Gdansk University of Technology Mathematics Entrance Exam Volume 3 Exhaustive Study Notes

General Information

• Document Title: Gdansk University of Technology Mathematics Entrance Exam — Volume 3
• Date: July 16, 2026
• Candidate: Ubongabasi Chrysanthus (Data Engineering, WETI), June 2026
• Volume Scope: Final volume of the Gdansk Tech prep series (Volume 3 of 3).
• Key Topics:
  • Inequalities (Linear, Quadratic, Rational)
  • Absolute Value (Equations and Inequalities)
  • Parametric Curves and Polar Coordinates
  • Mixed Calculus and Harder Algebra
  • Number Theory and Mathematical Proof
  • Statistics and Data (Relevant to Data Engineering)
  • Exam Simulation Questions

SECTION 1 — INEQUALITIES & ABSOLUTE VALUE

1.1 Linear and Quadratic Inequalities

• Question 1: Linear Inequality

  • Problem: Solve $3x - 7 > 2x + 1$
  • Answer: $x > 8 \rightarrow x \in (8, +\infty)$
  • Hint: Subtract $2x$ and add $7$ to both sides to isolate $x$.

• Question 2: Quadratic Inequality (Factored Form)

  • Problem: Solve $x(x - 4) \le 0$
  • Answer: $B) \, x \in [0, 4]$
  • Hint: Roots are at $x=0$ and $x=4$. Because the parabola opens upward (coefficient of $x^2$ is positive), the product is $\le 0$ between the roots.

• Question 3: Rational Inequality

  • Problem: Solve $\frac{x - 1}{x + 2} > 0$
  • Answer: $x \in (-\infty, -2) \cup (1, +\infty)$
  • Hint: Use a sign chart. The fraction is positive when both the numerator and denominator share the same sign (both positive or both negative).

• Question 4: Quadratic Inequality

  • Problem: Find the solution set of $x^2 - 5x + 6 \le 0$
  • Answer: $A) \, x \in [2, 3]$
  • Hint: Factor the expression into $(x-2)(x-3) \le 0$. The parabola opens upward, so the values are $\le 0$ between the roots of $2$ and $3$.

1.2 Absolute Value Equations and Inequalities

• Question 5: Absolute Value Equation

  • Problem: Solve $|2x - 3| = 7$
  • Answer:
  • Case 1: $2x - 3 = 7 \rightarrow 2x = 10 \rightarrow x = 5$
  • Case 2: $2x - 3 = -7 \rightarrow 2x = -4 \rightarrow x = -2$
  • Solutions: $x = 5$ or $x = -2$
  • Hint: Recall that $|A| = k$ means $A = k$ or $A = -k$.

• Question 6: Absolute Value Inequality (Less Than)

  • Problem: Solve $|x + 1| < 4$
  • Answer: $A) \, x \in (-5, 3)$
  • Hint: $|A| < k \leftrightarrow -k < A < k$. Thus, $-4 < x + 1 < 4$. Subtracting $1$ from all parts yields $-5 < x < 3$.

• Question 7: Absolute Value Inequality (Greater Than)

  • Problem: Solve $|3x - 2| \ge 5$
  • Answer:
  • Case 1: $3x - 2 \ge 5 \rightarrow 3x \ge 7 \rightarrow x \ge \frac{7}{3}$
  • Case 2: $3x - 2 \le -5 \rightarrow 3x \le -3 \rightarrow x \le -1$
  • Solution: $x \in (-\infty, -1] \cup [\frac{7}{3}, +\infty)$
  • Hint: $|A| \ge k \leftrightarrow A \ge k$ or $A \le -k$.

• Question 8: Absolute Value Equality

  • Problem: Solve $|x - 2| = |x + 4|$
  • Answer: Square both sides to eliminate the absolute values: $(x-2)^2 = (x+4)^2$. This expands to $x^2 - 4x + 4 = x^2 + 8x + 16$. Simplifying gives $-12x = 12 \rightarrow x = -1$.
  • Hint: Alternatively, consider $x-2 = x+4$ (which has no solution) or $x-2 = -(x+4)$.

SECTION 2 — EXPONENTIAL & LOGARITHMIC EQUATIONS (HARDER)

2.1 Harder Exponential Equations

• Question 9: Quadratic-Form Exponential

  • Problem: Solve $9^x - 4 \cdot 3^x + 3 = 0$
  • Answer: Let $u = 3^x$, which implies $9^x = (3^x)^2 = u^2$. The equation becomes $u^2 - 4u + 3 = 0$. Factoring gives $(u-1)(u-3) = 0$.
  • If $u=1$, then $3^x=1 \rightarrow x=0$
  • If $u=3$, then $3^x=3 \rightarrow x=1$
  • Solutions: $x=0, x=1$
  • Hint: Recognize $9^x$ as $(3^x)^2$ and substitute to solve for $u$.

