Relations and Functions: Mixed Practice — Worked Solutions

Ques­tion 1: Equal ordered pairs#

The prob­lem#

Find $a$ and $b$ if $(a + 3, 2b - 1) = (5, a + b)$.

Under­stand­ing the prob­lem#

Two ordered pairs are equal only when their first entries are equal and their sec­ond entries are equal. This gives you two equa­tions.

The idea#

Match first with first and sec­ond with sec­ond, then solve the two equa­tions.

Step-by-step solu­tion#

Step 1. First entries equal.

$$a + 3 = 5 \quad\Longrightarrow\quad a = 2.$$

Step 2. Sec­ond entries equal.

$$2b - 1 = a + b.$$

Step 3. Put $a = 2$ and solve for $b$.

$$\begin{aligned} 2b - 1 &= 2 + b\\ 2b - b &= 2 + 1\\ b &= 3. \end{aligned}$$

Check­ing the answer#

Left pair: $(2 + 3, 6 - 1) = (5, 5)$. Right pair: $(5, 2 + 3) = (5, 5)$. ✓

Answer#

$a = 2$, $b = 3$

Ques­tion 2: $A \times A$ and the num­ber of rela­tions#

The prob­lem#

$A = \{a, b\}$. Write $A \times A$ and count the rela­tions on $A$.

Under­stand­ing the prob­lem#

$A \times A$ is the set of all ordered pairs $(x, y)$ with both $x$ and $y$ in $A$. A rela­tion on $A$ is any sub­set of $A \times A$. So count­ing rela­tions means count­ing sub­sets.

The idea#

List all pairs. If a set has $n$ ele­ments it has $2^n$ sub­sets.

Step-by-step solu­tion#

Step 1. Pair each ele­ment of $A$ with each ele­ment of $A$.

$$A \times A = \{(a, a),\ (a, b),\ (b, a),\ (b, b)\}.$$

Step 2. Count: $n(A \times A) = 2 \times 2 = 4$.

Step 3. Each rela­tion is a sub­set of these $4$ pairs, and each pair is either in or out of the sub­set — two choices per pair.

$$\text{number of relations} = 2^4 = 16.$$

Check­ing the answer#

The gen­eral rule: if $n(A) = m$, there are $2^{m^2}$ rela­tions on $A$. With $m = 2$: $2^4 = 16$. ✓ (This count includes the empty rela­tion and $A \times A$ itself.)

Answer#

$A \times A = \{(a, a), (a, b), (b, a), (b, b)\}$; there are $16$ rela­tions on $A$.

Ques­tion 3: Carte­sian prod­uct dis­trib­utes over union#

The prob­lem#

$A = \{1, 2, 3\}$, $B = \{2, 3\}$, $C = \{3, 4\}$. Ver­ify $A \times (B \cup C) = (A \times B) \cup (A \times C)$.

Under­stand­ing the prob­lem#

"Ver­ify" means: work out both sides for these par­tic­u­lar sets and show that they are the same set.

The idea#

Com­pute the left side by first find­ing $B \cup C$. Com­pute the right side by find­ing the two prod­ucts and tak­ing their union. Com­pare.

Step-by-step solu­tion#

Step 1. Find $B \cup C$ (all ele­ments in either set, with­out repeats).

$$B \cup C = \{2, 3, 4\}.$$

Step 2. Left side: pair each of $1, 2, 3$ with each of $2, 3, 4$.

$$A \times (B \cup C) = \{(1,2), (1,3), (1,4), (2,2), (2,3), (2,4), (3,2), (3,3), (3,4)\}.$$

Step 3. Right side, first part.

$$A \times B = \{(1,2), (1,3), (2,2), (2,3), (3,2), (3,3)\}.$$

Step 4. Right side, sec­ond part.

$$A \times C = \{(1,3), (1,4), (2,3), (2,4), (3,3), (3,4)\}.$$

Step 5. Take the union, writ­ing each pair once (the pairs $(1,3), (2,3), (3,3)$ appear in both).

$$(A \times B) \cup (A \times C) = \{(1,2), (1,3), (1,4), (2,2), (2,3), (2,4), (3,2), (3,3), (3,4)\}.$$

Step 6. The sets in Steps 2 and 5 con­tain exactly the same nine pairs, so the two sides are equal.

