Functions Question 53
Question: Let $ f(x)=\frac{x^{2}-4}{x^{2}+4} $ for $ |x|\ >2 $ , then the function $ f:(-\infty ,\ -2]\cup [2,\ \infty )\to (-1,\ 1) $ is [Orissa JEE 2002]
Options:
A) One-one into
B) One-one onto
C) Many one into
D) Many one onto
Show Answer
Answer:
Correct Answer: C
Solution:
Let $ L{f}’,(2)\ne R{f}’,(2) $
Þ $ \frac{x^{2}-4}{x^{2}+4}=\frac{y^{2}-4}{y^{2}+4} $
Þ $ \frac{x^{2}-4}{x^{2}+4}-1=\frac{y^{2}-4}{y^{2}+4}-1,\Rightarrow x^{2}+4=y^{2}+4 $
Þ $ x=\pm y $ , \ $ f(x) $ is many-one. Now for each $ y\in (-1,,1), $ there does not exist $ x\in X $ such that $ f(x)=y $ . Hence f is into.
BETA
