Basic Mathematics - Lecture 10: Modulus Inequality

Topics

• Modulus Inequality
• Graph of Mod
• Properties of Mod
• Question Practice

Last Class Revision

• True/False Questions:
  1. If $|x| = \alpha$, then $x = \pm a$ - False
     • If $|x| = a$, then $x = \pm a$
  2. If $x^2 + 5|x| + 6 = 0$, then $x = \pm 2, \pm 3$ - False
     • Case I: $x \geq 0$, then $x^2 + 5x + 6 = 0$ implies $(x+2)(x+3) = 0$ so $x = -2, -3$ (rejected since $x \geq 0$)
     • Case II: x < 0, then $x^2 - 5x + 6 = 0$ implies $(x-2)(x-3) = 0$ so $x = 2, 3$ (rejected since x < 0)
     • No solutions.
  3. If $\sqrt{x^2 - 3x} = 3x - x^2$, then $x \in [0, 3]$ - True
     • The expression implies $x^2 - 3x \geq 0$.
     • Also, it implies $3x - x^2 \geq 0$ or $x^2 - 3x \leq 0$.
     • Thus, $x(x-3) \leq 0$, which means $x \in [0, 3]$.

Clarity

• $|x| = x$ if x > 0
• $|x| = 0$ if $x = 0$
• $|x| = -x$ if x < 0
• $|x| = x$ if $x \geq 0$
• $|x| = -x$ if $x \leq 0$

Question

• Sum of all non-negative integral values of 'x' satisfying $||x-1| - 2| - 3 = 5$

• $||x-1| - 2| = 8$

  • Case 1: $|x-1| - 2 = 8$ implies $|x-1| = 10$ which means $x-1 = 10$ or $x-1 = -10$. Thus $x = 11$ or $x = -9$.
  • Case 2: $|x-1| - 2 = -8$ implies $|x-1| = -6$, which is impossible.

• The non-negative integral value is $x = 11$.

• Since $x=11$ is the only non-negative integral value. The prompt requests the SUM of non-negative integral values. Thus the sum would be 11 as well.

• Sum of non-negative integral values: $11$

Question

• Solve for x: $|x - 1| - 2|x - 3| + 3|x + 1| = 10$
• Cases:
  • Case I: $x \leq -1$ (all moduli are negative)
    • $-(x - 1) - 2(3 - x) + 3(-x - 1) = 10$
    • $-x + 1 - 6 + 2x - 3x - 3 = 10$
    • $-2x - 8 = 10$
    • $-2x = 18$
    • $x = -9$, which satisfies $x \leq -1$
  • Case II: -1 < x < 1
    • $-(x - 1) - 2(3 - x) + 3(x + 1) = 10$
    • $-x + 1 - 6 + 2x + 3x + 3 = 10$
    • $4x - 2 = 10$
    • $4x = 12$
    • $x = 3$, which does not satisfy -1 < x < 1, thus rejected.
  • Case III: 1 \leq x < 3
    • $(x - 1) - 2(3 - x) + 3(x + 1) = 10$
    • $x - 1 - 6 + 2x + 3x + 3 = 10$
    • $6x - 4 = 10$
    • $6x = 14$
    • $x = \frac{7}{3}$, which satisfies 1 \leq x < 3
  • Case IV: $x \geq 3$
    • $(x - 1) - 2(x - 3) + 3(x + 1) = 10$
    • $x - 1 - 2x + 6 + 3x + 3 = 10$
    • $2x + 8 = 10$
    • $2x = 2$
    • $x = 1$, which doesn't satisfy $x \geq 3$, thus rejected.
• Solutions: $x = -9, \frac{7}{3}$

Question [JEE Mains-2020]

