Rational Inequalities Calculator

Inequalities are mathematical expressions showing the relationship between two values, indicating that one value is greater than, less than, or not equal to another. Understanding inequalities is crucial for solving various mathematical problems, from basic arithmetic to advanced calculus.

In this article, we will cover the concepts of the rational inequalities calculator. This concept falls under the broader category of complex numbers., a crucial Chapter in class 11 Mathematics. It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination (JEE Main), and other entrance exams such as SRMJEE, BITSAT, WBJEE, BCECE, and more. Over the last ten years of the JEE Main exam (from 2013 to 2023), a total of two questions have been asked on this concept, including one in 2013, and one in 2023.

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Inequalities

Inequalities are the relationship between two expressions that are not equal to one another. Symbols denoting the inequalities are <, >, ≤, ≥, and ≠.

• $x<4$, "is read as $x$ less than $4^", x \leq 4$, is read as $x$ less than or equal to $4^{\prime}$.
• Similarly $x>4$, "is read as $x$ greater than $4^{\prime}$ and $x \geq 4$, is read as $x$ greater than or equal to $4^4$.

The process of solving inequalities is the same as of equality but instead of equality symbol inequality symbol is used throughout the process.

Types of Inequalities

• Linear Inequalities: Involve linear expressions.
  • Example: $2 x+3 \leq 7$
• Quadratic Inequalities: Involve quadratic expressions.
  • Example: $x^2-4 x+3>0$
• Polynomial Inequalities: Involve polynomials of degree greater than two.
  • Example: $x^3-2 x^2+x-5<0$
• Rational Inequalities: Involve ratios of polynomials.
  • Example: $\frac{x+1}{x-3} \geq 2$
• Absolute Value Inequalities: Involve absolute value expressions.
  • Example: $|x-2| \leq 5$

A few rules that are different from equality rules

• If we multiply or divide both sides of the inequality by a negative number, then we reverse the inequality (reversing inequality means > gets converted to < and vice versa, and ≥ gets converted to ≤ and vice versa)(eg $4>3$ means $-4<-3$ )
• If we cross multiply a negative quantity in an inequality, then we reverse the inequality (eg 3 > -2 means -3/2 < 1)
• If we cancel the minus sign from both sides of an inequality, then we reverse the inequality.$(\mathrm{eg}-3>-4$ means $3<4)$
• As we usually do not know the sign of a variable term like $\mathrm{x}_1(\mathrm{x}-2)$ etc, so we do not cross-multiply them, as we cannot decide if we have to reverse the sign of inequality or not.

We get a range of solutions while solving inequality which satisfies the inequality,

Graphically inequalities can be shown as a region belonging to one side of the line or between lines, for example, inequality $-3<x \leq 5$can be represented as below, a region belonging to -3 and 5 are the region of possible x including 5 and excluding -3.

Frequently Used Inequalities

1. $(\mathrm{x}-\mathrm{a})(\mathrm{x}-\mathrm{b})<0 \Rightarrow \mathrm{x} \in(\mathrm{a}, \mathrm{b})$, where $\mathrm{a}<\mathrm{b}$
2. $(x-a)(x-b)>0 \Rightarrow x \in(-\infty, a) \cup(b, \infty)$, where $a<b$
3. $x^2 \leq a^2 \Rightarrow x \in[-a, a]$
4. $x^2 \geq a^2 \Rightarrow x \in(-\infty,-a] \cup[a, \infty)$

Rational Inequalities

We consider the algebraic inequalities of the following types

$\begin{aligned} & \frac{p(x)}{q(x)}<0, \frac{p(x)}{q(x)}>0 \\ & \frac{p(x)}{q(x)} \leq 0, \frac{p(x)}{q(x)} \geq 0\end{aligned}$

