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Question: Classify the following function \(F(x)\) defined in \(R \to R\) as injective, surjective, both or no...

Classify the following function F(x)F(x) defined in RRR \to R as injective, surjective, both or none
F(x) = x2+x+1x2x+1\dfrac{{{x^2} + x + 1}}{{{x^2} - x + 1}}
A) Injective but not surjective
B) Surjective but not injective
C) Neither surjective nor injective
D) Both injective and surjective

Explanation

Solution

The function F(x)F(x) is said to be injective, that is F(a)=F(b)F(a) = F(b) if and only if a=ba = b. So, here we have to examine whether the given function is injective or not by putting values of x at respective points and then equating it. The function F(x)F(x) is said to be surjective if for every y in the co-domain there exists x in the domain such that F(x)=yF(x) = y.

Complete step-by-step answer:
Firstly, we will check the given function is injective or not:
Let F(a)=F(b)F(a) = F(b) a,bR\forall a,b \in R
a2+a+1a2a+1=b2+b+1b2b+1\Rightarrow \dfrac{{{a^2} + a + 1}}{{{a^2} - a + 1}} = \dfrac{{{b^2} + b + 1}}{{{b^2} - b + 1}}
Since it cannot be further solved, so, the given function is not injective.
Take a=1a = 1 and b=1b = - 1 then F(1)F(1)F(1) \ne F( - 1), then also it is not injective.
Secondly, we will check that given function is surjective or not:
Let x2+x+1x2x+1=t\dfrac{{{x^2} + x + 1}}{{{x^2} - x + 1}} = t (say)
On further solving it, we will get the equation:
x2+x+1=t(x2x+1){x^2} + x + 1 = t({x^2} - x + 1)
(t1)x2+(t+1)x+(t1)=0\Rightarrow (t - 1){x^2} + (t + 1)x + (t - 1) = 0
Now, finding discriminant of the above, we will get:
(3t1)(3t)0(3t - 1)(3 - t) \geqslant 0
So, it means t[13,3]t \in \left[ {\dfrac{1}{3},3} \right], which is the range of F(x)F(x).
The co-domain and range of F is not the same, hence, the given function is not surjective.

Since, the given function is neither surjective nor injective, hence, the correct answer is option C.

Note: 1) Any function is said to be injective if and only if F(a)=F(b)F(a) = F(b) whenever a=ba = b. If this is not the case, the given function is not injective.
2) Any function is said to be surjective if and only if F(x)=yF(x) = y for all x, y belonging to the given domain.
3) Another way to find if a function is surjective or not is that if the co-domain and range of the function is the same, it is surjective.
4) A function is called bijective if it is both one-one and onto. 5) Injective function is also called as one one function.
6) Surjective function is also called as onto function.