Let A = [−1, 1]. Then, discuss whether the following functions from A to itself are one-one, onto or bijective:
Let A = [−1, 1]. Then, discuss whether the following functions from A to itself are one-one, onto or bijective:
(i) $f(x)=\frac{x}{2}$
(ii) $g(x)=|x|$
(iii) $h(x)=x^{2}$
[NCERT EXEMPLAR]
(i) $f: A \rightarrow A$, given by $f(x)=\frac{x}{2}$\
Injection test:
Let x and y be any two elements in the domain (A), such that f(x) = f(y).
$f(x)=f(y)$
$\frac{x}{2}=\frac{y}{2}$
$x=y$
So, f is one-one.
Surjection test:
Let y be any element in the co-domain (A), such that f(x) = y for some element x in A (domain)
x = 2y, which may not be in A.
For example, if y = 1, then
x = 2, which is not in A.
So, f is not onto.
So, f is not bijective.
(ii) g(x) = |x|
Injection test:
Let x and y be any two elements in the domain (A), such that f(x) = f(
$f(x)=f(y)$
$|x|=|y|$
$x=\pm y$
So, f is not one-one.
Surjection test:
For y = −1, there is no value of x in A.
So, f is not onto.
So, f is not bijective.
(iii) $h(x)=x^{2}$
Injection test:
Let x and y be any two elements in the domain (A), such that f(x) = f(y).
$f(x)=f(y)$
$x^{2}=y^{2}$
$x=\pm y$
So, f is not one-one.
Surjection test:
For y = −1, there is no value of x in A.
So, f is not onto.
So, f is not bijective.