#### Explain solution RD Sharma class 12 chapter Functions exercise 2.1 question 5 sub question (vi) maths

Neither injective nor surjective.

Given:

$f:Z \rightarrow Z$ defined by $f(x)=x^2+x$.

Hint:

One-one function means every element in the domain has a distinct image in the  co-domain.

Onto function means every element in the co-domain has at least one  pre image in the domain of function.

Bijective ⇒ Function should fulfill the injective, surjective condition.

Solution:

Let us check if the given function is an injective, surjective, bijective.

Injection:

Let x,y be any two elements in domain $f(x)=f(y)$.

$f(x)=x^2+x, f(y )=y^2+y$

Here we cannot say that x=y.

For example,

$x=2$ and $y=-3$

Then    $x^2+x=2^2+2 =6$

$y^2+y=(-3)^2-3 =6$

Therefore, we have two number 2 and -3  in the domain z whose image is same as 6.

So f  is not an injective.

Surjection:

Let y  be any element in co-domain (R) such that $f(x)=(y)$ for some element x in (R)

$f(x)=(y)$

$x^2+x=y$

Here we can’t say $x\in Z$.

\begin{aligned} &y=-4\\ &x^{2}+x=-4\\ &x^{2}+x+4=0 \text { , on solving this }\\ &x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a} \text { (formula) }\\ &a=1, b=1, c=4\\ &x=\frac{-1 \pm \sqrt{1^{2}-4 \cdot 1 \cdot 4}}{2 \cdot 1}\\ &x=\frac{-1 \pm \sqrt{-15}}{2 \cdot 1}\\ &x=\frac{-1 \pm i \sqrt{15}}{2 \cdot 1} \text { which is not in } Z \end{aligned}

So, f  is not  surjective and  f is not a bijective.