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Question: If \(f:R \to R\) be given by \(f\left( x \right) = \dfrac{{{x^2} + 2x + 5}}{{{x^2} + x + 1}}\) , the...

If f:R→Rf:R \to R be given by f(x)=x2+2x+5x2+x+1f\left( x \right) = \dfrac{{{x^2} + 2x + 5}}{{{x^2} + x + 1}} , then ff is
(1)One-one and onto
(2)Many one and onto
(3)One-one and into
(4)Many-one and into

Explanation

Solution

Here we need to check whether the given function is into, onto or many-one function. For that, we will use the definition and condition for a function to be onto, into or many-one function one by one and from there, we will find the type of the given function.

Complete step-by-step answer:
The given function is
f(x)=x2+2x+5x2+x+1f\left( x \right) = \dfrac{{{x^2} + 2x + 5}}{{{x^2} + x + 1}}
Let’s first consider the numerator of the given function.
x2+2x+5{x^2} + 2x + 5
Now, we will find the discriminant of this quadratic equation i.e.
⇒D=22−4⋅1⋅5\Rightarrow D = {2^2} - 4 \cdot 1 \cdot 5
On further calculation, we get
⇒D=4−20=−16<0\Rightarrow D = 4 - 20 = - 16 < 0
Therefore,
x2+2x+5>0{x^2} + 2x + 5 > 0
Let’s consider the denominator of the given function.
x2+x+1{x^2} + x + 1
Now, we will find the discriminant of this quadratic equation i.e.
⇒D=12−4⋅1⋅1\Rightarrow D = {1^2} - 4 \cdot 1 \cdot 1
On further calculation, we get
⇒D=1−4=−3<0\Rightarrow D = 1 - 4 = - 3 < 0
Therefore,
x2+x+1>0{x^2} + x + 1 > 0
As the numerator and denominator of the function is positive. So the given function is always greater than zero i.e. f(x)>0f\left( x \right) > 0
We can say that the range of a given function does not include all real numbers i.e. range≠Rrange \ne R . Hence, the range of a given function is not equal to the co domain of the function.
Therefore, the given function is an into function.
Now, we will differentiate the given function with respect to xx.
We can write the given function is
f(x)=x2+x+1x2+x+1+x+4x2+x+1 f(x)=1+x+4x2+x+1  f\left( x \right) = \dfrac{{{x^2} + x + 1}}{{{x^2} + x + 1}} + \dfrac{{x + 4}}{{{x^2} + x + 1}} \\\ f\left( x \right) = 1 + \dfrac{{x + 4}}{{{x^2} + x + 1}} \\\
⇒df(x)dx=f′(x)=ddx(1+x+4x2+x+1)\Rightarrow \dfrac{{df\left( x \right)}}{{dx}} = f'\left( x \right) = \dfrac{d}{{dx}}\left( {1 + \dfrac{{x + 4}}{{{x^2} + x + 1}}} \right)
On differentiating the function, we get
⇒f′(x)=0+(x2+x+1)−(x+4)(2x+1)(x2+x+1)2\Rightarrow f'\left( x \right) = 0 + \dfrac{{\left( {{x^2} + x + 1} \right) - \left( {x + 4} \right)\left( {2x + 1} \right)}}{{{{\left( {{x^2} + x + 1} \right)}^2}}}
On multiplying the terms, we get
⇒f′(x)=(x2+x+1−2x2−8x−x−4)(x2+x+1)2 ⇒f′(x)=−x2−8x−3(x2+x+1)2  \Rightarrow f'\left( x \right) = \dfrac{{\left( {{x^2} + x + 1 - 2{x^2} - 8x - x - 4} \right)}}{{{{\left( {{x^2} + x + 1} \right)}^2}}} \\\ \Rightarrow f'\left( x \right) = \dfrac{{ - {x^2} - 8x - 3}}{{{{\left( {{x^2} + x + 1} \right)}^2}}} \\\
For f′(x)=0 x∈R  f'\left( x \right) = 0 \\\ x \in R \\\
Therefore, the given function is many one function.
Hence, the given function is many one and into function.
Thus, the correct option is option 4.

Note: Since, the given function is onto and many one function. Function sets the relationship between domain and range of the function. Remember that a function cannot be into and one-one function at the same time and also many-one and one-one functions at the same time.