Relations and Functions — Class 12 Maths Solution

(i) If e is an identity element, then

a $*$ e $=$ a$=$ e $*$ a $\forall a \in Q$

$\Rightarrow$ a$-$e$=$ e$-$a $=$ $\forall a \in Q$
$\Rightarrow$ a$-$ e $=$ a and e $-$ a$=$ a

$\Rightarrow$ e $=$ 0 and e$=$ 2a $\forall a \in Q$

Which is not possible. Hence, identity element does not exist,

(ii) If e is an identity element, then

a $*$ e$=$ a$=$ e$*$ a $\forall a \in Q$
$\Rightarrow$ $a^2 + e^2 = e^2 + a^2 =$ a $\forall a \in Q$.

$\Rightarrow$ e $=$ $\sqrt {a - {a^2}}$ $\forall a \in Q$

Which is not possible. Hence, identity element does not exist,

(iil) If e is an identity element, then

a$*$ e$=$ a$=$ e$*$ $\forall a \in Q$

$\Rightarrow$ a + ae$=$ e + ae$=$ a $\forall a \in Q$.

$\Rightarrow$ a + ae$=$ a and e + ae $=$ a

$\Rightarrow$ e $=$ 0 and e $=$ $\cfrac{a}{{1 + a}}$ $\forall a \in Q$

Which is not possible. Hence, identity element does not exist,

(iv) If e is an identity element, then

a$*$e$=$ a$=$e$*$ a $\forall a \in Q$

$\Rightarrow$ $(a$ - $e)^2 =$ (e$- a)^2 =$ a $\forall a \in Q$.
$\Rightarrow$ (a$- e)^2 =$ a and (e$- a)^2 =$ a

$\Rightarrow$ a $- e = \pm \sqrt a$ and e $-$ a $= \pm \sqrt a$
$\Rightarrow$ e $= a \pm \sqrt a$ and e $= a \pm \sqrt a$ $\forall a \in Q$

Which is not possible. Hence, there is no identity element.

(v) If e is an identity element, then

a $*$ e$=$ a$=$ e $*$ a $\forall a \in Q$.

$\Rightarrow$ $\cfrac{{ae}}{4} = \cfrac{{ea}}{4} = a$ $\forall a \in Q$.

So, 4 is the identity element.

(vi) If e an identity element, then
a$*$ e$=$ a$=$e $*$ a $\forall a \in Q$.
$\Rightarrow$ $e^2 = ea^2 =$ a .
$\Rightarrow$ $ae^2 =$ a and $ea^2 =$ a

$\Rightarrow$ $e = \pm 1$ and e $= \cfrac{1}{a}$

Which is not possible. Hence, identity element does not exist.