Browse Questions

# Let A=[-1,1].Then,dicuss whether the following functions defined on A are one-one,onto or bijective$(iii)\quad h(x)\;=\;x|x|$

Note: This is the 3rd part of a  4 part question, which is split as 4 separate questions here.

Toolbox:
• 1.A function f on set A is one-one if $f(x)=f(y) =>x,y \in A$
• 2.A function f on set A is onto if for every $y \in A$ then exists $x \in A$ such that $f(x)=y$
• 3.A function is bijective if it is both one-one and onto
(iii)$h(x)=x |x| \qquad x \in [-1,1]$

case I when x takes -ve values

Let $x_1=-p_1\qquad x_2=-p_2\qquad p_1p_2 +ve$

Let $h(x_1)=h(x_2)$

$h(x_1)=-p_1|p_1|=h(x_2)=-p_2|p_2|$

$-p_1 \times p_1=-p_2p_2$

$p_1^2=p_2^2$

$=>p_1=p_2 \qquad (p_1,p_2 +ve)$

Let $h(x_1)=h(x_2)$

case I when x takes +ve values

$x_1=q_1 \qquad x_2=q_2$

$h(x_1)=q_1|q_1|=h(x_2)=q_2|q_2|$

$q_1^2=q_2^2$

$q_1=q_2$

Hence h is a one one function

Let $y \in [-1,1]$

$y=x|x|$

$=-x^2$ if y is -ve

$=x^2$ when y is +ve

and since $(-y)=x^2\; for\;y -ve$

$x =\sqrt {-y} \qquad \in [-1,1]\; since\; y \;is\; -ve [-1,1]$

and $y=x^2$

$x=\sqrt y$ when y is +ve [-1,1]

Hence for every $y \in [-1,1]$ then exists an element x in [-1,1] such that $R(x)=y$

R is onto function

Hence R is both one-one and onto