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Answer
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Hint: Equate the above expression to y. Use the fact that the discriminant must be greater than or equal to 0 if the roots of the equation are real. Find the range of y and the minimum and maximum value of the expression can be found using this range.
Formula used: Discriminant of the standard quadratic equation $a{x^2} + bx + c = 0$ is ${b^2} - 4ac$, ${\left( {p + q} \right)^2} = {p^2} + {q^2} + 2pq$, ${\left( {p - q} \right)^2} = {p^2} + {q^2} - 2pq$
Complete step by step solution:
Let $y = \dfrac{{{x^2} - 3x + 4}}{{{x^2} + 3x + 4}}$
${x^2}y + 3xy + 4y = {x^2} - 3x + 4$
$\left( {y - 1} \right){x^2} + 3\left( {y + 1} \right)x + 4\left( {y - 1} \right) = 0$
(Discriminant of the standard quadratic equation $a{x^2} + bx + c = 0$ is ${b^2} - 4ac$).
It is given to us that x is real, therefore, the discriminant must be greater than or equal to 0.
Therefore, $9{\left( {y + 1} \right)^2} - 4\left( {y - 1} \right)\left( {4\left( {y - 1} \right)} \right) \geqslant 0$
$9{y^2} + 18y + 9 - 16{\left( {y - 1} \right)^2} \geqslant 0$
\[9{y^2} + 18y + 9 - 16{y^2} + 32y - 16 \geqslant 0\]
$ - 7{y^2} + 50y - 7 \geqslant 0$
Changing the sign of inequality by multiplying both sides by -1,
$7{y^2} - 50y + 7 \leqslant 0$
$7{y^2} - 49y - y + 7 \leqslant 0$
$7y\left( {y - 7} \right) - 1\left( {y - 7} \right) \leqslant 0$
$\left( {7y - 1} \right)\left( {y - 7} \right) \leqslant 0$
Calculate the critical points, i.e., the values of y at which $\left( {7y - 1} \right)\left( {y - 7} \right) = 0$. Therefore, the critical points are $\dfrac{1}{7},7$. The value of $\left( {7y - 1} \right)\left( {y - 7} \right)$ will always be more than $0$ whenever $y$ is lesser than $\dfrac{1}{7}$ or greater than $7$.
Therefore,$y \in \left[ {\dfrac{1}{7},7} \right]$. These are the minimum and maximum values of the expression $\dfrac{{{x^2} - 3x + 4}}{{{x^2} + 3x + 4}}$.
The correct answer is option C. $7,\dfrac{1}{7}$
Note: Ensure that the sign of inequality is changed from $ \geqslant $ to $ \leqslant $ while changing the expression from $ - 7{y^2} + 50y - 7$ to $7{y^2} - 50y + 7$. $y \in \left[ {\dfrac{1}{7},7} \right]$ because if $y < \dfrac{1}{7}$ or $y > 7$, $\left( {7y - 1} \right)\left( {y - 7} \right)$ will not be lesser than or equal to 0.
Formula used: Discriminant of the standard quadratic equation $a{x^2} + bx + c = 0$ is ${b^2} - 4ac$, ${\left( {p + q} \right)^2} = {p^2} + {q^2} + 2pq$, ${\left( {p - q} \right)^2} = {p^2} + {q^2} - 2pq$
Complete step by step solution:
Let $y = \dfrac{{{x^2} - 3x + 4}}{{{x^2} + 3x + 4}}$
${x^2}y + 3xy + 4y = {x^2} - 3x + 4$
$\left( {y - 1} \right){x^2} + 3\left( {y + 1} \right)x + 4\left( {y - 1} \right) = 0$
(Discriminant of the standard quadratic equation $a{x^2} + bx + c = 0$ is ${b^2} - 4ac$).
It is given to us that x is real, therefore, the discriminant must be greater than or equal to 0.
Therefore, $9{\left( {y + 1} \right)^2} - 4\left( {y - 1} \right)\left( {4\left( {y - 1} \right)} \right) \geqslant 0$
$9{y^2} + 18y + 9 - 16{\left( {y - 1} \right)^2} \geqslant 0$
\[9{y^2} + 18y + 9 - 16{y^2} + 32y - 16 \geqslant 0\]
$ - 7{y^2} + 50y - 7 \geqslant 0$
Changing the sign of inequality by multiplying both sides by -1,
$7{y^2} - 50y + 7 \leqslant 0$
$7{y^2} - 49y - y + 7 \leqslant 0$
$7y\left( {y - 7} \right) - 1\left( {y - 7} \right) \leqslant 0$
$\left( {7y - 1} \right)\left( {y - 7} \right) \leqslant 0$
Calculate the critical points, i.e., the values of y at which $\left( {7y - 1} \right)\left( {y - 7} \right) = 0$. Therefore, the critical points are $\dfrac{1}{7},7$. The value of $\left( {7y - 1} \right)\left( {y - 7} \right)$ will always be more than $0$ whenever $y$ is lesser than $\dfrac{1}{7}$ or greater than $7$.
Therefore,$y \in \left[ {\dfrac{1}{7},7} \right]$. These are the minimum and maximum values of the expression $\dfrac{{{x^2} - 3x + 4}}{{{x^2} + 3x + 4}}$.
The correct answer is option C. $7,\dfrac{1}{7}$
Note: Ensure that the sign of inequality is changed from $ \geqslant $ to $ \leqslant $ while changing the expression from $ - 7{y^2} + 50y - 7$ to $7{y^2} - 50y + 7$. $y \in \left[ {\dfrac{1}{7},7} \right]$ because if $y < \dfrac{1}{7}$ or $y > 7$, $\left( {7y - 1} \right)\left( {y - 7} \right)$ will not be lesser than or equal to 0.
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