Question

Let

\[A = \left\{ x : - 1 \leq x \leq 1 \right\} \text{and} f : A \to \text{A such that f}\left( x \right) = x|x|\]

 

Options

• a bijection

• injective but not surjective

• surjective but not injective

• neither injective nor surjective

Answer 1

Injectivity:
Let x and y be any two elements in the domain A.
Case-1: Let x and y be two positive numbers, such that\[f\left( x \right) = f\left( y \right)\] 
\[ \Rightarrow x\left| x \right| = y\left| y \right|\] 
\[ \Rightarrow x\left( x \right) = y\left( y \right)\] 
\[ \Rightarrow x^2 = y^2 \] 
\[ \Rightarrow x = y\] 

Case-2: Let x and y be two negative numbers, such that

\[f\left( x \right) = f\left( y \right)\] 
\[ \Rightarrow x\left| x \right| = y\left| y \right|\] 
\[ \Rightarrow x\left( - x \right) = y\left( - y \right)\] 
\[ \Rightarrow - x^2 = - y^2 \] 
\[ \Rightarrow x^2 = y^2 \] 
\[ \Rightarrow x = y\] 

Case-3: Let x be positive and y be negative.

\[\text{Then},x \neq y\] 
\[ \Rightarrow f\left( x \right) = x\left| x \right| \text{is positive and }f\left( y \right) = y\left| y \right| \text{is negative}\] 
\[ \Rightarrow f\left( x \right) \neq f\left( y \right)\] 
\[So, x \neq y\] 
\[ \Rightarrow f\left( x \right) \neq f\left( y \right)\]

So, f is one-one.

Surjectivity:
Let y be an element in the co-domain, such that y = f (x)

\[Case-1: \text{Let } y>0. \text{Then}, 0<y\leq1\] 
\[y = f\left( x \right) = x\left| x \right| > 0\] 
\[ \Rightarrow x > 0\] 
\[ \Rightarrow \left| x \right| = x\] 
\[ \Rightarrow f\left( x \right) = y\] 
\[ \Rightarrow x\left| x \right| = y\] 
\[ \Rightarrow x\left( x \right) = y\] 
\[ \Rightarrow x^2 = y\] 
\[ \Rightarrow x = \sqrt{y} \in A \left( \text{We do not get}\pm, \text{as } x > 0 \right)\] \[\] \[Case-2: Lety<0.\text{Then},-1\leq y<0\] 
\[y = f\left( x \right) = x\left| x \right| < 0\] 
\[ \Rightarrow x < 0\] 
\[ \Rightarrow \left| x \right| = - x\] 
\[ \Rightarrow f\left( x \right) = y\] 
\[ \Rightarrow x\left| x \right| = y\] 
\[ \Rightarrow x\left( - x \right) = y\] 
\[ \Rightarrow - x^2 = y\] 
\[ \Rightarrow x^2 = - y\] 
\[ \Rightarrow x = - \sqrt{- y} \in A \left( \text{We do not get}\pm, \text{ as } x>0 \right)\]

\[\Rightarrow\]f is onto

\[\Rightarrow\] f is a bijection.
So, the answer is (a).