Relations and Functions — Class 12 Maths Solution

It is given that,, $A = [ - 1,1]$

(i) $f(x) = \frac{x}{2}$

Let $f\left( {{x_1}} \right) = f\left( {{x_2}} \right)$

$\Rightarrow$ $\frac{{{x_1}}}{2} = \frac{{{x_2}}}{2} \Rightarrow {x_1} = {x_2}$

So, $f(x)$ is one-one.

Now, let $y = \frac{x}{2}$

$\Rightarrow$ $x = 2y \notin A,\forall y \in A$

As for $y = 1 \in A,x = 2 \notin A$

So, $f(x)$ is not onto.

Also, $f(x)$ is not bijective as it is not onto.
(ii) $g(x) = |x|$

Let $g\left( {{x_1}} \right) = g\left( {{x_2}} \right)$

$\Rightarrow$ $\left| {{x_1}} \right| = \left| {{x_2}} \right| \Rightarrow {x_1} = \pm {x_2}$

So, $g(x)$ is not one-one.

Now, $y = |x| \Rightarrow x = \pm y \notin A,\forall y \in A$

So, $g(x)$ is not onto, also, $g(x)$ is not bijective.

(iii) $h(x) = x|x|$

Let $h\left( {{x_1}} \right) = h\left( {{x_2}} \right)$

$\Rightarrow$ ${x_1}\left| {{x_1}} \right| = {x_2}\left| {{x_2}} \right| \Rightarrow {x_1} = {x_2}$

So, $h(x)$ is one-one.
Now, let $y = x|x|$

$\Rightarrow$ $y = {x^2} \in A,\forall x \in A$

So, $h(x)$ is onto also, $h(x)$ is a bijective.
(iv) $k(x) = {x^2}$

Let $k\left( {{x_1}} \right) = k\left( {{x_2}} \right)$

$\Rightarrow$ $x_1^2 = x_2^2 \Rightarrow {x_1} = \pm {x_2}$

Thus, $k(x)$ is not one-one.
Now, let $y = {x^2}$

$\Rightarrow$ $x = \sqrt y \notin A,\forall y \in A$

As for $y = - 1,x = \sqrt { - 1} \notin A$

Hence we can say that, $k(x)$ is neither one-one nor onto.