Question

Let 

\[A = \left\{ x \in R : - 1 \leq x \leq 1 \right\} = B\] Then, the mapping\[f : A \to \text{B given by} f\left( x \right) = x\left| x \right|\] is 

 

पर्याय

• injective but not surjective

• surjective but not injective

• bijective

• none of these

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.

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

From the 3 cases, we can conclude that  f is one-one.
Surjectivity:
Let y be an element in the co-domain, such that y = f (x)

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

⇒ f is onto.

⇒ f is a bijection.

So, the answer is (c).