Check­ing the answer#

Count: $n(A) \times n(B \cup C) = 3 \times 3 = 9$ pairs on the left, and $9$ dis­tinct pairs on the right. ✓

Answer#

Both sides equal $\{(1,2), (1,3), (1,4), (2,2), (2,3), (2,4), (3,2), (3,3), (3,4)\}$, so the iden­tity holds.

Ques­tion 4: List­ing a rela­tion defined by a rule#

The prob­lem#

On $\{1, 2, 3, 4, 5\}$, $R = \{(a, b) : a < b \text{ and } a + b \text{ is odd}\}$. List $R$; give its domain and range.

Under­stand­ing the prob­lem#

You must find every pair $(a, b)$ from the set that meets both con­di­tions: $a$ is smaller than $b$, and $a + b$ is odd. The domain of $R$ is the set of first entries that actu­ally occur; the range is the set of sec­ond entries that occur.

The idea#

$a + b$ is odd exactly when one of $a, b$ is odd and the other even. Go through $a = 1, 2, 3, 4$ in turn and list the larger $b$ of oppo­site par­ity.

Step-by-step solu­tion#

Step 1. $a = 1$ (odd): need even $b > 1$: $b = 2, 4$. Pairs $(1, 2), (1, 4)$.

Step 2. $a = 2$ (even): need odd $b > 2$: $b = 3, 5$. Pairs $(2, 3), (2, 5)$.

Step 3. $a = 3$ (odd): need even $b > 3$: $b = 4$. Pair $(3, 4)$.

Step 4. $a = 4$ (even): need odd $b > 4$: $b = 5$. Pair $(4, 5)$.

Step 5. $a = 5$: there is no $b > 5$ in the set, so no pair.

$$R = \{(1, 2), (1, 4), (2, 3), (2, 5), (3, 4), (4, 5)\}.$$

Step 6. Read off the first entries and the sec­ond entries.

$$\text{domain} = \{1, 2, 3, 4\}, \qquad \text{range} = \{2, 3, 4, 5\}.$$

Check­ing the answer#

Each pair: $1+2=3$, $1+4=5$, $2+3=5$, $2+5=7$, $3+4=7$, $4+5=9$ — all odd, and in each pair the first is smaller. ✓

Answer#

$R = \{(1, 2), (1, 4), (2, 3), (2, 5), (3, 4), (4, 5)\}$; domain $\{1, 2, 3, 4\}$; range $\{2, 3, 4, 5\}$.

Com­mon mis­take to avoid#

The domain of $R$ is not the whole set $\{1, 2, 3, 4, 5\}$ — only the first entries that actu­ally appear. Here $5$ never appears first.

Ques­tion 5: Is this rela­tion a func­tion?#

The prob­lem#

Is $\{(2, 1), (5, 1), (8, 1), (11, 1)\}$ a func­tion? Its domain and range?

Under­stand­ing the prob­lem#

A rela­tion is a func­tion if no first entry is repeated with two dif­fer­ent sec­ond entries — each input has exactly one out­put.

The idea#

Look at the first entries. Dif­fer­ent inputs are allowed to share the same out­put.

Step-by-step solu­tion#

Step 1. The first entries are $2, 5, 8, 11$ — all dif­fer­ent. So each input has exactly one image. It is a func­tion.

Step 2. Every input goes to $1$. That is allowed (it is a con­stant func­tion on this domain).

Step 3. Domain and range.

$$\text{domain} = \{2, 5, 8, 11\}, \qquad \text{range} = \{1\}.$$

Check­ing the answer#

No input appears twice, so the def­i­n­i­tion of a func­tion is met. ✓

Answer#

Yes, it is a func­tion; domain $\{2, 5, 8, 11\}$, range $\{1\}$.

