Mock Exam

Mathematics for Master Students

This is the voluntary mock exam for the tutorial. It is adapted from the original KLU course exam and covers all six sessions: notation and logic, sets, functions, differentiation, optimization, and linear systems. Nothing here is graded and no result is tracked; the exam only exists so you can rehearse under exam-style conditions before your M.Sc. courses lean on this material.

Suggested conditions. Work through the ten problems in about 90 minutes without notes first, as if it were the real thing. Only then open the collapsible solutions to mark yourself, and pull out the cheatsheet that matches the problem you are stuck on. The pointer list at the end of this page maps every problem to its session. On the website every problem carries a fully worked solution you can expand; the PDF version prints only the problems, so you can use it as a clean exam sheet.

The problems are numbered as in the source exam, not in session order. Each is followed by a worked solution showing the named method, the steps, and a check where the topic bundles one.

Problem 1: Functions (Domain, Composition, Inverse)

A logistics cost model chains two functions, \[f(x) = 3x - 6 \qquad \text{and} \qquad g(x) = \sqrt{x}.\]

  1. State the domain of \(g\), then the domain of the composition \(g(f(x))\).
  2. Find the inverse \(f^{-1}(x)\).
  3. Is \(f\) convex, concave, or linear?

Collect the domain restrictions, then invert. Each part reuses one Session 3 tool: red-flag the root, chain the restriction through the composition, swap and solve for the inverse.

Step 1 - Domain of \(g\). A square root needs a non-negative argument, so \(g(x) = \sqrt{x}\) is defined for \(x \ge 0\), i.e. the domain is \([0, \infty)\).

Step 2 - Domain of the composition. First substitute \(f\) into \(g\): \[g(f(x)) = \sqrt{f(x)} = \sqrt{3x - 6}.\] The root is defined only where its argument is non-negative: \[3x - 6 \ge 0 \;\Rightarrow\; 3x \ge 6 \;\Rightarrow\; x \ge 2.\] So the domain of \(g \circ f\) is \([2, \infty)\).

Step 3 - Inverse of \(f\). Set \(y = 3x - 6\) and solve for \(x\): \[y = 3x - 6 \;\Rightarrow\; 3x = y + 6 \;\Rightarrow\; x = \frac{y + 6}{3}.\] Renaming the input, \(f^{-1}(x) = \dfrac{x + 6}{3} = \dfrac{x}{3} + 2\).

Step 4 - Convex, concave, or linear. \(f(x) = 3x - 6\) is a straight line (constant slope \(3\)), so it is linear: both weakly convex and weakly concave, but neither strictly. Its second derivative is \(f''(x) = 0\).

Check. Composing \(f\) with its inverse must return the input: \[f\big(f^{-1}(x)\big) = 3\left(\frac{x}{3} + 2\right) - 6 = x + 6 - 6 = x \ \checkmark.\]

Problem 2: Sum Equals Product

Find natural numbers \(x_1, x_2\) (with \(1 \le i \le 2\)) that solve \[\sum_{i=1}^{2} x_i = \prod_{i=1}^{2} x_i.\]

Step 1 - Unpack the notation. The sum \(\sum_{i=1}^{2} x_i = x_1 + x_2\) and the product \(\prod_{i=1}^{2} x_i = x_1 \cdot x_2\), so the condition is \[x_1 + x_2 = x_1 \cdot x_2.\]

Step 2 - Rearrange to factor. Bring everything to one side and add \(1\) to both sides: \[x_1 x_2 - x_1 - x_2 = 0 \;\Rightarrow\; x_1 x_2 - x_1 - x_2 + 1 = 1 \;\Rightarrow\; (x_1 - 1)(x_2 - 1) = 1.\]

Step 3 - Solve over the natural numbers. For natural numbers, \(x_1 - 1\) and \(x_2 - 1\) are non-negative integers whose product is \(1\), which forces \(x_1 - 1 = x_2 - 1 = 1\), hence \[x_1 = x_2 = 2.\]

Check. Sum \(= 2 + 2 = 4\); product \(= 2 \cdot 2 = 4\): they agree, and \((2, 2)\) is the only solution. \(\checkmark\)

Problem 3: When Is Total Profit Positive?