If p(x) and q(x) can be resolved in factor then we can solve these types of inequalities using a wavy curved method otherwise we use the following method to solve them.
$
\text { (1) } \begin{aligned}
& \frac{\mathrm{p}(\mathrm{x})}{\mathrm{q}(\mathrm{x})}>0 \Rightarrow \mathrm{p}(\mathrm{x}) \mathrm{q}(\mathrm{x})>0 \\
\Rightarrow & \mathrm{p}(\mathrm{x})>0, \mathrm{q}(\mathrm{x})>0 \text { or } \mathrm{p}(\mathrm{x})<0, \mathrm{q}(\mathrm{x})<0
\end{aligned}
$
$
\text { (2) } \begin{aligned}
& \frac{\mathrm{p}(\mathrm{x})}{\mathrm{q}(\mathrm{x})}<0 \Rightarrow \mathrm{p}(\mathrm{x}) \mathrm{q}(\mathrm{x})<0 \\
\Rightarrow & \mathrm{p}(\mathrm{x})>0, \mathrm{q}(\mathrm{x})<0 \text { or } \mathrm{p}(\mathrm{x})<0, \mathrm{q}(\mathrm{x})>0
\end{aligned}
$
$
\begin{aligned}
& \text { (3) } \frac{\mathrm{p}(\mathrm{x})}{\mathrm{q}(\mathrm{x})} \geq 0 \Rightarrow \mathrm{p}(\mathrm{x}) \mathrm{q}(\mathrm{x}) \geq 0 \text { and } \mathrm{q}(\mathrm{x}) \neq 0 \\
& \Rightarrow \mathrm{p}(\mathrm{x}) \geq 0, \mathrm{q}(\mathrm{x})>0 \text { or } \mathrm{p}(\mathrm{x}) \leq 0, \mathrm{q}(\mathrm{x})<0 \\
& \text { (4) } \frac{\mathrm{p}(\mathrm{x})}{\mathrm{q}(\mathrm{x})} \leq 0 \Rightarrow \mathrm{p}(\mathrm{x}) \mathrm{q}(\mathrm{x}) \leq 0 \text { and } \mathrm{q}(\mathrm{x}) \neq 0 \\
& \Rightarrow \mathrm{p}(\mathrm{x})>0, \mathrm{~g}(\mathrm{x})<0 \text { or } \mathrm{p}(\mathrm{x})<0 . \mathrm{g}(\mathrm{x}) \geq 0
\end{aligned}
$

Summary

Inequalities are a fundamental part of mathematics, providing a way to describe and solve problems involving ranges and constraints. Mastery of inequalities is essential for progressing in algebra, calculus, and applied mathematics, offering valuable tools for both theoretical and practical problem-solving.

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Solved Examples Based On Rational Inequalities Calculator:

Example 1: Find all the values of a for the inequality, $\frac{x^2+a x-2}{x^2-x+1}<2$ holds true for all values of x.
$
\begin{aligned}
& \text { 1) } a \in(-\infty,-6] \cup(2, \infty) \\
& \text { 2) } a \in[-6,2 \\
& \text { 3) } a \in(-\infty,-6) \cup(2, \infty)
\end{aligned}
$
$
\text { 4) } a \in(-6,2)
$

Solution:

This equation can be written as

$\begin{aligned} & \frac{\mathrm{x}^2+\mathrm{ax}-2-2\left(\mathrm{x}^2-\mathrm{x}+1\right)}{\mathrm{x}^2-\mathrm{x}+1}<0 \\ & \Rightarrow \frac{-x^2+(a+2) x-4}{x^2-x+1}<0 \\ & \mathrm{x}^2-\mathrm{x}+1, \mathrm{D}=1-4=-3 \text { also } \mathrm{a}>0, \text { it is always positive, so it can be cross multiplied } \\ & \Rightarrow-x^2+(a+2) x-4<0 \\ & \Rightarrow x^2-(a+2) x+4>0 \\ & \Rightarrow \text { for this to be always positive discriminant must be }-\mathrm{ve} \text {, hence } \\ & (\mathrm{a}+2)^2-16<0 \\ & (a+2+4)(a+2-4)<0 \\ & (a+6)(a-2)<0 \\ & \Rightarrow a \in(-6,2)\end{aligned}$

Hence, the answer is option (2).