Ques­tion 6: Domain, a value, and solv­ing $f(x) = 3$#

The prob­lem#

$\displaystyle f(x) = \frac{x - 1}{x + 2}$. Find the domain, $f(0)$ and $x$ with $f(x) = 3$.

Under­stand­ing the prob­lem#

Three tasks: (a) find all real $x$ where the for­mula works; (b) eval­u­ate at $x = 0$; (c) find the input that gives out­put $3$.

The idea#

A frac­tion is unde­fined only where its denom­i­na­tor is zero. For (c), set the for­mula equal to $3$ and cross-mul­ti­ply.

Step-by-step solu­tion#

Part (a): domain

Step 1. The denom­i­na­tor is zero when $x + 2 = 0$, that is $x = -2$. Every other real num­ber works.

$$\text{domain} = \mathbf{R} - \{-2\}.$$

Part (b): $f(0)$

Step 1. Sub­sti­tute $x = 0$.

$$\displaystyle f(0) = \frac{0 - 1}{0 + 2} = -\frac{1}{2}.$$

Part (c): $f(x) = 3$

Step 1. Set up the equa­tion.

$$\displaystyle \frac{x - 1}{x + 2} = 3.$$

Step 2. Mul­ti­ply both sides by $(x + 2)$, which is non-zero in the domain.

$$x - 1 = 3(x + 2) = 3x + 6.$$

Step 3. Col­lect terms.

$$\displaystyle \begin{aligned} -1 - 6 &= 3x - x\\ -7 &= 2x\\ x &= -\frac{7}{2}. \end{aligned}$$

Check­ing the answer#

$\displaystyle f\!\left(-\tfrac{7}{2}\right) = \frac{-\tfrac{7}{2} - 1}{-\tfrac{7}{2} + 2} = \frac{-\tfrac{9}{2}}{-\tfrac{3}{2}} = 3$. ✓ And $\displaystyle -\tfrac{7}{2} \ne -2$, so it is in the domain.

Answer#

Domain $\mathbf{R} - \{-2\}$; $\displaystyle f(0) = -\tfrac{1}{2}$; $f(x) = 3$ when $\displaystyle x = -\tfrac{7}{2}$.

Ques­tion 7: Domain and range of a square root#

The prob­lem#

Find the domain and range of $f(x) = \sqrt{16 - x^2}$.

Under­stand­ing the prob­lem#

A square root of a neg­a­tive num­ber is not real, so the expres­sion under the root must be $\ge 0$. That gives the domain. Then find all the val­ues the func­tion can actu­ally take.

The idea#

Solve $16 - x^2 \ge 0$ for the domain. For the range, see how large and how small $16 - x^2$ gets on that domain, then take square roots.

Step-by-step solu­tion#

Step 1. Domain con­di­tion.

$$16 - x^2 \ge 0 \quad\Longrightarrow\quad x^2 \le 16 \quad\Longrightarrow\quad -4 \le x \le 4.$$

So the domain is $[-4, 4]$.

Step 2. On $[-4, 4]$, $x^2$ runs from $0$ (at $x = 0$) up to $16$ (at $x = \pm 4$). So

$$0 \le 16 - x^2 \le 16.$$

Step 3. Take square roots (the square root keeps order for non-neg­a­tive num­bers).

$$0 \le \sqrt{16 - x^2} \le 4.$$

Both ends are reached: $f(\pm 4) = 0$ and $f(0) = 4$. So the range is $[0, 4]$.

Check­ing the answer#

The graph $y = \sqrt{16 - x^2}$ is the upper half of the cir­cle $x^2 + y^2 = 16$: it runs from $x = -4$ to $x = 4$ and rises from height $0$ to $4$. ✓

Answer#

Domain $[-4, 4]$; range $[0, 4]$.

Ques­tion 8: Range of a sum of two mod­uli#

The prob­lem#

Find the range of $f(x) = \lvert x - 3 \rvert + \lvert x + 1 \rvert$.