• Question 10: Advanced Substitution

  • Problem: Solve $2^{2x+1} - 5 \cdot 2^x + 2 = 0$
  • Answer: Rewrite the first term: $2^{2x+1} = 2^1 \cdot 2^{2x} = 2 \cdot (2^x)^2$. Let $u = 2^x$, substituting gives $2u^2 - 5u + 2 = 0$. Factoring gives $(2u-1)(u-2) = 0$.
  • If $u = \frac{1}{2}$, then $2^x = 2^{-1} \rightarrow x = -1$
  • If $u = 2$, then $2^x = 2^1 \rightarrow x = 1$
  • Hint: Rewrite $2^{2x+1}$ as $2 \cdot 4^x$ or $2 \cdot (2^x)^2$.

2.2 Harder Logarithmic Equations

• Question 11: Product Property of Logs

  • Problem: Solve $\log_{3}(x) + \log_{3}(x-8) = 2$
  • Answer: Combine logs into $\log_{3}[x(x-8)] = 2$. Convert to exponential form: $x(x-8) = 3^2 = 9$. Solve the quadratic $x^2 - 8x - 9 = 0 \rightarrow (x-9)(x+1) = 0$.
  • $x=9$ is valid because it makes the arguments of both original logs positive.
  • $x=-1$ is rejected as logs cannot have negative arguments.
  • Solution: $x=9$
  • Hint: Always verify the domain of logarithmic equations after solving.

• Question 12: Combination and Domain Verification

  • Problem: Solve $\log_{2}(x+3) = 2 - \log_{2}(x+1)$
  • Answer: Move to one side: $\log_{2}(x+3) + \log_{2}(x+1) = 2$. Combine: $\log_{2}[(x+3)(x+1)] = 2$. Convert to exponential: $(x+3)(x+1) = 2^2 = 4$. This leads to $x^2 + 4x - 1 = 0$. Using the quadratic formula, $x = \frac{-4 \pm \sqrt{20}}{2} = -2 \pm \sqrt{5}$.
  • $x = -2 + \sqrt{5} \approx 0.24$ is valid.
  • $x = -2 - \sqrt{5} \approx -4.24$ is rejected.
  • Hint: Check results against original domains (e.g., $x > -3$ and $x > -1$).

SECTION 3 — TRIGONOMETRY (HARDER)

3.1 Sum and Product Formulas

• Question 13: Compound Angle Formula

  • Problem: Show the formula for $\sin(A+B)$ and evaluate $\sin(75^{\circ})$ exactly.
  • Formula: $\sin(A+B) = \sin(A)\cos(B) + \cos(A)\sin(B)$.
  • Answer: $\sin(75^{\circ}) = \sin(45^{\circ} + 30^{\circ}) = \sin(45^{\circ})\cos(30^{\circ}) + \cos(45^{\circ})\sin(30^{\circ})$.
  • Substitute values: $(\frac{\sqrt{2}}{2})(\frac{\sqrt{3}}{2}) + (\frac{\sqrt{2}}{2})(\frac{1}{2}) = \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4} = \frac{\sqrt{6} + \sqrt{2}}{4}$.
  • Hint: Use exact values for $30^{\circ}$ and $45^{\circ}$.

• Question 14: Solving Equations with Double Angles

  • Problem: Solve $\sin(2x) = \cos(x)$ for $x \in [0^{\circ}, 360^{\circ})$.
  • Answer: Replace $\sin(2x)$ with $2\sin(x)\cos(x)$. Rearrange: $2\sin(x)\cos(x) - \cos(x) = 0$. Factor: $\cos(x)(2\sin(x) - 1) = 0$.
  • $\cos(x) = 0 \rightarrow x = 90^{\circ}, 270^{\circ}$
  • $\sin(x) = \frac{1}{2} \rightarrow x = 30^{\circ}, 150^{\circ}$
  • Solutions: $x = 30^{\circ}, 90^{\circ}, 150^{\circ}, 270^{\circ}$
  • Hint: Use the double angle identity before factoring.

• Question 15: Harmonic Form Transformation

  • Problem: Express $3\sin(x) + 4\cos(x)$ in the form $R\sin(x + \alpha)$.
  • Answer:
  • $R = \sqrt{3^2 + 4^2} = \sqrt{25} = 5$.
  • $\tan(\alpha) = \frac{4}{3} \rightarrow \alpha = 53.13^{\circ}$.
  • Form: $5\sin(x + 53.13^{\circ})$.
  • Hint: Match coefficients with the expanded form $R\sin(x)\cos(\alpha) + R\cos(x)\sin(\alpha)$.

