Unit FP2: Further Pure Mathematics 2 - Chapter Notes: Inequalities
Hello mathematicians! Welcome to the Inequalities chapter of FP2. While you’ve seen inequalities before, Further Pure Maths takes them up a level. We’re going to tackle situations where the variables are hidden inside fractions or locked behind modulus signs.
Why is this important? Inequalities are essential for defining domains, ranges, and understanding the bounds of mathematical functions. Mastering these techniques—especially the safe methods for handling variables in denominators—is crucial for success in the rest of FP2 and FP3.
1. Mastering Rational Inequalities (The Fraction Problem)
Rational inequalities involve fractions where the variable (\(x\)) appears in the denominator, for example, \(\frac{2x}{x-1} < 3\).
Why standard algebra fails here (The Danger!)
In standard algebra, if you have \(A < B\), you can multiply both sides by a positive number \(C\) to get \(AC < BC\). But if \(C\) is negative, you must flip the inequality sign.
In FP2, when we have an expression like \((x-1)\), we don't know if it's positive or negative! If we cross-multiply, we might forget to flip the sign, leading to incorrect solutions.
The Golden Rule of FP2 Inequalities: Never cross-multiply by an expression containing the variable!
Common Mistake Alert!
If you have \(\frac{1}{x} < 2\), DO NOT multiply by \(x\) to get \(1 < 2x\). This step assumes \(x > 0\) and misses half the solution set!
The Safe Method: Using Critical Values
The safest and most reliable method is the Critical Values Method. This relies on the fact that an algebraic expression only changes its sign (from positive to negative or vice versa) at points where the expression equals zero, or where it is undefined (division by zero).
Step-by-Step Process:
- Step 1: Move Everything Left
Ensure the Right Hand Side (RHS) is zero.\(\frac{2x}{x-1} < 3 \quad \implies \quad \frac{2x}{x-1} - 3 < 0\)
- Step 2: Combine into a Single Fraction
Find a common denominator and simplify the numerator.\(\frac{2x - 3(x-1)}{x-1} < 0 \quad \implies \quad \frac{2x - 3x + 3}{x-1} < 0 \quad \implies \quad \frac{3 - x}{x-1} < 0\)
- Step 3: Find the Critical Values (CVs)
The CVs are the values of \(x\) that make the numerator zero OR the denominator zero.- Numerator CV: \(3 - x = 0 \implies x = 3\)
- Denominator CV: \(x - 1 = 0 \implies x = 1\)
- Step 4: Test Intervals (The Sign Detective)
Plot the CVs (1 and 3) on a number line. They divide the line into intervals: \(x < 1\), \(1 < x < 3\), and \(x > 3\). Choose a test value in each interval and substitute it into the simplified fraction \(\frac{3 - x}{x-1}\) to determine the overall sign.- Test \(x=0\) (Interval \(x < 1\)): \(\frac{3 - 0}{0 - 1} = \frac{3}{-1} = -3\) (Negative)
- Test \(x=2\) (Interval \(1 < x < 3\)): \(\frac{3 - 2}{2 - 1} = \frac{1}{1} = 1\) (Positive)
- Test \(x=4\) (Interval \(x > 3\)): \(\frac{3 - 4}{4 - 1} = \frac{-1}{3}\) (Negative)
- Step 5: State the Solution
We wanted \(\frac{3 - x}{x-1} < 0\) (i.e., where the expression is negative).The solution is \(x < 1\) OR \(x > 3\).
Quick Review: Rational inequalities rely on finding the points where the expression changes sign (CVs) and then testing the regions in between.
2. Inequalities Involving Modulus (Absolute Value)
The modulus function, denoted \(|x|\), gives the non-negative value of \(x\). Think of it as distance from zero.
Understanding the Two Types
When solving modulus inequalities, we generally use two main algebraic techniques or a graphical approach.
Technique A: Squaring Both Sides
This is often the fastest and cleanest method in FP2, especially when comparing two modulus expressions, such as \(|P(x)| > |Q(x)|\). Since both sides are guaranteed to be positive (or zero), squaring them preserves the inequality sign.
Example: Solve \(|2x - 1| \ge |x + 5|\)
- Step 1: Square Both Sides
\((2x - 1)^2 \ge (x + 5)^2\)
- Step 2: Expand and Simplify
\(4x^2 - 4x + 1 \ge x^2 + 10x + 25\)
- Step 3: Rearrange to \(RHS = 0\)
\(3x^2 - 14x - 24 \ge 0\)
- Step 4: Find the Roots (Critical Points for the quadratic)
Using the quadratic formula or factoring: \((3x + 4)(x - 6) \ge 0\)
The roots are \(x = -\frac{4}{3}\) and \(x = 6\).
- Step 5: Sketch the Quadratic
Since the coefficient of \(x^2\) (3) is positive, the parabola is U-shaped (happy face). We want the regions where the parabola is above or touching the x-axis (\(\ge 0\)).
- Step 6: State the Solution
The solution is \(x \le -\frac{4}{3}\) OR \(x \ge 6\).
Did you know? Squaring is mathematically equivalent to writing \((P(x))^2 - (Q(x))^2 \ge 0\), which is a difference of two squares: \((P(x) - Q(x))(P(x) + Q(x)) \ge 0\). If you are comfortable with this identity, it saves expansion steps!
