Solving Modulus and Greatest Integer Functions

1. The Modulus (Absolute Value) Function

The modulus, or absolute value, of a number is one of the most fundamental concepts in this area.

Definition and Graph

Think of the modulus of a number as its distance from zero on the number line. Since distance cannot be negative, the absolute value of any number is always non-negative.

• The absolute value of 5 is 5 (since 5 is 5 units from 0).
• The absolute value of -5 is also 5 (since -5 is also 5 units from 0).

We denote the absolute value of $x$ as $|x|$. The formal definition is:
$|x|=\{\begin{array}{ll}x,& \text{if }x\ge 0\\ -x,& \text{if }x<0\end{array}$
Note that if $x$ is negative (e.g., $x=-5$), then $-x$ is $-(-5)=5$, which is positive.

The graph of $y=|x|$ has a characteristic 'V' shape, which visually confirms that the output is never negative.

Solving Modulus Equations

Solving equations with modulus functions is a crucial skill. The core principle is to consider the two possibilities that arise from the definition. The following video provides an excellent, step-by-step guide.

Key takeaways from the video:

1. The Golden Rule: If you have an equation $|f(x)|=c$ (where $c\ge 0$), you must solve two separate equations:
   • $f(x)=c$
   • $f(x)=-c$
2. Isolate First: Always simplify the equation to get the absolute value expression by itself on one side before applying the golden rule. For example, in $2|3x-4|+5=27$, you must first subtract 5 and then divide by 2 to get $|3x-4|=11$.
3. No Solution Case: An absolute value can never be equal to a negative number. If you simplify an equation to the form $|f(x)|=\text{negative value}$, there is no solution.

Application: Minimum Value of a Sum of Moduli

A classic CAT problem asks for the minimum value of an expression like $f(x)=|x-a|+|x-b|+|x-c|$.

Recalling that $|x-a|$ represents the distance between $x$ and $a$, the expression is the sum of distances from $x$ to the points $a$, $b$, and $c$. To minimize this sum of distances, you should choose $x$ to be the median of the numbers $a$, $b$, and $c$.

• If you have an odd number of terms (e.g., $a,b,c$), the minimum occurs when $x$ is the middle value.
• If you have an even number of terms (e.g., $a,b,c,d$), the minimum occurs for any $x$ between the two middle values (inclusive).

Let's look at an example from a provided resource.

In problem 84, the roots are 0, -4, -8, ..., -32. The median value is -16. The minimum value is found by substituting $x=-16$ into the expression.

2. The Greatest Integer Function (GIF)

The second special function for today is the Greatest Integer Function, also known as the floor function.

Definition and Graph

The greatest integer function, denoted by $[x]$, gives the greatest integer that is less than or equal to x. It essentially rounds any number down to the nearest integer.

• $[3.14]=3$
• $[7.99]=7$
• $[5]=5$ (since 5 is an integer)
• $[-2.7]$ = -3 (Be careful with negatives! -3 is the first integer to the left of -2.7 on the number line)

To get a more visual and dynamic explanation, let's watch a short video.

The graph of the GIF looks like a staircase, as shown in the video and the image below.

Properties and Solving Equations

Solving equations with the GIF relies on two key properties:

1. The Definition as an Inequality: If $[x]=n$ (where $n$ is an integer), this is equivalent to the inequality $n\le x<n+1$.
2. Decomposition of x: Any real number $x$ can be expressed as the sum of its integer part and its fractional part: $x=[x]+\{x\}$, where $0\le \{x\}<1$.

Let's see how to apply the second property to solve a common type of GIF equation.

Example: Solve $7[x]+4\{x\}=12+x$.

This problem is present in the Functions handbook. Let's see how to approach it.
We know $x=[x]+\{x\}$. Let's substitute this into the equation:
$7[x]+4\{x\}=12+([x]+\{x\})$
Now, group the $[x]$ terms on one side and the $\{x\}$ terms on the other.
$7[x]-[x]=12+\{x\}-4\{x\}$
$6[x]=12-3\{x\}$
Let's isolate $\{x\}$:
$3\{x\}=12-6[x]$
$\{x\}=\frac{12-6[x]}{3}=4-2[x]$
Now we use the fundamental property of the fractional part: $0\le \{x\}<1$.
$0\le 4-2[x]<1$
We can solve this compound inequality:

• Part 1: $0\le 4-2[x]⟹2[x]\le 4⟹[x]\le 2$.
• Part 2: $4-2[x]<1⟹3<2[x]⟹[x]>1.5$.

Combining these, we find that the only integer value that $[x]$ can take is 2.
So, $[x]=2$.
Now, we find $\{x\}$ using our derived formula: $\{x\}=4-2[x]=4-2(2)=0$.
Finally, we find $x$: $x=[x]+\{x\}=2+0=2$.
There is one real value of x that satisfies this equation.

Conclusion

In this lesson, we've explored two important functions that require special handling: the modulus and the greatest integer function. Mastering them is not just about memorizing formulas, but about understanding their core definitions and how those definitions lead to specific problem-solving strategies.

Key Takeaways:

• Modulus Function $|x|$: Represents distance from zero. When solving $|f(x)|=c$, you create two distinct equations. For sums like $|x-a|+|x-b|$, the minimum value is found by choosing $x$ as the median of the roots $a$ and $b$.
• Greatest Integer Function $[x]$: Represents the largest integer less than or equal to $x$. When solving equations, the key is to use the inequality $[x]\le x<[x]+1$ or the decomposition $x=[x]+\{x\}$.

Preview of the Next Lesson:
We have now completed the module on Progressions, Functions, and Logarithms. Congratulations! We will now move on to a completely new area of quantitative aptitude: Geometry. Our first lesson in this new module will cover the foundational concepts of plane geometry, where we will learn how to apply the properties of lines, angles, and triangles to find unknown values.