Example 2: For what values of 'a', the inequality $\frac{x^2+a x-2}{x^2-x+1}>-3$ holds true for all real values of x?

1) (-1,0)

2) (-1,7)

3) (-7, 0)

4) (0, 7)

Solution:

This equation can be written as

$\frac{x^2+a x-2+3\left(x^2-x+1\right)}{x^2-x+1}>0 \Rightarrow \frac{4 x^2+(a-3) x+1}{x^2-x+1}>0$ since denominator is always + ve, hence numerator must be + ve $\Rightarrow 4 x^2+(a-3) x+1>0 \Rightarrow D<0$

where, 10 is the discriminant of the polynomial equation $4 x^2+(a-3) x+1$ $1>0$ and $D=b^2-4 a$ $\Rightarrow(a-3)^2-16<0 \Rightarrow a \in(-1.7)$

Hence, the answer is the option 2.

Example 3: The last integral value $\alpha$ of $x$ , such that $\frac{x-5}{x^2+5 x-14}>0$ satisfies

$\begin{aligned} & \text { 1) } a^2+3 a-4=0 \\ & \text { 2) } a^2-5 a+4=0 \\ & \text { 3) } a^2-7 a+6=0 \\ & \text { 4) } a^2+5 a-6=0\end{aligned}$

Solution:

$\begin{aligned} & \frac{x-5}{x^2+5 x-14}>0 \\ & \frac{x-5}{(x-2)(x+7)}>0 \\ & -7<x<2 \quad \text { or } \quad x>5\end{aligned}$

$\alpha=-6$ satisfies the given condition. Also, it satisfies the equation:

$\alpha^2+5 \alpha-6=0$

Hence, the answer is the option (4).

Example 4: The interval of x for the inequality $\frac{x}{x-1} \geq 0$ is

$\begin{aligned} & \text { 1) } x \in(0,1 \\ & \text { 2) } x \in[0,1) \\ & \text { 3) } x \in(-\infty, 0] \cup(1, \infty) \\ & \text { 4) } x \in(-\infty, 0) \cup[1, \infty)\end{aligned}$

Solution

Here, critical points are $x=0,1$

The critical points are marked on the real number line. Starting with a positive sign in the rightmost interval, we denote signs of adjacent intervals by the alternating sign.

Hence, $x \in(-\infty, 0] \cup(1, \infty)$

correct option is (c)

Example 5: For $a \in C$, let $A=\{z \in C: \operatorname{Re}(a+\bar{z})>\operatorname{Im}(\bar{a}+z)\}$ and $B=\{z \in C: \operatorname{Re}(a+\bar{z})<\operatorname{Im}(\bar{a}+z)\}$.The among two statements:
(S1): If Re (a), Im (a) 0, then the set A contains all the real numbers
(S2): If Re (a), Im (a) 0, then the set B contains all the real numbers,

1) only (S1) is true

2) both are false

3) only (S2) is true

4) both are true

Solution:

$\begin{aligned} & \text { Let }{ }^a=\mathrm{x}_1+\mathrm{i} \mathrm{y}_1 z=\mathrm{x}+\mathrm{iy} \\ & \text { Now } \operatorname{Re}(a+z)>\operatorname{Im}(\bar{a}+z) \\ & \therefore \mathrm{x}_1+\mathrm{x}>-\mathrm{y}_1+\mathrm{y} \\ & \mathrm{x}_1=2, \mathrm{y}_1=10, \mathrm{x}=-12, \mathrm{y}=0\end{aligned}$
Given inequality is not valid for these values.
S1 is false.$\begin{aligned} & \text { Now } \operatorname{Re}(a+\bar{z})<\operatorname{lm}(\bar{a}+z) \\ & \mathrm{x}_1+\mathrm{x}<-\mathrm{y}_1+\mathrm{y} \\ & \mathrm{x}_1=-2, \mathrm{y}_1=-10, \mathrm{x}=12, \mathrm{y}=0\end{aligned}$
Given inequality is not valid for these values.
S2 is false.

Hence, the answer is the option (2).