Basic Mathematics: Wavy Curve Method, Modulus, Logarithms, and Irrational Equations Study Guide

Golden Rules of Inequality

• Addition and Subtraction: If $a \geq b$, then for any constant $k$:

  • $a + k \geq b + k$

  • $a - k \geq b - k$

• Multiplication and Division:

  • If $k > 0$ (positive constant) and $a \geq b$, then $ka \geq kb$. The sign of inequality remains unchanged.

  • If $k < 0$ (negative constant) and $a \geq b$, then $ka \leq kb$. The sign of inequality is reversed.

Wavy Curve Method (WCM)

• Objective: To solve inequalities involving polynomial and rational expressions.

• Standard Process:

  1. Ensure the right-hand side of the inequality is zero.

  2. Factorize the numerator and denominator completely into linear factors. Ensure the coefficient of $x$ in each factor is positive (if it is negative, multiply by $-1$ and reverse the inequality sign).

  3. Identify the roots of each factor (critical points).

  4. Plot these roots on a number line in increasing order.

  5. Starting from the rightmost interval (greater than the largest root), mark it with a positive sign ($+$).

  6. Move from right to left through each root:

     • If the root came from a factor with an odd power, the sign changes.

     • If the root came from a factor with an even power, the sign remains the same.

  7. Select the intervals that satisfy the inequality (e.g., intervals marked $+$ for $> 0$ or intervals marked $-$ for $< 0$).

• Tricks for Inclusion/Exclusion:

  • Dark Circle (\bullet): Used for inequalities including equality ($\geq$ or $\leq$) to include the root.

  • Hollow Circle ($\circ$): Used for strict inequalities ($>$ or $<$) or for any roots coming from the denominator, as division by zero is undefined.

• Examples:

  • Solve: $(x - 1)(x - 3) \geq 0$

    • Roots: $x = 1$, $x = 3$.

    • Number line: $(+\infty, 3]$ is $(+)$, $[1, 3]$ is $(-)$, $(-\infty, 1]$ is $(+)$.

    • Solution: $x \in (-\infty, 1] \cup [3, \infty)$.

  • Solve: $\frac{(x+1)(x-2)}{(x+3)} \geq 0$

    • Roots: $x = -3$ (denominator), $x = -1$, $x = 2$.

    • The root $-3$ must be excluded (hollow circle).

    • Solution: $x \in (-3, -1] \cup [2, \infty)$.

  • Complex Case: $\frac{(x-5)^{275}(x+1)}{(x^2-16)(x+11)^2} > 0$

    • Numerator factors: $(x-5)^{275}$, $(x+1)^1$.

    • Denominator factors: $(x-4)(x+4)$, $(x+11)^2$.

    • Apply WCM considering powers: sign change at $5$, sign change at $1$, sign change at $4$, sign change at $-4$, but no sign change at $-11$.

Handling Common Factors in Rational Inequalities

• Scenario: Common Factor with Cancellation:

  • Example: $\frac{(x-3)(x+2)}{(x+2)(x+4)} \leq 0$

  • One might cancel $(x+2)$, but the condition $x \neq -2$ must be maintained because the original expression is undefined at that point.

  • After cancellation, the simplified expression is $\frac{x-3}{x+4} \leq 0$.

  • Solution from simplified form: $x \in (-4, 3]$. Excluding the hole: $x \in (-4, 3] - \{-2\}$.

• Scenario: Common Factor without Cancellation:

  • Technically, the WCM handles this if you treat it as $(x+2)^2$ in the denominator. Since the power is even, the sign will not change at $x = -2$, and $-2$ must be excluded.

Common Intersection (Double Inequalities)

• To solve expressions like $f(x) < g(x) \leq h(x)$, split them into two parts: $f(x) < g(x)$ and $g(x) \leq h(x)$.

• Solve both inequalities separately and find the intersection of their solution sets.

• Example (IIT-1978): Find all integers $x$ for which $(5x - 1) < (x + 1)^2 < (7x - 3)$.

  • Part 1: $5x - 1 < (x+1)^2 \implies 0 < x^2 - 3x + 2 \implies 0 < (x-1)(x-2)$. Roots: $1, 2$. Intervals: $(-\infty, 1) \cup (2, \infty)$.

  • Part 2: $(x+1)^2 < 7x - 3 \implies x^2 - 5x + 4 < 0 \implies (x-1)(x-4) < 0$. Roots: $1, 4$. Intervals: $(1, 4)$.

  • Common Solution: $(2, 4)$. Since we need integers, the only integer in this interval is $x = 3$.

Irrational Equations and Inequalities

• Irrational Equations:

  • Steps: Identify the domain (all expressions under even roots must be $\geq 0$), then Square Both Sides (SBS).

  • Mandatory Recheck: Since squaring can introduce extraneous roots (e.g., $-3 = 3$ becomes $9 = 9$), always plug roots back into the original equation.

  • Example (IIT-1997): $\sqrt{x+1} - \sqrt{x-1} = \sqrt{4x-1}$

    • After SBS twice, the solution found is $x = \frac{5}{4}$.