Under­stand­ing the prob­lem#

You need every value $f(x)$ can take as $x$ runs over all real num­bers. Each mod­u­lus changes its for­mula at the point where the inside is zero: $x = 3$ and $x = -1$.

The idea#

Split the real line at $-1$ and $3$, remove the mod­u­lus signs on each piece, and see what val­ues each piece pro­duces. (You can also think of $f(x)$ as the dis­tance from $x$ to $3$ plus the dis­tance from $x$ to $-1$ on the num­ber line.)

Step-by-step solu­tion#

Step 1. For $x < -1$: both $x - 3$ and $x + 1$ are neg­a­tive.

$$f(x) = -(x - 3) - (x + 1) = -2x + 2.$$

As $x$ goes from $-1$ down to $-\infty$, this goes from $4$ up to $\infty$ (val­ues greater than $4$).

Step 2. For $-1 \le x \le 3$: $x - 3 \le 0$ but $x + 1 \ge 0$.

$$f(x) = -(x - 3) + (x + 1) = 4.$$

The func­tion is con­stant, equal to $4$, on this whole inter­val.

Step 3. For $x > 3$: both are pos­i­tive.

$$f(x) = (x - 3) + (x + 1) = 2x - 2.$$

As $x$ goes from $3$ to $\infty$, this goes from $4$ up to $\infty$.

Step 4. Put the pieces together: the small­est value is $4$, and every value above $4$ is reached.

$$\text{range} = [4, \infty).$$

Check­ing the answer#

$f(0) = 3 + 1 = 4$; $f(5) = 2 + 6 = 8$; $f(-3) = 6 + 2 = 8$. Noth­ing below $4$ appears. On the num­ber line, the dis­tance from $-1$ to $3$ is $4$, and a point between them has dis­tances adding to exactly $4$. ✓

Answer#

Range $= [4, \infty)$.

Ques­tion 9: Quo­tient of two func­tions#

The prob­lem#

$f(x) = x + 1$, $g(x) = x^2 - 1$. Find $\displaystyle \left(\tfrac{g}{f}\right)(x)$ and its domain; sim­plify.

Under­stand­ing the prob­lem#

$\displaystyle \left(\tfrac{g}{f}\right)(x)$ means $\displaystyle \frac{g(x)}{f(x)}$. Its domain is where both $f$ and $g$ are defined and $f(x) \ne 0$.

The idea#

Find the domain from the orig­i­nal quo­tient before can­celling any­thing; then sim­plify.

Step-by-step solu­tion#

Step 1. Write the quo­tient.

$$\displaystyle \left(\frac{g}{f}\right)(x) = \frac{x^2 - 1}{x + 1}.$$

Step 2. Domain: $f$ and $g$ are defined for all real $x$; exclude $f(x) = 0$, i.e. $x + 1 = 0$, $x = -1$.

$$\text{domain} = \mathbf{R} - \{-1\}.$$

Step 3. Sim­plify using $x^2 - 1 = (x - 1)(x + 1)$ and can­cel $(x + 1)$, which is non-zero on the domain.

$$\displaystyle \frac{(x - 1)(x + 1)}{x + 1} = x - 1, \qquad x \ne -1.$$

Check­ing the answer#

At $x = 2$: $\displaystyle \tfrac{g(2)}{f(2)} = \tfrac{3}{3} = 1$ and $x - 1 = 1$. ✓

Answer#

$\displaystyle \left(\tfrac{g}{f}\right)(x) = x - 1$, with domain $\mathbf{R} - \{-1\}$.

Com­mon mis­take to avoid#

Do not say the domain is all of $\mathbf{R}$ because $x - 1$ is defined every­where. The domain comes from the orig­i­nal quo­tient, so $x = -1$ stays excluded.

Ques­tion 10: Graph of the signum func­tion#

The prob­lem#

Draw the graph of the signum func­tion and state its range.

Under­stand­ing the prob­lem#

The signum func­tion gives the sign of a num­ber:

$$f(x) = \begin{cases} -1, & x < 0 \\ 0, & x = 0 \\ 1, & x > 0. \end{cases}$$

You must describe its graph and list the val­ues it takes.