A firm makes products \(p \in P\) from resources \(r \in R\). The notation \(\sum_{p \in P}\) sums over every product \(p\) in the set \(P\), the set-membership cousin of Session 1’s indexed sum \(\sum_{i=k}^{m}\) (and \(\sum_{r \in R}\) likewise runs over all resources \(r\)). With quantity \(q_p\), selling price \(s_p\), resource requirement \(a_{r,p}\) (units of resource \(r\) per unit of product \(p\)), and resource cost \(c_r\), total profit is \[\pi = \sum_{p \in P} q_p\left(s_p - \sum_{r \in R} a_{r,p}\, c_r\right).\]

Assume every quantity \(q_p > 0\) and every resource cost \(c_r > 0\). Decide which of the following conditions guarantee \(\pi > 0\), and justify each:

  1. \(a_{r,p} = 0\) for all \(r, p\), and \(s_p > 0\) for all \(p\).
  2. \(s_p - \sum_{r \in R} a_{r,p}\, c_r < 0\) for all \(p\).
  3. \(a_{r,p} > 0\) and \(s_p > 0\) for all \(r, p\).

Read the structure first. The inner bracket \(s_p - \sum_{r} a_{r,p} c_r\) is the per-unit margin of product \(p\) (price minus the resource cost of making one unit). Multiplying by the quantity \(q_p\) and summing over all products gives total profit. Since every \(q_p > 0\), the sign of each term follows the sign of its margin.

Condition i: guarantees \(\pi > 0\). With \(a_{r,p} = 0\) the inner sum vanishes, so each margin is just \(s_p > 0\). Every term \(q_p s_p > 0\), and a sum of positive numbers is positive: \[\pi = \sum_{p} q_p\, s_p > 0 \ \checkmark.\]

Condition ii: guarantees the opposite. Here every margin is negative, so every term \(q_p \cdot (\text{negative}) < 0\), and a sum of negative numbers is negative: \(\pi < 0\). This does not guarantee \(\pi > 0\): it guarantees \(\pi < 0\).

Condition iii: does not guarantee anything. Knowing only that \(a_{r,p} > 0\), \(s_p > 0\), \(c_r > 0\) leaves the margin \(s_p - \sum_r a_{r,p} c_r\) of unknown sign: a high price beats the resource cost, a low one does not. The sign of \(\pi\) is indeterminate, so this condition is not sufficient.

Answer: only condition i guarantees \(\pi > 0\).

Problem 4: Fuel Consumption Conversion

A delivery van imported from the US is rated at 20 miles per gallon. Your European fleet software needs the figure in litres per 100 km. Using \(1 \text{ mile} = 1.6 \text{ km}\) and \(1 \text{ gallon} = 3.8 \text{ litres}\), set up the conversion. A formula is sufficient, but give the value as well.

Step 1 - Convert the distance per gallon to km. One gallon takes the van \[20 \text{ miles} \cdot 1.6 \ \tfrac{\text{km}}{\text{mile}} = 32 \text{ km per gallon}.\]

Step 2 - Turn “km per gallon” into “litres per km”. One gallon is \(3.8\) litres, so \(32\) km cost \(3.8\) litres, giving \[\frac{3.8 \text{ litres}}{32 \text{ km}} \text{ litres per km}.\]

Step 3 - Scale to 100 km. Multiply by \(100\): \[\text{litres per 100 km} = \frac{3.8}{20 \cdot 1.6} \cdot 100 = \frac{380}{32}.\]

Value. \(\dfrac{380}{32} = 11.875\), so the van uses about 11.9 litres per 100 km.

Problem 5: Logical Implications

Let \(p(x)\) be the proposition \(x^2 = 16\), let \(q(x)\) be \(x = 4\), and let \(r(x)\) be \(x = \sqrt{x^2}\), all for real \(x\). Decide which of the following implications hold for every real \(x\):

  1. \(p(x) \Rightarrow q(x)\)
  2. \(q(x) \Rightarrow r(x)\)
  3. \(r(x) \Rightarrow p(x)\)
  4. \(p(x) \text{ and } r(x) \text{ together} \Rightarrow q(x)\)
  5. \(q(x) \Rightarrow p(x)\)

Translate each proposition first.

