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

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Neither injective nor surjective.


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


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.


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


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.


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



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.

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