Harder modulus problems

This les­son gath­ers the harder mod­u­lus prob­lems that com­pet­i­tive exam­i­na­tions like: equa­tions solved in one stroke by a prop­erty of the mod­u­lus, equa­tions with a mod­u­lus inside a mod­u­lus, inequal­i­ties with frac­tions, and ques­tions that ask "how many inte­gers sat­isfy...". Every one of them uses tools you already have: the prop­er­ties of $|x|$, the crit­i­cal point method, and the sign rules for qua­drat­ics. The skill is in choos­ing the right tool.

Using the equal­ity case of the tri­an­gle inequal­ity#

From the prop­er­ties of the mod­u­lus:

$$|x + y| = |x| + |y| \quad \text{exactly when } xy \geq 0$$

$$|x - y| = |x| + |y| \quad \text{exactly when } xy \leq 0$$

The sec­ond fol­lows from the first by replac­ing $y$ with $-y$. When an equa­tion has the shape $|A| + |B| = |A - B|$, it is true exactly when $AB \leq 0$, and no cases are needed.

Exam­ple 1#

Solve $|x^2 - 2x - 8| + |x^2 + x - 2| = 3|x + 2|$.

Let $A = x^2 - 2x - 8$ and $B = x^2 + x - 2$. Then

$$A - B = (x^2 - 2x - 8) - (x^2 + x - 2) = -3x - 6$$

so $|A - B| = |-3x - 6| = |3x + 6| = 3|x + 2|$, which is the right-hand side. The equa­tion is $|A| + |B| = |A - B|$, which holds exactly when $AB \leq 0$:

$$(x^2 - 2x - 8)(x^2 + x - 2) \leq 0$$

$$(x - 4)(x + 2)(x + 2)(x - 1) \leq 0$$

$$(x + 2)^2(x - 1)(x - 4) \leq 0$$

The fac­tor $(x + 2)^2$ is never neg­a­tive, and it is zero at $x = -2$. So the prod­uct is $\leq 0$ either when $x = -2$, or when $(x - 1)(x - 4) \leq 0$, that is, $x \in [1, 4]$. A sign chart across the crit­i­cal points $-2$, $1$, $4$ reads $+$, $+$, $-$, $+$, with zeros at all three points.

Answer: $x \in \{-2\} \cup [1, 4]$.

Count­ing inte­ger solu­tions#

Exam­ple 2#

Find the num­ber of inte­gers sat­is­fy­ing $\displaystyle |x| + \left|\frac{4 - x^2}{x}\right| = \left|\frac{4}{x}\right|$.

The expres­sion con­tains $\displaystyle \frac{1}{x}$, so first note that $x \neq 0$. Using $\displaystyle \left|\frac{p}{q}\right| = \frac{|p|}{|q|}$,

$$\displaystyle |x| + \frac{|4 - x^2|}{|x|} = \frac{4}{|x|}$$

Mul­ti­ply through by $|x|$, which is pos­i­tive, and use $|x|^2 = x^2$ and $|4 - x^2| = |x^2 - 4|$:

$$x^2 + |(x + 2)(x - 2)| = 4 \qquad (1)$$

The crit­i­cal points are $-2$ and $2$.

Case (i): $x \in (-\infty, -2] \cup [2, \infty)$. Here $(x + 2)(x - 2) \geq 0$, so (1) becomes

$$x^2 + x^2 - 4 = 4 \implies 2x^2 = 8 \implies x^2 = 4 \implies x = \pm 2$$

Both lie in this case: $\{-2, 2\}$.

Case (ii): $x \in (-2, 2)$. Here $(x + 2)(x - 2) < 0$, so (1) becomes

$$x^2 - (x^2 - 4) = 4 \implies 4 = 4$$

This is true for every $x$ in the inter­val. So all of $(-2, 2)$ sat­is­fies (1).

Com­bin­ing, (1) holds on $[-2, 2]$. But the orig­i­nal equa­tion needs $x \neq 0$, so the solu­tion set is $[-2, 2] - \{0\}$. The inte­gers in it are $-2, -1, 1, 2$.

Answer: $4$ inte­gers. (Leav­ing out the con­di­tion $x \neq 0$ would wrongly give $5$.)

A mod­u­lus inside a mod­u­lus#

Exam­ple 3#

Find the inte­gral val­ues of $x$ sat­is­fy­ing $|x - |x - 4|| = 4$.

Work from the out­side in. By $|T| = k \implies T = \pm k$:

$$x - |x - 4| = 4 \quad \text{or} \quad x - |x - 4| = -4$$

First: $x - |x - 4| = 4$ gives $|x - 4| = x - 4$. A mod­u­lus equals the quan­tity itself exactly when the quan­tity is non-neg­a­tive, so $x - 4 \geq 0$, that is, $x \in [4, \infty)$.

Sec­ond: $x - |x - 4| = -4$ gives $|x - 4| = x + 4$. Use the crit­i­cal point $4$.

• If $x < 4$: $4 - x = x + 4$, so $2x = 0$ and $x = 0$. This is less than $4$, so it is valid.
• If $x \geq 4$: $x - 4 = x + 4$, which would need $-4$ and $4$ to be equal. Impos­si­ble, so no solu­tion.