  • \(p(x): x^2 = 16\) holds exactly for \(x = 4\) or \(x = -4\).
  • \(q(x): x = 4\) holds only for \(x = 4\).
  • \(r(x): x = \sqrt{x^2}\). Since \(\sqrt{x^2} = |x|\), this says \(x = |x|\), which holds exactly for \(x \ge 0\).

i) \(p(x) \Rightarrow q(x)\) is false. \(x = -4\) makes \(x^2 = 16\) true but \(x = 4\) false. Counterexample \(x = -4\).

ii) \(q(x) \Rightarrow r(x)\) is true. If \(x = 4\) then \(x \ge 0\), so \(x = |x|\) holds.

iii) \(r(x) \Rightarrow p(x)\) is false. \(x = 1\) satisfies \(x \ge 0\) but \(1^2 = 1 \neq 16\). Counterexample \(x = 1\).

iv) \(p(x) \text{ and } r(x) \Rightarrow q(x)\) is true. \(p(x)\) gives \(x \in \{4, -4\}\); adding \(r(x)\), i.e. \(x \ge 0\), rules out \(-4\) and leaves \(x = 4\), which is \(q(x)\).

v) \(q(x) \Rightarrow p(x)\) is true. If \(x = 4\) then \(x^2 = 16\), so \(p(x)\) holds.

Answer: the implications that hold are ii, iv, and v.

Problem 6: Union and Cartesian Product

For the transport modes \(A = \{\text{road}, \text{rail}, \text{air}\}\) and \(B = \{1, 2, 3\}\), decide which of the following are correct:

  1. \(A \cup B = \{(\text{road}, 1), (\text{rail}, 2), (\text{air}, 3)\}\)
  2. \(A \cup B = \{\text{road}, \text{rail}, \text{air}, 1, 2, 3\}\)
  3. \(A \times B = \{(\text{road}, 1), (\text{rail}, 2), (\text{air}, 3)\}\)
  4. \(A \times B = \{(\text{road}, 1), (\text{road}, 2), (\text{road}, 3), (\text{rail}, 1), (\text{rail}, 2), (\text{rail}, 3), (\text{air}, 1), (\text{air}, 2), (\text{air}, 3)\}\)
  5. \(\{i \in \mathbb{N} \mid \text{letter number } i \text{ is 'r' in some } a \in A\} \cap \{b \in B \mid b \text{ is even}\} = \emptyset\)

i) is false. The union collects the elements of both sets; it does not pair them. Pairs belong to the Cartesian product, not the union.

ii) is true. \(A \cup B\) is every element that appears in \(A\) or in \(B\): \(\{\text{road}, \text{rail}, \text{air}, 1, 2, 3\}\).

iii) is false. The Cartesian product \(A \times B\) contains every ordered pair \((a, b)\) with \(a \in A\) and \(b \in B\): that is \(3 \times 3 = 9\) pairs, not just three.

iv) is true. This lists all nine pairs, one for each combination of a mode with a number.

v) is true. Look at where the letter ‘r’ sits in the words of \(A\): “road” (position \(1\)), “rail” (position \(1\)), “air” (position \(3\)). The set of positions is \(\{1, 3\}\). The even numbers in \(B\) form \(\{2\}\). Their intersection \(\{1, 3\} \cap \{2\} = \emptyset\), since no position equals \(2\).

Answer: correct statements are ii, iv, and v.

Problem 7: Set Identities

Let \(A\), \(B\), \(C\) be arbitrary sets and let \(A^c\) denote a complement. Decide which identities hold for all choices of \(A\), \(B\), \(C\):

  1. \(C \cap (A \cap B) = A \cap (B \cap C)\)
  2. \(C \cap (A \cup B) = A \cup (B \cap C)\)
  3. \(C \cup (A \cap B) = A \cup (B \cap C)\)
  4. \((A \cap B)^c = A^c \cup B^c\)
  5. \((A \cup B)^c = A^c \cap B^c\)

i) is true. Intersection is commutative and associative, so the order and grouping of \(A\), \(B\), \(C\) do not matter: both sides are “elements in all three sets”.

ii) is false. The left side, by distributivity, is \((C \cap A) \cup (C \cap B)\), which is not the same as the right side. Counterexample: \(A = \{1\}\), \(B = \{2\}\), \(C = \{2\}\) give left \(= \{2\} \cap \{1, 2\} = \{2\}\) but right \(= \{1\} \cup (\{2\} \cap \{2\}) = \{1, 2\}\).

iii) is false. Same counterexample \(A = \{1\}\), \(B = \{2\}\), \(C = \{2\}\): left \(= \{2\} \cup (\{1\} \cap \{2\}) = \{2\}\), right \(= \{1\} \cup \{2\} = \{1, 2\}\).

iv) is true. This is De Morgan’s law: the complement of an intersection is the union of the complements.

v) is true. The other De Morgan’s law: the complement of a union is the intersection of the complements.

