Sets, Relations and Functions — Practice

Question 1

In a group of $60$ people, $27$ like tea, $42$ like coffee and $30$ like both. How many like exactly one of the two drinks?

Accepted Answer

$S_1=27+42=69$ and $S_2=30$.

$n_{	ext{exactly }1}=69-2(30)=9$.

As a check the union is $69-30=39$, and $39-30=9$ agrees.

Question 2

How many relations on a four-element set are symmetric but not reflexive?

Accepted Answer

Symmetric relations: the four diagonal cells are free and the six off-diagonal pairs are tied, giving $2^{10}=1024$.

Symmetric and reflexive: only the six pairs are free, giving $2^6=64$.

$1024-64=960$.

Question 3

How many onto functions are there from a four-element set to a three-element set?

Accepted Answer

$3^4-\binom{3}{1}2^4+\binom{3}{2}1^4=81-48+3=36$.

Check: an onto map from four elements to three must send exactly one pair together, so choose the pair in $\binom{4}{2}=6$ ways and arrange the three resulting blocks in $3!=6$ ways, giving $36$.

Question 4

How many equivalence relations on $\{1,2,3,4,5\}$ have $1$, $2$ and $3$ all in the same class?

Accepted Answer

Fusing $1$, $2$ and $3$ leaves three objects to partition, so the answer is the Bell number $B_3=5$.

Explicitly: all together; the fused block alone with $4$ and $5$ separate; the fused block with $4$; the fused block with $5$; and the fused block alone with $\{4,5\}$ together.

Question 5

Find the range of $f(x)=\dfrac{x+2}{x^2+3x+3}$ for real $x$.

Accepted Answer

The denominator has discriminant $9-12<0$, so the domain is all of $\mathbb{R}$.

$yx^2+(3y-1)x+(3y-2)=0$.

For $y=0$: $-x-2=0$ gives $x=-2$, so $y=0$ is attained.

Otherwise $D=(3y-1)^2-4y(3y-2)=-3y^2+2y+1\ge0$, that is $(3y+1)(y-1)\le0$, so $-	frac13\le y\le1$.

Range: $\left[-	frac13,\,1ight]$.

Question 6

Let $f(x)=x^2+4x+7$ on $x\ge-2$. Solve $f(x)=f^{-1}(x)$.

Accepted Answer

On $x\ge-2$ the vertex is at the left endpoint, so $f$ is increasing and the equation reduces to $f(x)=x$.

$x^2+4x+7=x$ gives $x^2+3x+7=0$, whose discriminant is $9-28<0$.

There is no solution: the graph of $f$ lies entirely above $y=x$, so it never meets its own reflection.

Question 7

Find the domain of $f(x)=\sqrt{\,\log_{1/2}\!\left(\dfrac{x-3}{x+4}\right)}$.

Accepted Answer

Argument positive: $\dfrac{x-3}{x+4}>0$ gives $x<-4$ or $x>3$. Non-negative square root: since the base is $ frac12<1$, $\log_{1/2}u\ge0$ needs $0-4$. Intersecting: $x>3$. Domain $(3,\infty)$.

Question 8

How many functions $f:\{1,2,3,4,5\}	o\{1,2,3,4,5\}$ satisfy $f(f(x))=x$ for every $x$?

Accepted Answer

By the number of two-cycles:

none gives $1$; one gives $\binom{5}{2}=10$; two gives $\dfrac{1}{2}\binom{5}{2}\binom{3}{2}=15$.

Total $1+10+15=26$.

This matches the recursion $I_5=I_4+4I_3=10+4(4)=26$.

Question 9

Find the fundamental period of $f(x)=\sin^4 x+\cos^4 x$.

Accepted Answer

$\sin^4x+\cos^4x=(\sin^2x+\cos^2x)^2-2\sin^2x\cos^2x=1-	frac12\sin^2 2x$.

Using $\sin^2 2x=\dfrac{1-\cos4x}{2}$ gives $\dfrac34+\dfrac{\cos4x}{4}$.

Period $\dfrac{2\pi}{4}=\dfrac{\pi}{2}$, well below the $2\pi$ each original term suggests.

Question 10

A continuous function satisfies $f(x+y)=f(x)f(y)$ for all real $x,y$, and $f(1)=3$. Find $f(4)$ and explain what the continuity is doing.

Accepted Answer

If $f(a)=0$ for some $a$ then $f(x)=f(x-a)f(a)=0$ everywhere, so assume not.

$f(n)=f(1)^n$ by repeated multiplication, so $f(4)=81$.

The same argument gives $f(r)=3^r$ for rational $r$ only. Continuity extends it to $f(x)=3^x$ for every real $x$; without it the values at irrational arguments are not determined.

Question 11

Solve $3f(x)-2f\!\left(\dfrac1xight)=\dfrac1x$ for $x
eq0$.

Accepted Answer

Original: $3f(x)-2f(1/x)=\dfrac1x$.

Substituted: $3f(1/x)-2f(x)=x$.

Multiply the first by $3$ and the second by $2$, then add: $9f(x)-4f(x)=\dfrac3x+2x$.

$f(x)=\dfrac{1}{5}\left(\dfrac3x+2xight)$. Substituting back confirms it, which also proves a solution exists.

Question 12

Solve $[x]^2=x[x]$ for real $x$, where $[\,\cdot\,]$ is the greatest integer function.

Accepted Answer

Substituting, $[x]^2=[x]^2+[x]\{x\}$, so $[x]\{x\}=0$.

Either $\{x\}=0$, giving every integer $x$; or $[x]=0$, giving $x\in[0,1)$.

Solution set: all integers together with the interval $[0,1)$. Note that dividing by $[x]$ at the start would silently discard the whole interval.

Practice — Sets, Relations and Functions

Sign in to save attempts to your dashboard and track your accuracy over time

Find a verified tutor

Join a live course

Browse coaching centres

About

Categories

Support

Documentation

Get Study Tips & Updates

Download Our App

Download for iOS

Download for Android

Popular Cities