JEE Main & Advanced Mathematics Functions Question Bank Self Evaluation Test - Relations and Functions-I

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?

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) \[f(E\cap F)=\{0\}\]

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 \[E\cap F=\{0\}\subset f(E)\cap f(F)\]

\[[since\,\,f(E)=f(F)\therefore f(E)\cap f(F)=f(E)=f(F)]\]

Also \[E\cap F=\{0\}\Rightarrow f(E\cap F)=\{0\}\]

Next, \[E\cup F=\{x\in R:-1\le x\le 1\}\]

and \[f(E)\cup f(F)=\{x\in R:0\le x\le 1\}\]

\[\therefore E\cup F\subset f(E)\cup f(F)\]