• Question 16: Quadratic in Sine/Cosine

  • Problem: Find all solutions of $2\cos^2(x) - \cos(x) - 1 = 0$ in $[0, 2\pi)$.
  • Answer: Factor as a quadratic: $(2\cos(x) + 1)(\cos(x) - 1) = 0$.
  • $\cos(x) = 1 \rightarrow x = 0$.
  • $\cos(x) = -\frac{1}{2} \rightarrow x = \frac{2\pi}{3}, \frac{4\pi}{3}$.
  • Solutions: $x = 0, \frac{2\pi}{3}, \frac{4\pi}{3}$

3.2 Inverse Trig Functions

• Question 17: Right Triangle Method

  • Problem: Evaluate $\sin(\arctan(\frac{3}{4}))$
  • Answer: Let $\theta = \arctan(\frac{3}{4})$. Then $\tan(\theta) = \frac{3}{4}$. Imagine a right triangle with opposite = $3$ and adjacent = $4$. By Pythagoras, the hypotenuse is $\sqrt{3^2 + 4^2} = 5$. Then $\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{3}{5}$.

• Question 18: Multiple Angle with Inverse

  • Problem: Simplify $\cos(2\arcsin(x))$
  • Answer: Let $\theta = \arcsin(x)$, so $\sin(\theta) = x$. Use the double angle identity: $\cos(2\theta) = 1 - 2\sin^2(\theta)$. Substituting gives $1 - 2x^2$.

SECTION 4 — ANALYTIC GEOMETRY (HARDER)

4.1 Parabolas and Conic Sections

• Question 19: Parabola Characteristics

  • Problem: Find vertex, focus, and directrix of $y = x^2 - 4x + 7$.
  • Answer: Complete the square: $y = (x-2)^2 + 3$.
  • Vertex: $(2, 3)$.
  • Standard Form: $(x-h)^2 = 4p(y-k)$ where $4p = 1 \rightarrow p = 0.25$.
  • Focus: Vertex shifted up by $p$: $(2, 3.25)$.
  • Directrix: Vertex shifted down by $p$: $y = 2.75$.

• Question 20: Standard Ellipse

  • Problem: Find the equation of the ellipse with center $(0,0)$, semi-major axis $a=5$ (along x-axis), and semi-minor axis $b=3$.
  • Answer: $\frac{x^2}{25} + \frac{y^2}{9} = 1$.

• Question 21: Classifying Conics

  • Problem: Classify the conic: $4x^2 - 9y^2 = 36$.
  • Answer: $D) \, \text{Hyperbola}$.
  • Hint: Divide by $36$ to get $\frac{x^2}{9} - \frac{y^2}{4} = 1$. The negative sign between terms indicates a hyperbola.

4.2 Parametric Equations

• Question 22: Cartesian Conversion (Ellipse)

  • Problem: Eliminate the parameter for $x = 3\cos(t), y = 2\sin(t)$.
  • Answer: $\cos(t) = \frac{x}{3}, \sin(t) = \frac{y}{2}$. via Identity $\cos^2(t) + \sin^2(t) = 1$, we get $\frac{x^2}{9} + \frac{y^2}{4} = 1$.

• Question 23: Cartesian Conversion (Parabola)

  • Problem: Convert $x = t^2 - 1, y = 2t + 1$ to Cartesian form.
  • Answer: Isolate $t$ from the y-equation: $t = \frac{y - 1}{2}$. Substitute into x: $x = (\frac{y - 1}{2})^2 - 1 = \frac{(y - 1)^2}{4} - 1$.

4.3 Polar Coordinates

• Question 24: Polar to Cartesian

  • Problem: Convert $(r, \theta) = (4, \frac{\pi}{3})$ to Cartesian.
  • Answer: $x = r\cos(\theta) = 4\cos(\frac{\pi}{3}) = 4(\frac{1}{2}) = 2$. $y = r\sin(\theta) = 4\sin(\frac{\pi}{3}) = 4(\frac{\sqrt{3}}{2}) = 2\sqrt{3}$. Point is $(2, 2\sqrt{3})$.

• Question 25: Cartesian to Polar

  • Problem: Convert Cartesian point $(-3, 3)$ to polar form.
  • Answer: $r = \sqrt{(-3)^2 + 3^2} = \sqrt{18} = 3\sqrt{2}$. $\tan(\theta) = \frac{3}{-3} = -1$. Since the point is in the second quadrant ($-x, +y$), $\theta = \frac{3\pi}{4}$. Result: $(3\sqrt{2}, \frac{3\pi}{4})$.

