3.5 Rational inequality  (Page 5/6)

Problem : Find solution of :

$$\frac{2}{1+x}+\frac{3}{1-x}<1$$

Solution : Rearranging, we have :

$$\Rightarrow \frac{2-2x+3+3x}{\left(1+x\right)\left(1-x\right)}-1<0$$ $$\Rightarrow \frac{5+x-\left(1-x^2\right)}{\left(1+x\right)\left(1-x\right)}<0$$ $$\Rightarrow \frac{x^2+x+4}{\left(1+x\right)\left(1-x\right)}<0$$

Now, polynomial in the numerator i.e. $x^2+x+4$ is positive for all real x as D<0 and a>0. Thus, dividing either side of the inequality by this polynomial does not change inequality. Now, we need to change the sign of x in one of the linear factors of the denominator positive in accordance with sign rule. This is required to be done in the factor (1-x). For this, we multiply each side of inequality by -1. This change in sign accompanies change in inequality as well :

$$\Rightarrow \frac{1}{\left(1+x\right)\left(1-x\right)}>0$$

Critical points are -1 and 1. Hence, solution of the inequality in x is :

$$x\in \left(-\infty ,-1\right)\cup \left(\mathrm{1,}\infty \right)$$

Problem : Find interval of x satisfying the inequality given by :

$$\frac{\left(2x+1\right)\left(x-1\right)}{\left(x^3-3x^2+2x\right)}\ge 0$$

Solution : We factorize each of the polynomials in numerator and denominator :

$$\Rightarrow \frac{\left(2x+1\right)\left(x-1\right)}{\left(x^3-3x^2+2x\right)}=\frac{\left(2x+1\right)\left(x-1\right)}{x\left(x-1\right)\left(x-2\right)}$$

It is important that we do not cancel common factors or terms. Here, critical points are -1/2,1,0,1 and 2. The critical point "1" is repeated even times. Hence, we do not change sign about "1" while drawing sign scheme.

While writing interval, we drop equality sign for critical points, which corresponds to denominator.

$$-1/2\le x<0\cup 2<x<\infty$$

$$[-1/2,0)\cup (2,\infty )$$

We do not include "-1" and "1" as they reduce denominator to zero.