Exercise 1 — level ★☆☆☆☆
\[ f:\mathbb{R}\to\mathbb{R}, \quad f(x)=2x+1 \]
Answer
The function is bijective.
Solution
Injectivity
To check whether the function is injective, suppose two elements share the same image:
\[ f(x_1)=f(x_2) \implies 2x_1+1=2x_2+1 \]
Cancelling the constant term gives:
\[ 2x_1=2x_2 \implies x_1=x_2 \]
Therefore distinct inputs always produce distinct outputs: the function is injective.
Surjectivity
To determine surjectivity, take any real number \(y\) and check whether there exists an \(x\) such that \(f(x)=y\):
\[ y=2x+1 \]
Solving for \(x\):
\[ x=\frac{y-1}{2} \]
This value is always real, so every real number is the image of at least one element of the domain. The function is surjective.
Conclusion
Being both injective and surjective, the function is bijective.
\[ \boxed{f \text{ is bijective}} \]
Exercise 2 — level ★☆☆☆☆
\[ f:\mathbb{R}\to\mathbb{R}, \quad f(x)=x^2 \]
Answer
Neither injective nor surjective (on \(\mathbb{R}\)).
Solution
Injectivity
It suffices to exhibit two distinct inputs with the same output. For example:
\[ f(1)=1, \qquad f(-1)=1 \]
Since \(1\neq -1\) yet their images coincide, the function is not injective.
Surjectivity
The square of any real number is always non-negative:
\[ x^2 \ge 0 \quad \forall x\in\mathbb{R} \]
Hence the range of the function is contained in \([0,+\infty)\). In particular, no negative number is the image of any \(x\).
Therefore the function is not surjective on \(\mathbb{R}\).
Conclusion
\[ \boxed{f \text{ is neither injective nor surjective}} \]
Exercise 3 — level ★☆☆☆☆
\[ f:\mathbb{R}\to\mathbb{R}, \quad f(x)=x^3 \]
Answer
The function is bijective.
Solution
Injectivity
Suppose:
\[ f(x_1)=f(x_2) \implies x_1^3=x_2^3 \]
Since the cubic function is strictly increasing, this equality necessarily implies:
\[ x_1=x_2 \]
Hence the function is injective.
Surjectivity
Take any \(y\in\mathbb{R}\) and consider the equation:
\[ y=x^3 \]
It always has a real solution:
\[ x=\sqrt[3]{y} \]
This shows that every real number is the image of some \(x\), so the function is surjective.
Conclusion
\[ \boxed{f \text{ is bijective}} \]
Exercise 4 — level ★☆☆☆☆
\[ f:\mathbb{N}\to\mathbb{N}, \quad f(n)=n+1 \]
Answer
Injective but not surjective.
Solution
Injectivity
Suppose two natural numbers share the same image:
\[ f(n_1)=f(n_2) \implies n_1+1=n_2+1 \]
It follows immediately that:
\[ n_1=n_2 \]
Hence the function is injective.
Surjectivity
The value \(0\) (or \(1\), depending on the convention for \(\mathbb{N}\)) is not the image of any natural number.
Indeed, there is no \(n\in\mathbb{N}\) such that \(n+1=0\).
Therefore the function is not surjective.
Conclusion
\[ \boxed{f \text{ is injective but not surjective}} \]
Exercise 5 — level ★☆☆☆☆
\[ f:\mathbb{R}\to\mathbb{R}, \quad f(x)=5 \]
Answer
Neither injective nor surjective.
Solution
Injectivity
The function assigns the same value to every real number:
\[ f(x)=5 \quad \forall x\in\mathbb{R} \]
So distinct inputs share the same output. The function is not injective.
Surjectivity
The range of the function consists of the single value \(5\):
\[ \operatorname{Im}(f)=\{5\} \]
Since this set does not coincide with \(\mathbb{R}\), the function is not surjective.
Conclusion
\[ \boxed{f \text{ is neither injective nor surjective}} \]
Exercise 6 — level ★★☆☆☆
\[ f:\mathbb{Z}\to\mathbb{Z}, \quad f(n)=2n \]
Answer
Injective but not surjective.
