpha x^{2}}{x+1},x\ne -1. $ The value of $ \alpha $ for which $ f(a)=a,(a\ne 0) $ is
Options:
A) $ 1-\frac{1}{a} $
B) $ \frac{1}{a} $
C) $ 1+\frac{1}{a} $
D) $ \frac{1}{a}-1 $
Show Answer
Answer:
Correct Answer: C
Solution:
[c] Given $ f(x)=\frac{\alpha x^{2}}{x+1},x\ne -1;f(a)=a $ $ \frac{\alpha a^{2}}{a+1}=a\Rightarrow \alpha =1+\frac{1}{a}. $
BETA
