Question

If x is real, then the minimum value of $y = \dfrac{{{x^2} - x + 1}}{{{x^2} + x + 1}}$.
(A) $3$
(B) $\dfrac{1}{3}$
(C) $\dfrac{1}{2}$
(D) $2$

Answer 1

Hint: In the given question, we have to find the range of the given rational function in the variable x for all possible real values of x. So, in order to find the range of the function, we will first consider the function as y, and then form a quadratic equation in x. Now, we know that the value of x belongs to the real number set. So, we put some conditions on discriminating the function accordingly. Then, we get a quadratic inequality in y. Solving the quadratic inequality, we get the value of range of the function.

Complete answer: So, we have the function, $y = \dfrac{{{x^2} - x + 1}}{{{x^2} + x + 1}}$.
We have to find the range of the function for x belonging to the set of real numbers.
Hence, $y = \dfrac{{{x^2} - x + 1}}{{{x^2} + x + 1}}$.
Cross multiplying the terms and simplifying the expression, we get,
$ \Rightarrow y\left( {{x^2} + x + 1} \right) = \left( {{x^2} - x + 1} \right)$
$ \Rightarrow y{x^2} + yx + y = {x^2} - x + 1$
Grouping the like terms together, we get,
$ \Rightarrow y{x^2} + yx + y = {x^2} - x + 1$
$ \Rightarrow \left( {y - 1} \right){x^2} + \left( {y + 1} \right)x + \left( {y - 1} \right) = 0$
Now, we know that the values of x belong to the real number set. Hence, the discriminant of the quadratic equation must be greater than or equal to zero for x to assume all the real values.
So, we first find the discriminant of the quadratic equation. So, comparing the equation $\left( {y - 1} \right){x^2} + \left( {y + 1} \right)x + \left( {y - 1} \right) = 0$ with the standard form of a quadratic equation $a{x^2} + bx + c = 0$, we get,
$a = y - 1$, $b = y + 1$ and $c = y - 1$
Now, $D = {b^2} - 4ac$
\[ \Rightarrow D = {\left( {y + 1} \right)^2} - 4\left( {y - 1} \right)\left( {y - 1} \right)\]
Opening the brackets and simplifying the expression, we get,
\[ \Rightarrow D = {\left( {y + 1} \right)^2} - 4{\left( {y - 1} \right)^2}\]
Using the algebraic identities ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$ and ${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$, we get,
\[ \Rightarrow D = {y^2} + 2y + 1 - 4\left( {{y^2} - 2y + 1} \right)\]
Opening the bracket, we get,
\[ \Rightarrow D = {y^2} + 2y + 1 - 4{y^2} + 8y - 4\]
Simplifying the equation by adding up like terms together, we get,
\[ \Rightarrow D = - 3{y^2} + 10y - 3\]
Now, expressing the discriminant is greater than or equal to zero, we get,
\[ \Rightarrow D = - 3{y^2} + 10y - 3 \geqslant 0\]
Now, we solve the above inequality by factoring the left side. So, we get,
\[ \Rightarrow - 3{y^2} + 9y + y - 3 \geqslant 0\]
Taking the common factors outside the bracket, we get,
\[ \Rightarrow - 3y\left( {y - 3} \right) + \left( {y - 3} \right) \geqslant 0\]
\[ \Rightarrow \left( {1 - 3y} \right)\left( {y - 3} \right) \geqslant 0\]
Multiplying both the sides of inequality by \[\left( { - 1} \right)\] and changing the sign of inequality, we get,
\[ \Rightarrow \left( {3y - 1} \right)\left( {y - 3} \right) \leqslant 0\]
Hence, the solution of the above inequality is $y \in \left[ {\dfrac{1}{3},3} \right]$.
So, the range of the function $y = \dfrac{{{x^2} - x + 1}}{{{x^2} + x + 1}}$ is $\left[ {\dfrac{1}{3},3} \right]$.
So, the minimum value of $y = \dfrac{{{x^2} - x + 1}}{{{x^2} + x + 1}}$ is $\dfrac{1}{3}$.
Hence, option (B) is correct.

Note:
Range of a function is the set of values assumed by the function for different values of variables in the domain. Each preimage in the domain has a unique image in the range for a function. So, for each value of variable in function, we get a unique value of function. There are different methods to find the range of different types of functions. The one used here can be used to find the range of any rational function in a variable.