Solution
Injectivity
Suppose:
\[ f(n_1)=f(n_2) \implies 2n_1=2n_2 \]
Dividing by \(2\) gives:
\[ n_1=n_2 \]
Hence the function is injective.
Surjectivity
The values assumed by the function are precisely the even integers:
\[ \operatorname{Im}(f)=\{2n \mid n\in\mathbb{Z}\} \]
Odd integers are not the image of any element of the domain, so the function is not surjective.
Conclusion
\[ \boxed{f \text{ is injective but not surjective}} \]
Exercise 7 — level ★★☆☆☆
\[ f:\mathbb{R}\to\mathbb{R}, \quad f(x)=x^2+1 \]
Answer
Neither injective nor surjective.
Solution
Injectivity
As in the case of \(x^2\), we have:
\[ f(1)=2,\qquad f(-1)=2 \]
Distinct inputs share the same image, so the function is not injective.
Surjectivity
We observe that:
\[ x^2 \ge 0 \implies x^2+1 \ge 1 \]
Therefore:
\[ \operatorname{Im}(f)=[1,+\infty) \]
Numbers less than \(1\) are never attained, so the function is not surjective.
Conclusion
\[ \boxed{f \text{ is neither injective nor surjective}} \]
Exercise 8 — level ★★☆☆☆
\[ f:\mathbb{R}\to[0,+\infty), \quad f(x)=x^2 \]
Answer
Surjective but not injective.
Solution
Injectivity
As already seen:
\[ f(1)=f(-1)=1 \]
with \(1\neq -1\), so the function is not injective.
Surjectivity
The codomain is now \([0,+\infty)\). Let \(y\ge 0\); we need to verify that there exists \(x\) such that:
\[ y=x^2 \]
This equation always has a real solution:
\[ x=\pm\sqrt{y} \]
So every element of the codomain is indeed attained, and the function is surjective.
Conclusion
\[ \boxed{f \text{ is surjective but not injective}} \]
Exercise 9 — level ★★☆☆☆
\[ f:\mathbb{R}\to\mathbb{R}, \quad f(x)= \begin{cases} x & x\ge 0 \\ -x & x<0 \end{cases} \]
Answer
Neither injective nor surjective.
Solution
Interpretation
This function coincides with the absolute value:
\[ f(x)=|x| \]
Injectivity
For example:
\[ f(1)=1,\qquad f(-1)=1 \]
with distinct inputs from the domain. Hence the function is not injective.
Surjectivity
Since \(|x|\ge 0\), the range is:
\[ \operatorname{Im}(f)=[0,+\infty) \]
Negative numbers are never attained, so the function is not surjective on \(\mathbb{R}\).
Conclusion
\[ \boxed{f \text{ is neither injective nor surjective}} \]
Exercise 10 — level ★★☆☆☆
\[ f:\mathbb{R}\to\mathbb{R}, \quad f(x)=e^x \]
Answer
Injective but not surjective.
Solution
Injectivity
The exponential function is strictly increasing on all of \(\mathbb{R}\). This means that whenever \(x_1<x_2\), we have:
\[ e^{x_1} < e^{x_2} \]
Consequently, no two distinct inputs can share the same output. The function is therefore injective.
Surjectivity
We observe that:
\[ e^x > 0 \quad \forall x\in\mathbb{R} \]
So the range of the function is:
\[ \operatorname{Im}(f)=(0,+\infty) \]
Negative numbers and zero are never attained, so not every real value is reached.
The function is not surjective on \(\mathbb{R}\).
Conclusion
\[ \boxed{f \text{ is injective but not surjective}} \]
Exercise 11 — level ★★☆☆☆
\[ f:\mathbb{R}\to(0,+\infty), \quad f(x)=e^x \]
Answer
The function is bijective.
Solution
Injectivity
As already noted, the exponential function is strictly increasing on \(\mathbb{R}\), which implies that distinct inputs yield distinct outputs.
Hence the function is injective.
Surjectivity
The codomain is \((0,+\infty)\). Let \(y>0\); we need to find \(x\) such that:
\[ y=e^x \]
Solving:
\[ x=\ln(y) \]
Since the natural logarithm is defined for every \(y>0\), every element of the codomain is the image of some \(x\).
