Question
Let $$f$$ be a function on $${\bf{R}}$$ given by $$f\left( x \right) = {x^2}$$ and let
$$E = \left\{ {x\, \in \,{\bf{R}}: - 1 \leqslant x \leqslant 0} \right\}$$ and
$$F = \left\{ {x\, \in \,{\bf{R}}:0 \leqslant x \leqslant 1} \right\}$$
then which of the following is false ?
A.
$$f\left( E \right) = f\left( F \right)$$
B.
$$E \cap F \subset f\left( E \right) \cap f\left( F \right)$$
C.
$$E \cup F \subset f\left( E \right) \cup f\left( F \right)$$
D.
$$f\left( {E \cap F} \right) = \left\{ 0 \right\}$$
Answer :
$$E \cup F \subset f\left( E \right) \cup f\left( F \right)$$
Solution :
$$\eqalign{
& {\text{We have }} - 1 \leqslant x \leqslant 0 \Rightarrow 0 \leqslant {x^2} \leqslant 1......({\text{i}}) \cr
& {\text{and }}0 \leqslant x \leqslant 1 \Rightarrow 0 \leqslant {x^2} \leqslant 1......({\text{ii}}) \cr
& \therefore \,E = \left\{ {x\, \in \,{\bf{R}}: - 1 \leqslant x \leqslant 0} \right\} \cr
& \Rightarrow f\left( E \right) = \left\{ {x\, \in \,{\bf{R}}:0 \leqslant x \leqslant 1} \right\}{\text{ from (i)}} \cr
& {\text{Also, }}F = \left\{ {x\, \in \,{\bf{R}}:0 \leqslant x \leqslant 1} \right\} \cr
& \Rightarrow f\left( F \right) = \left\{ {x\, \in \,{\bf{R}}:0 \leqslant x \leqslant 1} \right\}{\text{ from (ii)}} \cr
& {\text{Hence, }}f\left( E \right) = f\left( F \right) \cr
& {\text{Again }}E \cap F = \left\{ 0 \right\} \subset f\left( E \right) \cap f\left( F \right) \cr
& \left[ {{\text{Since }}f\left( E \right) = f\left( F \right)\,\therefore \,f\left( E \right) \cap f\left( F \right) = f\left( E \right) = f\left( F \right)} \right]{\text{ }} \cr
& {\text{Also }}E \cap F = \left\{ 0 \right\} \Rightarrow f\left( {E \cap F} \right) = \left\{ 0 \right\} \cr
& {\text{Next }}E \cup F = \left\{ {x\, \in \,{\bf{R}}: - 1 \leqslant x \leqslant 1} \right\} \cr
& {\text{and }}f\left( E \right) \cup f\left( F \right) = \left\{ {x\, \in \,{\bf{R}}:0 \leqslant x \leqslant 1} \right\} \cr
& \therefore \,E \cup F \subset f\left( E \right) \cup f\left( F \right) \cr} $$