Question

If $x$ is real, then the maximum and minimum values of the expression $\dfrac{{{x^2} + 14x + 9}}{{{x^2} + 2x + 3}}$ will be
A. $4, - 5$
B. $5, - 4$
C. $ - 4,5$
D. $ - 4, - 5$

Answer 1

Hint: Suppose that the given expression is equal to variable $y$ . Now, solve the equation till we get an equation in the form of quadratics with respect to the variable $x$ . Now, all we have to do is use the discriminant approach here to get the maximum and minimum value of the expression. For this, compare the obtained quadratic with the general quadratic equation to find the coefficients $a$ , $b$ and $c$ .

Formula Used: For a given real $x$ , the discriminant i.e., $D = {b^2} - 4ac \geqslant 0$ .

Complete step-by-step solution:
We have the expression $\dfrac{{{x^2} + 14x + 9}}{{{x^2} + 2x + 3}}$ .
Suppose the given expression is $y = \dfrac{{{x^2} + 14x + 9}}{{{x^2} + 2x + 3}}$ .
Cross-multiplying both the sides
$ \Rightarrow y({x^2} + 2x + 3) = {x^2} + 14x + 9$
$ \Rightarrow y{x^2} + 2xy + 3y = {x^2} + 14x + 9$
On further solving,
$ \Rightarrow y{x^2} - {x^2} - 14x + 2xy + 3y - 9 = 0$
$ \Rightarrow (y - 1){x^2} + (2y - 14)x + (3y - 9) = 0$
If we compare this equation with the standard equation of quadratic form $a{x^2} + bx + c = 0$ , we will get
$a = (y - 1)$ , $b = (2y - 14)$ and $c = (3y - 9)$
To find the maximum and minimum point we use the concept of discriminant here. We now know that $D \geqslant 0$ , for a given real-value of $x$ . Hence, we have ${b^2} - 4ac \geqslant 0$ .
Now, substitute the values of $a$ , $b$ and $c$ in the above inequality.
${(2y - 14)^2} - 4(y - 1)(3y - 9) \geqslant 0$
$ \Rightarrow - {y^2} - y + 20 \geqslant 0$
Multiply both the sides of the inequality with $( - )$ sign.
$ \Rightarrow {y^2} + y - 20 \leqslant 0$
Factoring the inequality to get the maximum and minimum points:
$(y - 4)(y + 5) \leqslant 0$
If these two factors have negative signs, then the product of them is negative. Consequently, the next two scenarios arise:
$(y - 4) \geqslant 0$ or $y \geqslant 4$ and $(y + 5) \leqslant 0$ or $y \leqslant - 5$
Else, we can write these as
$(y - 4) \leqslant 0$ or $y \leqslant 4$ and $(y + 5) \geqslant 0$ or $y \geqslant - 5$
Both the cases can be satisfied for $ - 5 \leqslant y \leqslant 4$ .
So, we get the maximum and the minimum value of the given expression be $4, - 5$ .
Hence, the correct option will be A.

Note: We can use an alternative approach to get the result for the given quadratics. We can easily equate the expression to the variable $y$ . Then, we will express the equation as a parabola and respective of the quadrants in which the parabola is, we could find the maximum and minimum point for that parabola i.e., the coordinates of the vertex.