• The product of the roots of the equation $9x^2 - 18|x| + 5 = 0$
• Case 1: $x \geq 0$, then $9x^2 - 18x + 5 = 0$
  • $9x^2 - 15x - 3x + 5 = 0$
  • $3x(3x - 5) - 1(3x - 5) = 0$
  • $(3x - 1)(3x - 5) = 0$
  • $x = \frac{1}{3}, \frac{5}{3}$
• Case 2: x < 0, then $9x^2 + 18x + 5 = 0$
  • $9x^2 + 15x + 3x + 5 = 0$
  • $3x(3x + 5) + 1(3x + 5) = 0$
  • $(3x + 1)(3x + 5) = 0$
  • $x = -\frac{1}{3}, -\frac{5}{3}$
• Product of roots: $\frac{1}{3} \cdot \frac{5}{3} \cdot (-\frac{1}{3}) \cdot (-\frac{5}{3}) = \frac{25}{81}$

Critical Thinking Question (CTQ)

• Solve: \frac{1}{x-1} + \frac{1}{x-2} + \frac{1}{x-3} + \frac{1}{x-4} < \frac{30}{4}

• Pairing terms: \frac{1}{x-1} + \frac{1}{x-4} + \frac{1}{x-2} + \frac{1}{x-3} < 30

  • \frac{x-4+x-1}{(x-1)(x-4)} + \frac{x-3+x-2}{(x-2)(x-3)} < 30
  • \frac{2x-5}{x^2-5x+4} + \frac{2x-5}{x^2-5x+6} < 30

• Let $t = x^2 - 5x$

  • \frac{2x-5}{t+4} + \frac{2x-5}{t+6} < 30
  • \frac{1}{t+4} + \frac{1}{t+6} < \frac{30}{2x-5}

• \frac{1}{x-1} + \frac{1}{x-2} + \frac{1}{x-3} + \frac{1}{x-4} < 30

• Pairing terms and adding:

  • \frac{x-4+x-1}{(x-1)(x-4)} + \frac{x-3+x-2}{(x-2)(x-3)} < 30
  • \frac{2x-5}{x^2-5x+4} + \frac{2x-5}{x^2-5x+6} < 30

• Let $t = x^2 - 5x$

  • \frac{1}{t+4} + \frac{1}{t+6} <30
  • \frac{t+6+t+4}{(t+4)(t+6)} < 30
  • \frac{2t+10}{(t+4)(t+6)} < 30
  • \frac{2(t+5)}{(t+4)(t+6)} < 30

• \frac{2}{t+4} + \frac{1}{t+6} < 30

• $x = (-\infty, -2) \cup (-1, 1) \cup (2, 3) \cup (4, 6) \cup (7, \infty)$

Question

• Value of x satisfying: $|x - 3| + |x + 5| = 8$
• Case 1: x < -5
  • $-(x - 3) - (x + 5) = 8$
  • $-x + 3 - x - 5 = 8$
  • $-2x - 2 = 8$
  • $-2x = 10$
  • $x = -5$
• Case 2: -5 \leq x < 3
  • $-(x - 3) + (x + 5) = 8$
  • $-x + 3 + x + 5 = 8$
  • $8 = 8$
  • $x \in [-5, 3)$
• Case 3: $x \geq 3$
  • $(x - 3) + (x + 5) = 8$
  • $2x + 2 = 8$
  • $2x = 6$
  • $x = 3$
• Therefore, $x \in [-5, 3]$

Question

• Moduli Inequality - No Trick!
• (i) $|x| \geq 2$
• (ii) |x| < 5
• (iii) 3 < |x| \leq 7
• $|x| \geq 2$
  • $x \in (-\infty, -2] \cup [2, \infty)$
• |x| < 5
  • $x \in (-5, 5)$
• 3 < |x|
  • $x < -3$ or $x > 3$
• $|x| \leq 7$
  • $-7 \leq x \leq 7$
• Combining 3 < |x| \leq 7
  • $x \in [-7, -3) \cup (3, 7]$

