Functions Question 668
Question: $\begin{aligned} & f(x)=\begin{cases} x, & \text { if } \mathrm{x} \text { is rational } \\ 0, & \text { if } \mathrm{x} \text { is irrational }\end{cases}, g(x) = \begin{cases}0, & \text { if } \mathrm{x} \text { is rational } \\ x, & \text { if } \mathrm{x} \text { is irrational }\end{cases} \end{aligned}$
[IIT Screening 2005]
Options:
A) One-one onto
B) One-one not onto
C) Not one-one but onto
D) Not one-one not onto
Show Answer
Answer:
Correct Answer: A
Solution:
$ (f-g)(x)= \begin{cases} & x,x\in Q \\ & -x,x\notin Q \\ \end{cases} . $
BETA