Answer: $x \in \{0\} \cup [4, \infty)$. This con­tains infi­nitely many inte­gers ($0, 4, 5, 6, \ldots$).

Inequal­i­ties with frac­tions#

When an inequal­ity has $x$ in a denom­i­na­tor, never mul­ti­ply across by an expres­sion whose sign you do not know. Instead, bring every­thing to one side, com­bine into a sin­gle frac­tion and study its sign. Remem­ber that $\displaystyle \frac{p}{q} < 0$ has the same solu­tion as $pq < 0$ (with $q \neq 0$).

Exam­ple 4#

Solve $\displaystyle \left|x + \frac{1}{x}\right| > 2$.

By $|T| > a \iff T < -a$ or $T > a$:

First part: $\displaystyle x + \frac{1}{x} < -2$.

$$\displaystyle \frac{x^2 + 1}{x} + 2 < 0 \implies \frac{x^2 + 2x + 1}{x} < 0 \implies \frac{(x + 1)^2}{x} < 0$$

The numer­a­tor is a square, so it is pos­i­tive except at $x = -1$, where the frac­tion is $0$ (not less than $0$). So we need $x < 0$ and $x \neq -1$.

Sec­ond part: $\displaystyle x + \frac{1}{x} > 2$.

$$\displaystyle \frac{x^2 + 1}{x} - 2 > 0 \implies \frac{(x - 1)^2}{x} > 0$$

This needs $x > 0$ and $x \neq 1$.

Answer: $x \in \mathbb{R} - \{-1, 0, 1\}$. (At $x = \pm 1$, $\displaystyle \left|x + \frac{1}{x}\right|$ equals exactly $2$, not more than $2$.)

Exam­ple 5#

Find the num­ber of inte­gers sat­is­fy­ing $\displaystyle \frac{|x - 1|}{x + 2} > 1$.

The crit­i­cal point of the mod­u­lus is $x = 1$. (Also $x \neq -2$.)

Case (i): $x < 1$. Then $|x - 1| = 1 - x$:

$$\displaystyle \frac{1 - x}{x + 2} - 1 > 0 \implies \frac{1 - x - x - 2}{x + 2} > 0 \implies \frac{-1 - 2x}{x + 2} > 0$$

Mul­ti­ply by $-1$ and reverse: $\displaystyle \frac{2x + 1}{x + 2} < 0$, which is the same as $(2x + 1)(x + 2) < 0$. So $\displaystyle x \in \left(-2, -\frac{1}{2}\right)$. All of these are less than $1$. The only inte­ger is $-1$.

Case (ii): $x \geq 1$. Then $|x - 1| = x - 1$:

$$\displaystyle \frac{x - 1}{x + 2} - 1 > 0 \implies \frac{-3}{x + 2} > 0 \implies x + 2 < 0 \implies x < -2$$

This has no com­mon value with $x \geq 1$.

Answer: the solu­tion set is $\displaystyle \left(-2, -\frac{1}{2}\right)$, and the num­ber of inte­gers is $1$ (namely $x = -1$).

Exam­ple 6#

Solve $\displaystyle \left|\frac{12x}{4x^2 + 9}\right| \leq 1$.

The denom­i­na­tor $4x^2 + 9$ is always pos­i­tive, so the inequal­ity is $\displaystyle \frac{12|x|}{4x^2 + 9} \leq 1$, that is, $12|x| \leq 4x^2 + 9$.

Case (i): $x < 0$. $-12x \leq 4x^2 + 9$, so $4x^2 + 12x + 9 \geq 0$, that is, $(2x + 3)^2 \geq 0$: true for all real $x$. Com­mon val­ues: $(-\infty, 0)$.

Case (ii): $x \geq 0$. $12x \leq 4x^2 + 9$, so $(2x - 3)^2 \geq 0$: always true. Com­mon val­ues: $[0, \infty)$.

Answer: $x \in (-\infty, 0) \cup [0, \infty) = \mathbb{R}$.

A sum of mod­uli equal to zero#

Exam­ple 7#

Solve $|x - 1| + |x - 2| = 0$.

Each mod­u­lus is at least $0$, so their sum is $0$ only if both are $0$: $x = 1$ and $x = 2$ at the same time. No num­ber does both. No real $x$ sat­is­fies the equa­tion.

Prac­tice#

1. Solve $|x^2 - 3x + 2| + |x^2 - 5x + 6| = 2|x - 2|$.
2. Solve $\displaystyle \left|x + \frac{1}{x}\right| \geq 2$.
3. Solve $|x - |x - 2|| = 2$.
4. Solve $\displaystyle \frac{|x + 2|}{x - 1} > 1$.
5. Solve $|x| + |x - 1| = 0$.
6. Solve $|x - 3| + |2x + 1| = 0$.
7. Solve $\displaystyle \left|\frac{2x}{x^2 + 1}\right| \leq 1$.
8. Find the num­ber of inte­gers sat­is­fy­ing $\displaystyle \frac{|x - 2|}{x + 1} > 1$.
9. Find the num­ber of inte­gers sat­is­fy­ing $\displaystyle |x| + \left|\frac{9 - x^2}{x}\right| = \left|\frac{9}{x}\right|$.

Answers#