Functions Question 673
Question: Range of the function $ f(x)=\frac{x^{2}}{x^{2}+1} $ is
[Orissa JEE 2005]
Options:
A) (?1, 0)
B) (?1, 1)
C) [0, 1)
D) (1, 1)
Show Answer
Answer:
Correct Answer: C
Solution:
Let $ y=\frac{x^{2}}{x^{2}+1} $
Þ $ (y-1)x^{2}+0x+y=1,y\ne 1 $ for real values of x, we have $ D\ge 0\Rightarrow -4y(y-1)\ge 0\Rightarrow y(y-1)\le 0\Rightarrow y\in [0,,1) $ $ 0\le \frac{x^{2}}{x^{2}+1}<1 $ .
BETA