    • Verification: LHS at $x = 5/4$ is $\sqrt{9/4} - \sqrt{1/4} = 3/2 - 1/2 = 1$. RHS is $\sqrt{4(5/4) - 1} = \sqrt{4} = 2$. Since $1 \neq 2$, there is no solution.

• Irrational Inequalities:

  • Critical Rule: Squaring both sides of an inequality is only permitted when both sides are non-negative.

  • If one side could be negative, the inequality must be split into cases based on sign.

  • Steps: Define the domain (Defining Part), check signs, SBS only if non-negative.

Modulus Function ($|x|$)

• Definition: $|x| = \begin{cases} x & x \geq 0 \\ -x & x < 0 \end{cases}$.

• Properties:

  • $|ab| = |a| \times |b|$

  • $\left| \frac{a}{b} \right| = \frac{|a|}{|b|}$

  • $|x^n| = |x|^n = x^n$ (if $n$ is even).

  • $\sqrt{x^2} = |x|$.

• Modulus Equations:

  • Case Method (Supreme Method): Define cases based on the critical points where the expression inside the modulus is zero. Solve for each case and verify if the result falls within the assumed range.

  • Example (JEE Mains 2023): Solve $|x^2 - 8x + 15| - 2x + 7 = 0$.

    • Critical points of $x^2 - 8x + 15 = (x-3)(x-5)$ are $x = 3$ and $x = 5$.

    • Case 1 ($x \leq 3$ or $x \geq 5$): $(x^2 - 8x + 15) - 2x + 7 = 0 \implies x^2 - 10x + 22 = 0 \implies x = 5 \pm \sqrt{3}$. Both are in range.

    • Case 2 ($3 < x < 5$): $-(x^2 - 8x + 15) - 2x + 7 = 0 \implies -x^2 + 6x - 8 = 0 \implies x^2 - 6x + 8 = 0 \implies x = 2, 4$. Only $x = 4$ is in the range $(3, 5)$.

    • Final solutions: $5+\sqrt{3}, 5-\sqrt{3}, 4$. Sum = $14$.

• Special Modulus Property: Triangle Inequality variants:

  • $|a+b| = |a| + |b| \iff ab \geq 0$ (same sign).

  • $|a-b| = |a| + |b| \iff ab \leq 0$ (opposite sign).

  • $|a+b| < |a| + |b| \iff ab < 0$ (opposite sign).

  • Example: Solve $|x^2 + 2x + 2| + |3x + 7| > |x^2 + 5x + 9|$

    • Let $a = x^2 + 2x + 2$ and $b = 3x + 7$. Note that $a + b = x^2 + 5x + 9$.

    • The inequality is $|a| + |b| > |a+b|$, which occurs when $ab < 0$.

    • $(x^2 + 2x + 2)(3x + 7) < 0$. Since $x^2 + 2x + 2$ has a negative discriminant ($D = -4$), it is always positive. We only need $3x + 7 < 0 \implies x < -\frac{7}{3}$.

Logarithms and Exponentials

• Fundamental Definition: $\log_B A = C \iff B^C = A$.

  • Domain Conditions: $A > 0$, $B > 0$, and $B \neq 1$.

• Logarithmic Properties:

  • P-1: $\log_B A = C \implies A = B^C$.

  • P-2: $\log_B m + \log_B n = \log_B (mn)$.

  • P-2b: $\log_B m - \log_B n = \log_B (m/n)$.

  • P-3 (Base Change): $\log_b a = \frac{\log_c a}{\log_c b}$. Also $\log_b a = \frac{1}{\log_a b}$.

  • P-4 (Power Rule): $\log_{b^n} a^m = \frac{m}{n} \log_b a$.

  • P-5 (Exponents): $a^{\log_a x} = x$, and $a^{\log_c b} = b^{\log_c a}$.

• Logarithmic Inequalities:

  • If Base > 1: The sign of inequality remains the same when removing the log (e.g., $\log_2 x > \log_2 y \implies x > y$).

  • If 0 < Base < 1: The sign of inequality is reversed when removing the log (e.g., $\log_{1/2} x > \log_{1/2} y \implies x < y$).

  • Always include the domain defining part ($\text{Argument} > 0$) in the intersection of solutions.

Algebraic Identities

• $a^2 + b^2 + c^2 - ab - bc - ca = \frac{1}{2} [(a-b)^2 + (b-c)^2 + (c-a)^2]$.

• $a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca)$.

  • If $a + b + c = 0$ or $a = b = c$, then $a^3 + b^3 + c^3 = 3abc$.

• Special Identity (Sophic-Germain): $a^4 + a^2 + 1 = (a^2 - a + 1)(a^2 + a + 1)$.

Questions & Discussion

• Q: Find the value of $((log_2 9)^2)^{\frac{1}{log_2(log_2 9)}} \times 7^{\frac{1}{log_4 7}}$.

• A: Let $\alpha = \log_2 9$. The first term is $(\alpha^2)^{\log_{\alpha} 2} = \alpha^{2 \log_{\alpha} 2} = \alpha^{\log_{\alpha} 4} = 4$.     The second term is $7^{\log_7 4} = 4$.     Final result: $4 \times 2 = 8$ (Note: specifically $4 \times 2$ because the transcript simplifies the power of 7 using base properties).