Technique B: Definition Method (Case Analysis)
This method is essential if you have a modulus expression compared to a non-modulus expression, e.g., \(|x - 2| < 2x\). You must consider the cases based on where the expression inside the modulus is positive or negative.
Standard Definitions:
- If \(|x| < a\), then \(-a < x < a\).
- If \(|x| > a\), then \(x < -a\) or \(x > a\).
Example: Solve \(|3x + 2| \le 5\)
- Use the definition: \(-5 \le 3x + 2 \le 5\)
- Subtract 2 from all parts: \(-7 \le 3x \le 3\)
- Divide by 3: \(-\frac{7}{3} \le x \le 1\)
This method works well when the constant \(a\) is positive. If \(a\) involves \(x\), you must use case analysis or the graphical method (Section 3).
Key Takeaway for Modulus: Squaring is usually best when comparing two modulus expressions. Case analysis/definition is best when comparing a modulus expression to a constant.
3. Solving Inequalities Graphically (Visual Confirmation)
Sometimes, complex inequalities are difficult to solve purely algebraically without risking sign errors. The graphical method is a powerful tool to verify your algebraic answers or solve problems directly by visualization.
The Principle of Graphical Solving
When asked to solve an inequality like \(f(x) > g(x)\), you are simply looking for the range of \(x\) values where the graph of \(y = f(x)\) is above the graph of \(y = g(x)\).
This method is particularly useful for modulus functions, as their V-shaped graphs are easy to sketch.
Example: Solve \(|x + 1| < |2x - 3|\) graphically.
- Step 1: Sketch \(y_1 = |x + 1|\) and \(y_2 = |2x - 3|\).
\(y_1\) (V-shape) has its vertex (turning point) at \(x = -1\).
\(y_2\) (V-shape) has its vertex at \(x = \frac{3}{2}\). (The gradient is steeper due to the \(2x\)). - Step 2: Find the Intersection Points
The intersection points define the boundaries (Critical Values). You must solve the equations algebraically to find the exact points, typically by squaring or case analysis. (The algebra confirms the critical values are \(x = \frac{2}{3}\) and \(x = 4\).) - Step 3: Identify the Required Region
We want \(y_1 < y_2\), meaning where the graph of \(y_1 = |x + 1|\) is below the graph of \(y_2 = |2x - 3|\). - Step 4: State the Solution
Based on the sketch, \(y_1\) is below \(y_2\) when \(x < \frac{2}{3}\) and when \(x > 4\).
Solution: \(x < \frac{2}{3}\) OR \(x > 4\).
Don't worry if sketching is tricky at first. Focus on accurately locating the vertices and the intercepts (where \(x=0\) and \(y=0\)).
Connection to Rational Functions (Brief Review)
When dealing with rational inequalities like \(\frac{f(x)}{g(x)} < 0\), sketching the graph of \(y = \frac{f(x)}{g(x)}\) is another way to determine the solution regions. You must remember to include:
- Roots: Where \(f(x) = 0\). The graph crosses the x-axis here.
- Vertical Asymptotes: Where \(g(x) = 0\). These correspond exactly to the denominator Critical Values!
The graph will flip sign around both the roots and the asymptotes, confirming the intervals found using the sign detective method.
4. Synthesis and Troubleshooting
Troubleshooting Complex Inequalities
When faced with an inequality that involves both modulus and a variable on the other side (e.g., \(|x^2 - 4| > 2x\)), always remember your options:
Option 1: Squaring (Always Safe for non-negative expressions):
If both sides are known to be non-negative (e.g., \(|P(x)| > 0\) and \(Q(x) > 0\)), squaring is best. If one side might be negative (like the right side in the example \(|x^2 - 4| > 2x\)), you MUST use Option 2 or 3.
Option 2: Case Analysis (The Definition Method):
This involves breaking the problem down based on the sign of the expression inside the modulus.
Example Case Analysis Structure:
Solve \(|x - 5| < 3x\).
- Case 1: \(x - 5 \ge 0\) (\(x \ge 5\))
The inequality becomes: \(x - 5 < 3x\). Solving gives \(-5 < 2x\), or \(x > -\frac{5}{2}\).
Intersection: We must satisfy both \(x \ge 5\) AND \(x > -\frac{5}{2}\). The solution for Case 1 is \(x \ge 5\). - Case 2: \(x - 5 < 0\) (\(x < 5\))
The inequality becomes: \(-(x - 5) < 3x\). Solving gives \(5 - x < 3x\), or \(5 < 4x\), or \(x > \frac{5}{4}\).
Intersection: We must satisfy both \(x < 5\) AND \(x > \frac{5}{4}\). The solution for Case 2 is \(\frac{5}{4} < x < 5\). - Final Solution: Combine the valid ranges from Case 1 and Case 2.
\(\left( \frac{5}{4} < x < 5 \right) \cup (x \ge 5)\).
Final Answer: \(x > \frac{5}{4}\).
Key Takeaways for FP2 Inequalities
- Rational: Always move everything to one side, combine into a single fraction, find CVs (roots and asymptotes), and test intervals. Do not cross-multiply.
- Modulus: Squaring is efficient if comparing two modulus terms. Case analysis/definition is required if the expression is compared to a variable term.
- Graphical Check: Use sketching to quickly identify intersection boundaries and confirm which side of the curve is "greater than" or "less than."
Phew! That covers the techniques you need. Remember, precision in checking your critical values and interval signs is key. Keep practising those sign checks, and you’ll master this topic!