Functions Question 386
Question: Let f be a function on R given by $ f(x)=x^{2} $ and let $ E={x\in R:-1\le x\le 0} $ And $ F={x\in R:0\le x\le 1} $ then which of the following is false?
Options:
A) $ f(E)=f(F) $
B) $ E\cap F\subset f(E)\cap f(F) $
C) $ E\cup F\subset f(E)\cup f(F) $
D) $ f(E\cap F)={0} $
Show Answer
Answer:
Correct Answer: C
Solution:
[c] We have $ -1\le x\le 0\Rightarrow 0\le x^{2}\le 1 $ (i) And $ 0\le x\le 1\Rightarrow 0\le x^{2}\le 1 $ (ii)
$ \therefore E={x\in R:-1\le x\le 0} $
$ \Rightarrow f(E)={x\in R:0\le x\le 1} $ From (i) Also $ F={x\in R:0\le x\le 1} $
$ \Rightarrow f(F)={x\in R:0\le x\le 1} $ From (ii) Hence, $ f(E)=f(F) $ Again $ E\cap F={0}\subset f(E)\cap f(F) $ $ [sincef(E)=f(F)\therefore f(E)\cap f(F)=f(E)=f(F)] $ Also $ E\cap F={0}\Rightarrow f(E\cap F)={0} $ Next, $ E\cup F={x\in R:-1\le x\le 1} $ and $ f(E)\cup f(F)={x\in R:0\le x\le 1} $
$ \therefore E\cup F\subset f(E)\cup f(F) $
BETA