The function is surjective.
Conclusion
\[ \boxed{f \text{ is bijective}} \]
Exercise 12 — level ★★☆☆☆
\[ f:(0,+\infty)\to\mathbb{R}, \quad f(x)=\ln(x) \]
Answer
The function is bijective.
Solution
Injectivity
The natural logarithm is strictly increasing on its domain \((0,+\infty)\), so distinct inputs produce distinct outputs.
The function is injective.
Surjectivity
Let \(y\in\mathbb{R}\). Consider:
\[ y=\ln(x) \]
Solving:
\[ x=e^y \]
Since \(e^y>0\), there is always a value in the domain whose image is \(y\).
Hence the function is surjective.
Conclusion
\[ \boxed{f \text{ is bijective}} \]
Exercise 13 — level ★★★☆☆
\[ f:[0,+\infty)\to\mathbb{R}, \quad f(x)=\sqrt{x} \]
Answer
Injective but not surjective.
Solution
Injectivity
The square root function is increasing on \([0,+\infty)\), so distinct inputs yield distinct outputs.
It is therefore injective.
Surjectivity
We have:
\[ \sqrt{x} \ge 0 \]
So the range is \([0,+\infty)\), which does not coincide with \(\mathbb{R}\).
Negative numbers are never attained, so the function is not surjective.
Conclusion
\[ \boxed{f \text{ is injective but not surjective}} \]
Exercise 14 — level ★★★☆☆
\[ f:\mathbb{R}\to\mathbb{R}, \quad f(x)=x^3-x \]
Answer
Surjective but not injective.
Solution
Injectivity
The function is not monotone on all of \(\mathbb{R}\). For instance, there exist distinct values with the same image (the graph has an "S"-shaped profile).
Hence the function is not injective.
Surjectivity
Being a polynomial of odd degree, we have:
\[ \lim_{x\to+\infty}f(x)=+\infty, \qquad \lim_{x\to-\infty}f(x)=-\infty \]
Moreover the function is continuous, so by the Intermediate Value Theorem it attains every real value.
It is therefore surjective.
Conclusion
\[ \boxed{f \text{ is surjective but not injective}} \]
Exercise 15 — level ★★★☆☆
\[ f:\mathbb{R}\to\mathbb{R}, \quad f(x)=ax+1 \]
Answer
Bijective if and only if \(a\neq 0\).
Solution
Injectivity
Suppose:
\[ ax_1+1=ax_2+1 \implies ax_1=ax_2 \]
If \(a\neq 0\), dividing gives \(x_1=x_2\), so the function is injective. If \(a=0\), the function is constant and therefore not injective.
Surjectivity
Solving \(y=ax+1\):
\[ x=\frac{y-1}{a} \]
This solution exists for every \(y\) only when \(a\neq 0\).
Conclusion
\[ \boxed{f \text{ is bijective} \iff a\neq 0} \]
Exercise 16 — level ★★★☆☆
\[ f:\mathbb{R}\to\mathbb{R}, \quad f(x)=x^2+ax \]
Answer
Never injective on \(\mathbb{R}\).
Solution
Injectivity
This is a degree-two polynomial whose graph is a parabola, hence it is not monotone on all of \(\mathbb{R}\).
Consequently, there always exist two distinct inputs with the same output.
The function is not injective for any value of \(a\).
Conclusion
\[ \boxed{f \text{ is never injective}} \]
Exercise 17 — level ★★★☆☆
\[ f:\mathbb{R}\setminus\{1\}\to\mathbb{R}, \quad f(x)=\frac{1}{x-1} \]
Answer
Injective but not surjective.
Solution
Injectivity
Suppose:
\[ \frac{1}{x_1-1}=\frac{1}{x_2-1} \]
It follows that:
\[ x_1-1=x_2-1 \implies x_1=x_2 \]
Hence the function is injective.
Surjectivity
The function never takes the value \(0\), since a fraction with numerator \(1\) cannot equal zero.
Therefore:
\[ 0 \notin \operatorname{Im}(f) \]
The function is not surjective on \(\mathbb{R}\).
