Question

Find the set of all x for which $\dfrac{2x}{2{{x}^{2}}+5x+2}>\dfrac{1}{x+1}$.
(a) $x\in (-2,-1)\bigcup (\dfrac{-2}{3},\dfrac{-1}{2})$
(b) $-2\ge x\ge -1$
(c) $-2\ge x<-1$
(d) $-2

Answer 1

Hint: We will solve the given expression and try to convert it to factors so that we get to know about the roots of the given expression. Finally, we will represent all the roots of the number and try to find the set of the value of x.

Complete step-by-step answer:

The given expression is $\dfrac{2x}{2{{x}^{2}}+5x+2}>\dfrac{1}{x+1}$, then we have to find the set of x. On solving the given expression, we get,


= $\dfrac{2x}{2{{x}^{2}}+5x+2}>\dfrac{1}{x+1}$


Transporting $\dfrac{1}{(x+1)}$ from RHS to LHS, we get the following inequality


= $\dfrac{2x}{2{{x}^{2}}+5x+2}-\dfrac{1}{x+1}>0$


Taking LCM of $(2{{x}^{2}}+5x+2)$ and $(x+1)$, we get,


= $\dfrac{(2x)(x+1)-(2{{x}^{2}}+5x+2)}{(2{{x}^{2}}+5x+2)(x+1)}>0$


Opening brackets, we get,


=$\dfrac{2{{x}^{2}}+2x-2{{x}^{2}}-5x-2}{(2{{x}^{2}}+5x+2)(x+1)}>0$


Now we can cancel the similar terms and take - sign outside from the numerator. Also, we can factorize the denominator. Then, we will get


= $\dfrac{(-1)(3x+2)}{(2{{x}^{2}}+4x+x+2)(x+1)}>0$


= $\dfrac{-(3x+2)}{\left[ 2x(x+2)+1(x+2) \right](x+1)}>0$


= $\dfrac{-(3x+2)}{(2x+1)(x+2)(x+1)}>0$


From the above equation, we get values of x to be, -2, -1,$\dfrac{-2}{3}$, $\dfrac{-1}{2}$. When we arrange these values on a number line, they are in ascending order. Starting from the negative values (for $x<-2$ interval), the sign of these terms will change after every value of x because all are linear terms. We will consider only those intervals from the number line for which values are positive. Therefore we get values of x as $x\in (-2,-1)\bigcup (\dfrac{-2}{3},\dfrac{-1}{2})$.


Therefore, option (a) is the correct answer for the given question.


Note: Many times students may get confused in transporting the terms from LHS to RHS when any inequality like < or > is given but always remember that we have to take care that how the value and sign are changing. So, be careful when transporting the terms. Assure that the sign has been changed after transposing.