Question

If $\dfrac{{2x}}{{2{x^2} + 5x + 2}} > \dfrac{1}{{x + 1}}$, then
a) $- 2 > x > - 1$
b) $- 2 \geqslant x \geqslant - 1$
c) $- 2 < x < \- 1$
d) $- 2 \leqslant x \leqslant - 1$

Answer 1

This is a question of linear inequations. We will first shift the term in the RHS to LHS and make the RHS equal to zero. Then we will simplify the terms and make them as a single term. Then we will find the roots by equating each term to zero. Further we will plot the roots on the number line and select the appropriate region accordingly.

Complete answer:
Given:
$\dfrac{{2x}}{{2{x^2} + 5x + 2}} > \dfrac{1}{{x + 1}}$
Now we will shift the RHS to LHS and make the RHS equal to zero. So, we get;
$\Rightarrow \dfrac{{2x}}{{2{x^2} + 5x + 2}} - \dfrac{1}{{x + 1}} > 0$
Now we will take the LCM of the denominator and simplify the given equation further.
$\Rightarrow \dfrac{{2x\left( {x + 1} \right) - \left( {2{x^2} + 5x + 2} \right)}}{{\left( {2{x^2} + 5x + 2} \right)\left( {x + 1} \right)}} > 0$
Here, $x \ne - 1$, $x \ne \dfrac{{ - 1}}{2}$, $x \ne - 2$, because at these points the function will be undefined.
Now we will simplify the numerator by expanding the terms in the bracket using distributive property.
$\Rightarrow \dfrac{{2{x^2} + 2x - 2{x^2} - 5x - 2}}{{\left( {2{x^2} + 5x + 2} \right)\left( {x + 1} \right)}} > 0$
Now we will cancel the terms in the numerators and we get;
$\Rightarrow \dfrac{{ - 3x - 2}}{{\left( {2{x^2} + 5x + 2} \right)\left( {x + 1} \right)}} > 0$
Now we will take the minus common in the numerator. So, we have;
$\Rightarrow \dfrac{{ - \left( {3x + 2} \right)}}{{\left( {2{x^2} + 5x + 2} \right)\left( {x + 1} \right)}} > 0$
Now we will simplify the quadratic term in the denominator by middle term factorisation.
$\Rightarrow \dfrac{{ - \left( {3x + 2} \right)}}{{\left( {2{x^2} + 4x + x + 2} \right)\left( {x + 1} \right)}} > 0$
Now we will group the terms by taking common. So, we have;
\Rightarrow \dfrac{{ - \left( {3x + 2} \right)}}{{\left\\{ {2x\left( {x + 2} \right) + \left( {x + 2} \right)} \right\\}\left( {x + 1} \right)}} > 0
On further simplification we get;
$\Rightarrow \dfrac{{ - \left( {3x + 2} \right)}}{{\left( {2x + 1} \right)\left( {x + 2} \right)\left( {x + 1} \right)}} > 0$
Now we will multiply both sides by minus one. So, we get;
$\Rightarrow \dfrac{{\left( {3x + 2} \right)}}{{\left( {2x + 1} \right)\left( {x + 2} \right)\left( {x + 1} \right)}} < 0 - - - - (1)$
Now we will equate each term to zero and find the roots.
If,
$3x + 2 = 0$
$\Rightarrow x = - \dfrac{2}{3}$
When,
$2x + 1 = 0$
$\Rightarrow x = - \dfrac{1}{2}$
When,
$x + 2 = 0$
$\Rightarrow x = - 2$
And if,
$x + 1 = 0$
$\Rightarrow x = - 1$

Now we will locate all the roots on the number line.

Now we will put any number in place of $x$, that belongs to region FB on the number line in equation $(1)$ and we will observe that we get a positive number. So, we mark the region FB as positive.
Now we can either check each region by putting the values or we can simply mark the region as positive and negative alternately. So, we get-

Now in equation $\left( 1 \right)$ we can see it is $< 0$, so we will choose the $- ve$ regions.
Here we can see that there are two $- ve$ regions so we will include both of them using the union sign. So,
$\Rightarrow \dfrac{{\left( {3x + 2} \right)}}{{\left( {2x + 1} \right)\left( {x + 2} \right)\left( {x + 1} \right)}} < 0 - - - - (1)$
$\Rightarrow x \in \left( { - 2, - 1} \right) \cup \left( { - \dfrac{2}{3}, - \dfrac{1}{2}} \right)$

So, option c is correct.

Note:
Here in these types of questions the most important thing to note is the use of different brackets. For example, if in the answer we had written $\Rightarrow x \in \left[ { - 2, - 1} \right] \cup \left( { - \dfrac{2}{3}, - \dfrac{1}{2}} \right)$, then the answer would get wrong. Another thing to note is that we cannot always write the $+ ve, - ve$ regions alternately.
We can only write when all the terms are linear expressions.