Answer: correct identities are i, iv, and v.

Problem 8: Profit Maximization

A manufacturer faces the price-demand relation \(p(q) = 100 - 2q\) (price in € when \(q\) units are sold) and the cost function \(C(q) = 20q + 100\) (in €). Find the quantity that maximizes profit, confirm it is a maximum, and state the price and the maximum profit.

Step 1 - Build the profit function. Profit is revenue minus cost, with revenue \(= p(q) \cdot q\): \[\pi(q) = (100 - 2q)\,q - (20q + 100) = 100q - 2q^2 - 20q - 100 = -2q^2 + 80q - 100.\]

Step 2 - First-order condition (FOC). Differentiate and set to zero: \[\pi'(q) = -4q + 80 = 0 \;\Rightarrow\; q = 20.\]

Step 3 - Second-order condition (SOC). \(\pi''(q) = -4 < 0\), so the profit function is concave and \(q = 20\) is a maximum.

Step 4 - Price and profit. The price at \(q = 20\) is \[p(20) = 100 - 2(20) = 60,\] and the maximum profit is \[\pi(20) = -2(20)^2 + 80(20) - 100 = -800 + 1600 - 100 = 700.\]

Answer: sell 20 units at a price of €60, for a maximum profit of €700.

Problem 9: Derivative and Monotonicity

For \(f(x) = x^2 - 4x + 3\), find the derivative \(f'(x)\), then decide which of the following are correct:

  1. \(f\) is decreasing on \((-\infty, 2]\)
  2. \(f\) is increasing on \((-\infty, 2]\)
  3. \(f\) is decreasing on \([2, \infty)\)
  4. \(f\) is increasing on \([2, \infty)\)
  5. \(f\) is increasing on the set of non-negative real numbers
  6. \(f\) is decreasing on the set of non-negative real numbers
  7. \(f\) is neither increasing nor decreasing on the whole set of real numbers

Step 1 - Differentiate. With the power rule, \(f'(x) = 2x - 4\).

Step 2 - Find the sign of the derivative. \(f'(x) = 2x - 4 = 0\) at \(x = 2\). For \(x < 2\), \(f'(x) < 0\) (decreasing); for \(x > 2\), \(f'(x) > 0\) (increasing). The parabola turns at its vertex \(x = 2\).

Step 3 - Test each statement.

  • i) true: on \((-\infty, 2]\) the derivative is \(\le 0\), so \(f\) decreases.
  • ii) false: \(f\) decreases, not increases, on that interval.
  • iii) false: on \([2, \infty)\) the derivative is \(\ge 0\), so \(f\) increases.
  • iv) true: matches the increasing behaviour on \([2, \infty)\).
  • v) false: the non-negative reals \([0, \infty)\) contain both sides of the vertex; \(f\) decreases on \([0, 2)\) then increases, so it is not increasing throughout.
  • vi) false: for the same reason it is not decreasing throughout.
  • vii) true: over all of \(\mathbb{R}\) the function first falls then rises, so it is neither increasing nor decreasing everywhere.

Answer: \(f'(x) = 2x - 4\); correct statements are i, iv, and vii.

Problem 10: Solve the Linear System

Solve the system \[x - 2y = 2, \qquad \frac{x}{3} - y = \frac{1}{3}.\]

Elimination: first clear the fraction, then cancel a variable.

Step 1 - Clear the fraction. Multiply the second equation by \(3\): \[\frac{x}{3} - y = \frac{1}{3} \;\xrightarrow{\times 3}\; x - 3y = 1.\]

Step 2 - Eliminate \(x\) by subtracting this from the first equation: \[(x - 2y) - (x - 3y) = 2 - 1 \;\Rightarrow\; y = 1.\]

Step 3 - Back-substitute into \(x - 2y = 2\): \[x - 2(1) = 2 \;\Rightarrow\; x = 4.\]

Solution: \((x, y) = (4, 1)\).

Check. First equation: \(4 - 2(1) = 2\) ✓. Second equation: \(\dfrac{4}{3} - 1 = \dfrac{4}{3} - \dfrac{3}{3} = \dfrac{1}{3}\) ✓.

After the Mock Exam

Whatever felt shaky points you straight back to the session that covers it. Each cheatsheet is a one-page refresher, and each self-test quiz gives you a few more reps:

Rework anything you missed, then you are ready for the math your M.Sc. courses will lean on.