Question

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

Answer 1

Hint: 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
$f\left( x \right) = \dfrac{{{x^2} + 2x + 5}}{{{x^2} + x + 1}}$
Let’s first consider the numerator of the given function.
${x^2} + 2x + 5$
Now, we will find the discriminant of this quadratic equation i.e.
$ \Rightarrow D = {2^2} - 4 \cdot 1 \cdot 5$
On further calculation, we get
 $ \Rightarrow D = 4 - 20 = - 16 < 0$
Therefore,
${x^2} + 2x + 5 > 0$
Let’s consider the denominator of the given function.
${x^2} + x + 1$
Now, we will find the discriminant of this quadratic equation i.e.
$ \Rightarrow D = {1^2} - 4 \cdot 1 \cdot 1$
On further calculation, we get
 $ \Rightarrow D = 1 - 4 = - 3 < 0$
Therefore,
${x^2} + x + 1 > 0$
As the numerator and denominator of the function is positive. So the given function is always greater than zero i.e. $f\left( x \right) > 0$
We can say that the range of a given function does not include all real numbers i.e. $range \ne R$ . 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 $x$.
We can write the given function is
$
  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}} \\
$
$ \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
$ \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
$
   \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'\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.