Solveeit Logo
Login

Question

Question: For real x, the greatest value of \(\dfrac{{{x^2} + 2x + 4}}{{2{x^2} + 4x + 9}}\) is A. 1 B. -1 ...

For real x, the greatest value of x2+2x+42x2+4x+9\dfrac{{{x^2} + 2x + 4}}{{2{x^2} + 4x + 9}} is
A. 1
B. -1
C. 12\dfrac{1}{2}
D. 14\dfrac{1}{4}

Explanation

Solution

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’
⇒y=x2+2x+42x2+4x+9\Rightarrow y = \dfrac{{{x^2} + 2x + 4}}{{2{x^2} + 4x + 9}}
⇒y(2x2+4x+9)=x2+2x+4\Rightarrow y\left( {2{x^2} + 4x + 9} \right) = {x^2} + 2x + 4
⇒2x2y+4xy+9y=x2+2x+4\Rightarrow 2{x^2}y + 4xy + 9y = {x^2} + 2x + 4
⇒2x2y−x2+4xy−2x+9y−4=0 ⇒x2(2y−1)+2x(2y−1)+9y−4=0  \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 x2(2y−1){x^2}\left( {2y - 1} \right) forms a quadratic equation so
D=>0D = > 0
b2−4ac=>0{b^2} - 4ac = > 0
Now, putting values
⇒[2(2y−1)]2−4(9y−4)(2y−1)=>0 ⇒4(2y−1)2−4(9y−4)(2y−1)=>0  \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 \\\
⇒4(2y−1)[(2y−1)−(9y−4)]=>0 ⇒4(2y−1)(2y−1−9y+4)=>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 \\\
⇒4(2y−1)(3−7y)=>0 ⇒(2y−1)(−)(7y−3)=>0 ⇒(2y−1)(7y−3)<=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,(2y−1)\left( {2y - 1} \right)
y=12y = \dfrac{1}{2}
For, (7y−3)\left( {7y - 3} \right)
y=37y = \dfrac{3}{7}
(Planting the above mentioned values on number line we get,)
37⩽y⩽12\dfrac{3}{7} \leqslant y \leqslant \dfrac{1}{2} (From here we can say that 12\dfrac{1}{2} is the greatest value.)
Therefore, 12\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.