Question

Find the greatest value of \[\dfrac{x+2}{2{{x}^{2}}+3x+6}\] for real values of x.

Answer 1

Hint: First of all, write the given expression as \[E=\dfrac{x+2}{2{{x}^{2}}+3x+6}\]. Now cross multiply this expression and get the quadratic equation in x. As x is real, \[D\ge 0\] or \[{{b}^{2}}-4ac\ge 0\]. From here, get an inequation in terms of E and find the values of E that satisfies that inequation.

Complete step-by-step answer:
In this question, we have to find the maximum value of \[\dfrac{x+2}{2{{x}^{2}}+3x+6}\] for the real values of x. Let us consider the expression given in the question.
\[E=\dfrac{x+2}{2{{x}^{2}}+3x+6}....\left( i \right)\]
First of all, let us cross multiply the above equation, we get
\[2E{{x}^{2}}+3Ex+6E-x+2\]
By transposing all the terms to LHS of the above equation, we get,
\[2E{{x}^{2}}+3x+6E-x-2=0\]
By rearranging the terms of the above equation, we get,
\[2E{{x}^{2}}+x\left( 3E-1 \right)+\left( 6E-2 \right)=0.....\left( ii \right)\]
The above equation is quadratic in x and we are given that x is real. We know that in a quadratic equation, where x is real discriminant (D) is greater than or equal to 0 or \[D\ge 0\]. We know that for the quadratic equation, \[a{{x}^{2}}+bx+c\], D is \[{{b}^{2}}-4ac\]. So by comparing equation (ii) by quadratic equation \[a{{x}^{2}}+bx+c\], we get,
a = 2E, b = (3E – 1), c = 6E – 2 = 0
So, we get,
\[D={{b}^{2}}-4ac\]
\[={{\left( 3E-1 \right)}^{2}}-4\times 2E\times \left( 6E-2 \right)\ge 0\]
By implying the above inequation, we get,
\[{{\left( 3E-1 \right)}^{2}}-8E\left( 6E-2 \right)\ge 0\]
By using the formula \[{{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab\], we get,
\[{{\left( 3E \right)}^{2}}+1-6E-48{{E}^{2}}+16E\ge 0\]
\[9{{E}^{2}}+1-6E-48{{E}^{2}}+16E\ge 0\]
\[-39{{E}^{2}}+10E+1\ge 0\]
We know that when (– 1) is multiplied to the inequation, the sign of inequality reverses. So, by multiplying (–1) in the above inequation, we get,
\[39{{E}^{2}}-10E-1\le 0\]
We can also write the above inequation as
\[39{{E}^{2}}-13E+3E-1\le 0\]
\[13E\left( 3E-1 \right)+1\left( 3E-1 \right)\le 0\]
By taking out (3E – 1) common, we get,
\[\left( 3E-1 \right)\left( 13E+1 \right)\le 0\]

Now, we will check the value of the above inequation, for the different values.
For, \[E\ge \dfrac{1}{3}\], \[3E-1\ge 0\]
And \[13E+1\ge 0\]
So, \[\left( 3E-1 \right)\left( 13+1 \right)\ge 0\]
For instance, if we take E = 2, we get,
\[\left[ \left( 3\left( 2 \right)-1 \right)\left( \left( 13\times 2 \right)+1 \right) \right]=\left( 5 \right)\left( 27 \right)\]
which is greater than 0, so, \[E\ge 1\] does not satisfy our inequality.
Now for, \[\dfrac{-1}{13}\le E<\dfrac{1}{3}\],
\[3E-1<0\]
\[13E+1>0\]
So, (3E – 1) (13E + 1) < 0
For instance, if we take E = 0, we get,
\[\left( 0-1 \right)\left( 13\left( 0 \right)+1 \right)=-1\left( 1 \right)=-1\]
which is less than 0. So, \[\dfrac{-1}{13}\le E<\dfrac{1}{3}\] satisfies the inequation. Hence, we get the value of E as \[\dfrac{-1}{13}\le E<\dfrac{1}{3}\].
Now, for \[E<\dfrac{-1}{13}\]
\[3E-1<0\]
\[13E+1<0\]
So, (3E – 1) (13E + 1) > 0
For instance, if we take E = – 1, we get,
\[\left( 3\left( -1 \right)-1 \right)\left( 13\left( -1 \right)+1 \right)\]
\[\left( -4 \right)\left( -11 \right)=44\]
which is greater than 0. So, \[E<\dfrac{-1}{13}\] does not satisfy the inequality.
So, we finally get the value of \[E=\dfrac{x+2}{2{{x}^{2}}+3x+6}\] between \[\left[ \dfrac{-1}{13},\dfrac{1}{3} \right]\].
Hence, the greatest value of the given expression is \[\dfrac{1}{3}\].

Note: In this question, students often make this mistake of multiplying (– 1) to the whole inequation but keeping the sign of inequality the same which is wrong because we know that whenever we multiply (– 1) to any inequation, the sign of inequality reverses. For the equation, we know that 2 > 1. Now, when we multiply – 1 to both the sides, we get – 2 < – 1 that is the sign of the inequality got reversed.