IMO Practice Test — Relations and Functions
6 Questions • 20 min • Olympiad level
20:00
Question 1 of 6
hard
Let $f: \mathbb{R} \to \mathbb{R}$ be defined by $f(x) = \frac{x}{1+|x|}$. Then $f$ is:
Injective but not surjective
Surjective but not injective
Bijective
Neither injective nor surjective
Explanation: For $x>0$: $f'(x)=\frac{1}{(1+x)^2}>0$ and for $x<0$: $f'(x)=\frac{1}{(1-x)^2}>0$. Strictly increasing so injective. Range $\in (-1,1) \ne \mathbb{R}$, so not surjective.
Question 2 of 6
hard
If $f(x) = \frac{x+1}{x-1}$ for $x \ne 1$, then $f(f(f(f(x))))$ equals:
$x$
$f(x)$
$\frac{1}{x}$
$-x$
Explanation: $f(f(x))=f\left(\frac{x+1}{x-1}\right)=\frac{\frac{x+1}{x-1}+1}{\frac{x+1}{x-1}-1}=\frac{2x}{2}=x$. So $f(f(x))=x$, meaning $f$ is its own inverse. $f^4(x)=f^2(f^2(x))=x$.
Question 3 of 6
medium
The number of onto functions from $\{1,2,3,4\}$ to $\{a,b\}$ is:
14
16
12
8
Explanation: Total functions = $2^4=16$. Non-onto (all to $a$ only or all to $b$ only) = 2. Onto = $16-2=14$.
Question 4 of 6
hard
How many equivalence relations can be defined on $A = \{1,2,3\}$?
5
6
4
8
Explanation: Equivalence relations on $\{1,2,3\}$ correspond to partitions of the set. Partitions: $\{\{1\},\{2\},\{3\}\}$, $\{\{1,2\},\{3\}\}$, $\{\{1,3\},\{2\}\}$, $\{\{2,3\},\{1\}\}$, $\{\{1,2,3\}\}$ — exactly 5.
Question 5 of 6
hard
If $*$ is defined on $\mathbb{R}-\{-1\}$ by $a*b=\frac{a+b}{1+ab}$ (like $\tan$ addition), then $* $ is:
Commutative and associative
Commutative but not associative
Associative but not commutative
Neither
Explanation: $a*b=\frac{a+b}{1+ab}=\frac{b+a}{1+ba}=b*a$ (commutative). Associativity follows from the tan addition formula being associative in its domain.
Question 6 of 6
hard
If $f: \mathbb{R} \to \mathbb{R}$ satisfies $f(x+y)=f(x)+f(y)$ for all $x,y$ and $f(1)=3$, then $\sum_{r=1}^{n}f(r)$ is:
$\frac{3n(n+1)}{2}$
$3n^2$
$\frac{n(n+1)}{2}$
$3n$
Explanation: From the functional equation, $f(n)=nf(1)=3n$. So $\sum_{r=1}^n f(r)=3\sum_{r=1}^n r=\frac{3n(n+1)}{2}$.