Mathematics for Data Science I · End Term · Sep 2025 · FN

**1. Review the exam instructions:** Question 2 presents the general instructions and standard mathematical notations ($\mathbb{R}, \mathbb{Q}, \mathbb{Z}, \mathbb{N}$) for the Mathematics for Data Science I paper. Selecting **"Instructions has been mentioned above."** acknowledges these instructions. **Answer: Option A** A: **Correct:** Acknowledges the exam instructions and notation definitions given above. B: **Incorrect:** Standard exam procedure requires selecting the first option to confirm reading the instructions.

**1. Determine the levels of the vertices in the BFS tree rooted at $A$:** - **Level 0:** $A$ - **Level 1:** $B, F$ - **Level 2:** $E, C$ - **Level 3:** $D, G$ In an undirected graph, any edge in the original graph can only connect vertices whose BFS levels differ by at most $1$ ($|\text{level}(u) - \text{level}(v)| \le 1$). **2. Check each option:** - **Option A `(A, D)`:** $A$ is at Level 0 and $D$ is at Level 3 ($|0 - 3| = 3 > 1$). If `(A, D)` were an edge in the original graph, $D$ would have been discovered at Level 1 directly from $A$. Thus, **`(A, D)` cannot be an edge in the original graph**. - **Option B `(E, C)`:** Both $E$ and $C$ are at Level 2 ($|2 - 2| = 0 \le 1$), so `(E, C)` can be a non-tree edge. - **Option C `(D, G)`:** Both $D$ and $G$ are at Level 3 ($|3 - 3| = 0 \le 1$), so `(D, G)` can be a non-tree edge. - **Option D `(B, F)`:** Both $B$ and $F$ are at Level 1 ($|1 - 1| = 0 \le 1$), so `(B, F)` can be a non-tree edge. **Answer: Option A** A: **Correct:** Since $A$ is at Level 0 and $D$ is at Level 3, an edge `(A, D)` would have placed $D$ at Level 1 in a BFS from $A$. B: **Incorrect:** $E$ and $C$ are both at Level 2, so `(E, C)` can exist in the original graph. C: **Incorrect:** $D$ and $G$ are both at Level 3, so `(D, G)` can exist in the original graph. D: **Incorrect:** $B$ and $F$ are both at Level 1, so `(B, F)` can exist in the original graph.

**1. Complete the square for $f(x) = x^2 + 2x - 8$:** $$f(x) = (x^2 + 2x + 1) - 9 = (x + 1)^2 - 9$$ **2. Evaluate each statement:** - **Options A and B:** Since $f(x)$ is a quadratic function defined on $\mathbb{R}$ (e.g., $f(-4) = f(2) = 0$), it is **not one-one** and **does not have an inverse** on its domain. - **Option C:** From the vertex form $f(x) = (x + 1)^2 - 9$, the vertex of the parabola is at **$(-1, -9)$**. (**Correct**) - **Option D:** The $y$-intercept is $f(0) = 0^2 + 2(0) - 8 = -8 \ne -1$. (**Incorrect**) **Answer: Option C** A: **Incorrect:** $f(-4) = f(2) = 0$, so $f$ is not one-one on $\mathbb{R}$. B: **Incorrect:** Because $f$ is not one-one on $\mathbb{R}$, it has no inverse on its domain. C: **Correct:** Completing the square gives $f(x) = (x + 1)^2 - 9$, so the vertex is $(-1, -9)$. D: **Incorrect:** The $y$-intercept is $f(0) = -8$, not $-1$.

**1. Find the domain of $(g \circ f)(x) = g(f(x))$:** - Given $f(x) = \sqrt{x + 4}$ and $g(x) = \log(1 + x^2)$: - The inner function $f(x) = \sqrt{x + 4}$ is defined when $x + 4 \ge 0 \iff x \ge -4$. - Since $g$ is defined on all of $\mathbb{R}$, the domain of $(g \circ f)(x)$ is **$[-4, \infty)$** (**Option B is correct**, Option C is incorrect). **2. Simplify $(g \circ f)(x)$ on $[-4, \infty)$:** $$(g \circ f)(x) = g\left(\sqrt{x + 4}\right) = \log\left(1 + \left(\sqrt{x + 4}\right)^2\right) = \log(1 + x + 4) = \log(x + 5)$$ (**Option D is correct**, Option A is incorrect). **Answer: Options B, D** A: **Incorrect:** $1 + (\sqrt{x + 4})^2 = x + 5$, not $2x + 5$. B: **Correct:** $\sqrt{x + 4}$ requires $x + 4 \ge 0$, so the domain is $[-4, \infty)$. C: **Incorrect:** The domain is $[-4, \infty)$, not $[-6, -1]$. D: **Correct:** $(g \circ f)(x) = \log(1 + (x + 4)) = \log(x + 5)$ on $[-4, \infty)$.

