# Let A = [−1, 1]. Then, discuss whether the following functions from A to itself are one-one, onto or bijective:

Question:

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]

Solution:

(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.