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