Question

• \frac{1}{|x| - 3} < 1
• \frac{1}{|x| - 3} - 1 < 0
• \frac{1 - (|x| - 3)}{|x| - 3} < 0
• \frac{4 - |x|}{|x| - 3} < 0
• $|x| - 3 = 0 \implies |x| = 3 \implies x = \pm 3$
• $4 - |x| = 0 \implies |x| = 4 \implies x = \pm 4$
• Therefore, $x \in (-\infty, -4) \cup (-3, 3) \cup (4, \infty)$
• Solve: \frac{1}{|x|-3} < 1
• \frac{1}{|x|+4} - 1 > 0
• \frac{1-|x|-4}{|x|+4} > 0
• \frac{-|x|-3}{|x|+4} > 0
• Since $|x|+4$ is always positive, we just need to examine $|x|-3$
• Since $|x| \geq 0$, $|x|+4$ is always positive, and $-|x|-3$ is always negative. The expression is always negative, which leads to no solution.

Question

• \frac{1}{|x-1|} > \frac{1}{x-5}
• \frac{1}{x-1} - \frac{1}{x-5} > 0
• \frac{x-5-x+1}{(x-1)(x-5)} > 0
• \frac{-4}{(x-1)(x-5)} > 0
• \frac{1}{(x-1)(x-5)} < 0
• (x-1)(x-5) < 0
• Which gives 1 < x < 5
• Final Answer: $x \in (- \infty, -2] \cup [2, 3] \cup (5, \infty)$

Inequalities

• |x| < a
  • -a < x < a
  • $x \in (-a, a)$
• $|x| \le a$
  • $-a \le x \le a$
  • $x \in [-a, a]$

More Inequalities

• |x| > a
  • $x < -a \cup x > a$
  • $x \in (-\infty, -a) \cup (a, \infty)$
• $|x| \geq a$
  • $x \leq -a \cup x \geq a$
  • $x \in (-\infty, -a] \cup [a, \infty)$

Question

• Number of solutions of the equation: $|x + 1| + |x - 3| = 11$
• * Case 1: x < -1
  • $-(x+1)-(x-3) = 11$
  • $-2x + 2 = 11$
  • $x = -\frac{9}{2}$
• Case 2: -1 \leq x < 3
  • $(x+1) - (x-3) = 11$
  • $4 = 11$ No Solution!
• Case 3: $x \ge 3$
  • $(x+1)+(x-3) = 11$
  • $2x = 13$
  • $x = \frac{12}{2}$

Do It By Yourself (DIBY)

• Challenger Problems leads us to New Learning

DIBY-26

• Solve following Inequalities:
  • (i) $\frac{x^2 + 1}{x^2 - 11x + 28} \geq 0$
  • (ii) \frac{x^2 - x + 1}{x^2 + 7x + 15} < 0
  • Ans. x = (-∞, 4) U (7,∞)
  • Ans. x = R

DIBY-27

• The number of real solution of the equation, $x^2- |x|-12=0$ is
  • [JEE Mains-2021]
  • (A) 2
  • (B) 3
  • (c) 1
  • (D) 4

DIBY-28

• Solve $|x^2 + 4x+3|+2x+5=0$

DIBY-29

• The sum of all the real roots of the equation $|x - 2|^2+|x-2|-2=0$ is ____
  • [JEE-1997, 2M]
  • Ans.: 4

DIBY-30

• $(x-3)^2+|x-3|-11 = 0$, the sum of solutions of this equation is?
  • (A) 5
  • (B) 6
  • (c) 7

DIBY-31

• The sum of all roots of the equation $|x^2 - 8x + 15|- 2x + 7 = 0$ is
  • (A) 9-√3
  • (B) 9+ √3
  • (c) 11-√3
  • (D) 11+ √3
  • Ans. (B)

Homework

• Solve DIBY & HW Questions
• DPP
• Module Homework:
  • Prarambh (Topicwise): 1,2,3,4,6,7,9,12,13,40
  • Prabal (JEE Mains): 1,23,38,39,40,45,52,53

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• It's always about the Journey
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