SECTION 5 — MIXED CALCULUS

5.1 More Integration Techniques

• Question 26: Tan Integral

  • Problem: Evaluate $\int \tan(x) \,dx$
  • Answer: Rewrite as $\int \frac{\sin(x)}{\cos(x)} \,dx$. Let $u = \cos(x), du = -\sin(x) \,dx$. Result is $-\ln|\cos(x)| + C = \ln|\sec(x)| + C$.

• Question 27: Inverse Trig Standard Form

  • Problem: Evaluate $\int \frac{1}{x^2 + 4} \,dx$
  • Answer: $\frac{1}{2}\arctan(\frac{x}{2}) + C$.
  • Hint: Formula $\int \frac{1}{x^2 + a^2} \,dx = \frac{1}{a}\arctan(\frac{x}{a}) + C$. Here $a=2$.

• Question 28: Logarithmic Substitution

  • Problem: Evaluate $\int \frac{x}{x^2 + 1} \,dx$
  • Answer: Let $u = x^2 + 1$, then $du = 2x \,dx$. Integral becomes $\frac{1}{2}\int \frac{1}{u} \,du = \frac{1}{2}\ln|x^2 + 1| + C$.

• Question 29: Secant Squared Integral

  • Problem: Evaluate $\int \sec^2(3x) \,dx$
  • Answer: $\frac{1}{3}\tan(3x) + C$.

• Question 30: Odd Power of Sine Integral

  • Problem: Evaluate $\int_{0}^{\pi} \sin^3(x) \,dx$
  • Answer: Rewrite as $\int_{0}^{\pi} \sin(x)(1 - \cos^2(x)) \,dx$. Let $u = \cos(x), du = -\sin(x) \,dx$. Limits change from $[0, \pi]$ to $[1, -1]$. Evaluating yields $[u - \frac{u^3}{3}]_{-1}^{1} = \frac{4}{3}$.

5.2 Differential Equations

• Question 31: Separable Variables

  • Problem: Solve $\frac{dy}{dx} = \frac{y}{x}$ with $y(1) = 2$.
  • Answer: $\frac{dy}{y} = \frac{dx}{x} \rightarrow \ln|y| = \ln|x| + C \rightarrow y = Ax$. Using $y(1)=2 \rightarrow 2 = A(1)$. Solution: $y = 2x$.

• Question 32: First-Order Linear ODE

  • Problem: Solve $\frac{dy}{dx} + y = e^x$
  • Answer: Integrating factor $\mu = e^{\int 1 \,dx} = e^x$. Multiply both sides: $e^x y' + e^x y = e^{2x}$. This is $\frac{d}{dx}[e^x y] = e^{2x}$. Integrate: $e^x y = \frac{1}{2}e^{2x} + C \rightarrow y = \frac{1}{2}e^x + Ce^{-x}$.

SECTION 6 — NUMBER THEORY & MATHEMATICAL PROOF

6.1 Divisibility and Modular Arithmetic

• Question 33: Product of Integers

  • Problem: Prove that $n^2 + n$ is always even for any integer $n$.
  • Proof: $n^2 + n = n(n+1)$. These are consecutive integers; one must be even. The product of any number and an even number is always even.

• Question 34: Power Modulo

  • Problem: What is $7^{100} \pmod 4$?
  • Answer: $B) \, 1$.
  • Logic: $7 \equiv 3 \equiv -1 \pmod 4$. Therefore $7^{100} \equiv (-1)^{100} \equiv 1 \pmod 4$.

• Question 35: Remainder Pattern

  • Problem: Find the remainder when $2^{50}$ is divided by $7$.
  • Answer: $4$.
  • Logic: $2^1 \equiv 2, 2^2 \equiv 4, 2^3 \equiv 1 \pmod 7$. The period is $3$. $50 = 3(16) + 2$. Thus, $2^{50} \equiv 2^2 \equiv 4 \pmod 7$.

6.2 Proof by Contradiction

• Question 36: Irrationality of Square Root of 2
  • Theorem: Prove $\sqrt{2}$ is irrational.
  • Proof: Assume $\sqrt{2} = \frac{p}{q}$ where $\text{gcd}(p, q) = 1$. Then $2 = \frac{p^2}{q^2} \rightarrow p^2 = 2q^2$, meaning $p$ is even ($p = 2k$).
  • Substituting gives: $(2k)^2 = 2q^2 \rightarrow 4k^2 = 2q^2 \rightarrow q^2 = 2k^2$, meaning $q$ is also even.
  • Contradiction: If $p$ and $q$ are both even, $\text{gcd}(p, q) \ge 2$, violating the assumption. Thus, $\sqrt{2}$ must be irrational.