The idea#

Draw each of the three pieces sep­a­rately, tak­ing care with open and filled dots at $x = 0$.

Step-by-step solu­tion#

Step 1. Draw the axes. Mark $1$ and $-1$ on the $y$-axis.

Step 2. For $x > 0$: draw a hor­i­zon­tal ray at height $y = 1$, start­ing just right of the $y$-axis and going to the right. Put an open cir­cle at $(0, 1)$, because $x = 0$ is not included.

Step 3. For $x < 0$: draw a hor­i­zon­tal ray at height $y = -1$ going to the left. Put an open cir­cle at $(0, -1)$.

Step 4. At $x = 0$: put a filled dot at the ori­gin $(0, 0)$, since $f(0) = 0$.

Step 5. The only out­put val­ues are $-1$, $0$ and $1$.

Check­ing the answer#

$\displaystyle f(5) = \tfrac{\lvert 5 \rvert}{5} = 1$, $\displaystyle f(-2) = \tfrac{2}{-2} = -1$, $f(0) = 0$. The graph passes the ver­ti­cal-line test (one point above each $x$). ✓

Answer#

Two hor­i­zon­tal rays, $y = 1$ for $x > 0$ and $y = -1$ for $x < 0$ (open cir­cles at $x = 0$), plus the point $(0, 0)$; range $\{-1, 0, 1\}$.

Ques­tion 11: Profit func­tion and break-even#

The prob­lem#

The cost of mak­ing $x$ pens is $C(x) = 300 + 8x$ rupees and each sells for ₹14. Write the profit $P(x)$ and find how many pens must be sold to break even.

Under­stand­ing the prob­lem#

Rev­enue is money received from sales; profit is rev­enue minus cost. "Break even" means there is no profit and no loss: $P(x) = 0$.

The idea#

Rev­enue $R(x) = 14x$. Then $P(x) = R(x) - C(x)$, and solve $P(x) = 0$.

Step-by-step solu­tion#

Step 1. Rev­enue from sell­ing $x$ pens at ₹14 each.

$$R(x) = 14x.$$

Step 2. Profit.

$$P(x) = 14x - (300 + 8x) = 14x - 300 - 8x = 6x - 300.$$

Step 3. Break even.

$$\displaystyle 6x - 300 = 0 \quad\Longrightarrow\quad x = \frac{300}{6} = 50.$$

Check­ing the answer#

For $50$ pens: cost $= 300 + 400 = ₹700$, rev­enue $= 14 \times 50 = ₹700$. Equal, so no profit and no loss. ✓ (Each pen brings in ₹6 above its mak­ing cost, and ₹300 of fixed cost needs $50$ such pens.)

Answer#

$P(x) = 6x - 300$ rupees; $50$ pens must be sold to break even.

Ques­tion 12: A lin­ear func­tion from two points#

The prob­lem#

A lin­ear func­tion on $\mathbf{Z}$ con­tains $(0, -2)$ and $(3, 7)$. Find $f(x)$.

Under­stand­ing the prob­lem#

A lin­ear func­tion has the form $f(x) = mx + c$. "Con­tains $(0, -2)$" means $f(0) = -2$; "con­tains $(3, 7)$" means $f(3) = 7$.

The idea#

Use each point to get an equa­tion in $m$ and $c$, and solve.

Step-by-step solu­tion#

Step 1. From $(0, -2)$:

$$f(0) = m \cdot 0 + c = -2 \quad\Longrightarrow\quad c = -2.$$

Step 2. From $(3, 7)$:

$$f(3) = 3m + c = 7 \quad\Longrightarrow\quad 3m - 2 = 7 \quad\Longrightarrow\quad 3m = 9 \quad\Longrightarrow\quad m = 3.$$

Step 3. Write the func­tion.

$$f(x) = 3x - 2.$$

Check­ing the answer#

$f(0) = -2$ ✓ and $f(3) = 9 - 2 = 7$ ✓.

Answer#

$f(x) = 3x - 2$