Conclusion
\[ \boxed{f \text{ is injective but not surjective}} \]
Exercise 18 — level ★★★☆☆
\[ f:\mathbb{R}\to\mathbb{R}, \quad f(x)=\arctan(x) \]
Result
Injective but not surjective.
Solution
Injectivity
The function \(\arctan(x)\) is strictly increasing on all of \(\mathbb{R}\). Therefore, distinct values in the domain have distinct images.
Hence, the function is injective.
Surjectivity
The image of the function is:
\[ \operatorname{Im}(f)=\left(-\frac{\pi}{2},\,\frac{\pi}{2}\right) \]
This interval does not coincide with the codomain \(\mathbb{R}\). For example, \(\pi\) is not the image of any real number.
Hence, the function is not surjective.
Conclusion
\[ \boxed{f \text{ is injective but not surjective}} \]
Exercise 19 — level ★★★☆☆
\[ f:A\to B,\quad A=\{1,2,3,4\},\quad B=\{a,b,c\} \] \[ f(1)=a,\quad f(2)=b,\quad f(3)=c,\quad f(4)=a \]
Answer
Surjective but not injective.
Solution
Injectivity
We observe that:
\[ f(1)=a,\qquad f(4)=a \]
with \(1\neq 4\), so the function is not injective.
Surjectivity
Every element of \(B\) is the image of at least one element of \(A\):
\[ a,b,c \in \operatorname{Im}(f) \]
Hence the function is surjective.
Conclusion
\[ \boxed{f \text{ is surjective but not injective}} \]
Exercise 20 — level ★★★★☆
\[ f:\mathbb{R}\to\mathbb{R}, \quad f(x)=x^3-3x \]
Answer
Surjective but not injective.
Solution
Injectivity
The function is not monotone on all of \(\mathbb{R}\); it has both a local maximum and a local minimum, so there exist distinct inputs with the same output.
It is therefore not injective.
Surjectivity
Being a polynomial of odd degree, we have:
\[ \lim_{x\to+\infty}f(x)=+\infty,\quad \lim_{x\to-\infty}f(x)=-\infty \]
By continuity, the function attains every real value.
Conclusion
\[ \boxed{f \text{ is surjective but not injective}} \]
Exercise 21 — level ★★★★☆
\[ f:\mathbb{R}\to\mathbb{R}, \quad f(x)=\frac{x}{1+x^2} \]
Answer
Neither injective nor surjective.
Solution
Injectivity
The function is not monotone on all of \(\mathbb{R}\) (it increases then decreases), so there exist distinct inputs with the same output.
Surjectivity
One can verify that:
\[ -\frac{1}{2} \le f(x) \le \frac{1}{2} \]
so the range is a bounded interval and does not coincide with \(\mathbb{R}\).
Conclusion
\[ \boxed{f \text{ is neither injective nor surjective}} \]
Exercise 22 — level ★★★★☆
\[ f:\mathbb{R}\to\mathbb{R}, \quad f(x)=\ln(x^2+1) \]
Answer
Neither injective nor surjective.
Solution
Injectivity
Since \(x^2\) is an even function, we have:
\[ f(x)=f(-x) \]
with \(x\neq -x\) (for \(x\neq 0\)), so the function is not injective.
Surjectivity
We have:
\[ x^2+1 \ge 1 \implies \ln(x^2+1) \ge 0 \]
Therefore:
\[ \operatorname{Im}(f)=[0,+\infty) \]
Negative values are never attained.
Conclusion
\[ \boxed{f \text{ is neither injective nor surjective}} \]
Exercise 23 — level ★★★★☆
\[ f:\mathbb{R}\to\mathbb{R}, \quad f(x)=e^x+x \]
Answer
The function is bijective.
Solution
Injectivity
The sum of two strictly increasing functions is again strictly increasing. Since both \(e^x\) and \(x\) are strictly increasing, \(f(x)\) is strictly increasing too.
Surjectivity
We have:
\[ \lim_{x\to+\infty}f(x)=+\infty,\quad \lim_{x\to-\infty}f(x)=-\infty \]
Being continuous with opposite limiting behaviour, the function attains every real value.
Conclusion
\[ \boxed{f \text{ is bijective}} \]