Question

For real x, the greatest value of $\dfrac{{{x^2} + 2x + 4}}{{2{x^2} + 4x + 9}}$ is
A. 1
B. -1
C. $\dfrac{1}{2}$
D. $\dfrac{1}{4}$

Answer 1

Hint- Convert the equation to quadratic form by equating the given equation to y. Apply a quadratic formula to find out the answer. (Be careful with calculations)

Complete step by step answer:

Now we know that for this we put the value equal of ‘y’
$ \Rightarrow y = \dfrac{{{x^2} + 2x + 4}}{{2{x^2} + 4x + 9}}$
$ \Rightarrow y\left( {2{x^2} + 4x + 9} \right) = {x^2} + 2x + 4$
$ \Rightarrow 2{x^2}y + 4xy + 9y = {x^2} + 2x + 4$
$
   \Rightarrow 2{x^2}y - {x^2} + 4xy - 2x + 9y - 4 = 0 \\
   \Rightarrow {x^2}\left( {2y - 1} \right) + 2x\left( {2y - 1} \right) + 9y - 4 = 0 \\
 $
Now, as we know that in ${x^2}\left( {2y - 1} \right)$ forms a quadratic equation so
$D = > 0$
${b^2} - 4ac = > 0$
Now, putting values
$
   \Rightarrow {\left[ {2\left( {2y - 1} \right)} \right]^2} - 4\left( {9y - 4} \right)\left( {2y - 1} \right) = > 0 \\
   \Rightarrow 4{\left( {2y - 1} \right)^2} - 4\left( {9y - 4} \right)\left( {2y - 1} \right) = > 0 \\
 $
$
   \Rightarrow 4\left( {2y - 1} \right)\left[ {\left( {2y - 1} \right) - \left( {9y - 4} \right)} \right] = > 0 \\
   \Rightarrow 4\left( {2y - 1} \right)\left( {2y - 1 - 9y + 4} \right) = > 0 \\
 $
$
   \Rightarrow 4\left( {2y - 1} \right)\left( {3 - 7y} \right) = > 0 \\
   \Rightarrow \left( {2y - 1} \right)\left( - \right)\left( {7y - 3} \right) = > 0 \\
   \Rightarrow \left( {2y - 1} \right)\left( {7y - 3} \right) < = 0 \\
 $
For,$\left( {2y - 1} \right)$
$y = \dfrac{1}{2}$
For, $\left( {7y - 3} \right)$
$y = \dfrac{3}{7}$
(Planting the above mentioned values on number line we get,)
$\dfrac{3}{7} \leqslant y \leqslant \dfrac{1}{2}$ (From here we can say that $\dfrac{1}{2}$ is the greatest value.)
Therefore, $\dfrac{1}{2}$ is our answer.
Hence, (C) is the correct option.

Note- The discriminant is the part of the quadratic formula underneath the square root symbol: b²-4ac. The discriminant tells us whether there are two solutions, one solution, or no solutions.