SECTION 7 — STATISTICS & DATA

7.1 Descriptive Statistics

• Question 37: Central Tendency

  • Dataset: ${3, 7, 7, 8, 10, 12, 7}$
  • Mean: $\frac{54}{7} \approx 7.71$
  • Median: Sorted: ${3, 7, 7, 7, 8, 10, 12}$. Middle value is $7$.
  • Mode: $7$ (most frequent).

• Question 38: Dispersion

  • Dataset: ${2, 4, 4, 4, 5, 5, 7, 9}$
  • Answer: Mean = $5$. Deviations squared: $9, 1, 1, 1, 0, 0, 4, 16$. Sum = $32$. Variance = $4$. Standard deviation = $\sqrt{4} = 2$.
  • Formula: $\sigma = \sqrt{\frac{1}{n}\sum(x_i - \bar{x})^2}$.

7.2 Probability Distributions

• Question 39: Binomial Distribution

  • Parameters: $X \sim \text{Binomial}(n=10, p=0.3)$.
  • Problem: Compute $P(X=3)$.
  • Answer: $C(10,3)(0.3)^3(0.7)^7 = 120 \times 0.027 \times 0.0824 \approx 0.2668$.

• Question 40: Normal Distribution

  • Parameters: $X \sim \text{Normal}(\mu=50, \sigma=10)$.
  • Problem: Find $P(40 < X < 60)$.
  • Answer: Standardize: $z_1 = \frac{40-50}{10} = -1, z_2 = \frac{60-50}{10} = 1$. By the $68-95-99.7$ rule, $\pm 1\sigma$ covers approximately $0.6827$ or $68.27\%$.

SECTION 8 — EXAM SIMULATION

• Q1: Solve $x^2 - 7x + 12 = 0$. Answer: $A) \, x=3, 4$.
• Q2: Domain of $f(x) = \ln(x - 2)$. Answer: $B) \, x > 2$.
• Q3: $\lim_{x \to 2} \frac{x^2 - 4}{x - 2}$. Answer: $C) \, 4$.
• Q4: Derivative of $5x^4 - 3x^2 + 7$. Answer: $20x^3 - 6x$.
• Q5: $\int (6x^2 + 2) \,dx$. Answer: $2x^3 + 2x + C$.
• Q6: Arithmetic sequence: $a_7 = 29, d = 4$. Find $a_1$. Answer: $a_1 = 29 - (6 \times 4) = 5$.
• Q7: $\sin(\theta) = \frac{5}{13}$. Find $\cos(\theta)$. Answer: $\frac{12}{13}$ (Pythagorean triple $5-12-13$).
• Q8: Circle equation (center $(-2, 3)$, radius $5$). Answer: $(x+2)^2 + (y-3)^2 = 25$.
• Q9: $\log_{5}(125)$. Answer: $3$ ($5^3 = 125$).
• Q10: No real roots for $kx^2 + 3x + 1 = 0$. Answer: $\Delta = 9 - 4k < 0 \rightarrow k > \frac{9}{4}$. (Note: $k \neq 0$).
• Q11: Area between $y = 4 - x^2$ and $y = x + 2$. Intersection points: $x=-2, x=1$. Area = $\int_{-2}^{1} [(4-x^2) - (x+2)] \,dx = 4.5$ or $\frac{9}{2}$.
• Q12: Parametric curve $x = 2t, y = t^2 + 1$. At $t=3$, $\frac{dy}{dx} = \frac{2t}{2} = t = 3$. Point is $(6, 10)$. Tangent: $y - 10 = 3(x - 6) \rightarrow y = 3x - 8$.
• Q13: System: $3x + y = 11, x - 2y = -3$. Answer: $x = \frac{19}{7}, y = \frac{20}{7}$.
• Q14: $\int_{0}^{\pi/4} (\sec^2(x) - \sin(x)) \,dx$. Answer: $[\tan(x) + \cos(x)]_{0}^{\pi/4} = (1 + \frac{\sqrt{2}}{2}) - (0 + 1) = \frac{\sqrt{2}}{2} \approx 0.707$.
• Q15: Geometric series $S_{\infty} = 9$, term 2 = $2$. Solve $a^2 - 9a + 18 = 0$. Possible first terms: $a=3$ (with $r=\frac{2}{3}$) or $a=6$ (with $r=\frac{1}{3}$).