**1. Analyze $R_1 = \{(x, y) \in \mathbb{Z}^2 \mid 7 \text{ divides } (x - y)\}$:** - **Symmetric:** If $7 \mid (x - y)$, then $x - y = 7k \implies y - x = 7(-k)$, so $7 \mid (y - x)$. Thus **$R_1$ is symmetric** (**Option C is correct**). - **Transitive:** If $7 \mid (x - y)$ and $7 \mid (y - z)$, then $7 \mid ((x - y) + (y - z)) = (x - z)$, so $R_1$ **is** transitive (Option A is incorrect). **2. Analyze $R_2 = \{(x, y) \in \mathbb{Z}^2 \mid x + y = 2\}$:** - **Symmetric:** If $x + y = 2$, then $y + x = 2$ by commutativity of addition, so **$R_2$ is symmetric** (**Option B is correct**). - **Transitive:** $(0, 2) \in R_2$ and $(2, 0) \in R_2$, but $(0, 0) \notin R_2$ since $0 + 0 = 0 \ne 2$. Thus $R_2$ is **not** transitive (Option D is incorrect). **Answer: Options B, C** A: **Incorrect:** Congruence modulo 7 ($R_1$) is an equivalence relation and is therefore transitive. B: **Correct:** Since $x + y = 2 \implies y + x = 2$, $R_2$ is symmetric. C: **Correct:** Since $7 \mid (x - y) \implies 7 \mid (y - x)$, $R_1$ is symmetric. D: **Incorrect:** $(0, 2) \in R_2$ and $(2, 0) \in R_2$, but $(0, 0) \notin R_2$, so $R_2$ is not transitive.

**1. Inspect the $5 \times 5$ symmetric adjacency matrix of $G$:** $$A = \begin{bmatrix} 0 & 1 & 0 & 0 & 1 \\ 1 & 0 & 1 & 0 & 0 \\ 0 & 1 & 0 & 1 & 0 \\ 0 & 0 & 1 & 0 & 1 \\ 1 & 0 & 0 & 1 & 0 \end{bmatrix}$$ - Since the matrix is $5 \times 5$, **the graph has $5$ nodes** (**Option A is correct**). - Each row has exactly two $1$'s, so the degree of each of the 5 nodes is $2$ (a 5-cycle $C_5$). - The sum of the degrees of all nodes is $2 + 2 + 2 + 2 + 2 = 10$ (so Option D, which says $20$, is incorrect). - By the Handshaking Lemma ($\sum \deg(v) = 2|E|$), the number of undirected edges is: $$|E| = \frac{10}{2} = 5$$ (**Option C is correct**, and Option B is incorrect). **Answer: Options A, C** A: **Correct:** The adjacency matrix is $5 \times 5$, so $G$ has 5 nodes. B: **Incorrect:** $10$ is the sum of the entries in the symmetric matrix ($2|E|$), so there are $10/2 = 5$ undirected edges. C: **Correct:** The graph is a 5-cycle with $10/2 = 5$ edges. D: **Incorrect:** Each of the 5 nodes has degree 2, so the sum of degrees is $10$, not $20$.

**1. List the directed edges in the graph:** From the diagram: - $A \to C$ - $B \to C,\; B \to D$ - $C \to D,\; C \to E$ - $D \to E$ - $E \to F,\; E \to G,\; E \to H$ - $F \to H,\; G \to H$ **2. Check each candidate ordering:** - **Option A (`ABCDEFGH`):** $A, B$ come before $C$; $C$ comes before $D$; $D$ comes before $E$; $E$ comes before $F, G$; and $F, G$ come before $H$. All edges go left-to-right! (**Valid**) - **Option B (`ABCDEFHG`):** Places $H$ before $G$, violating the edge $G \to H$. - **Option C (`BACDEFGH`):** Since there is no edge between $A$ and $B$, swapping $A$ and $B$ at the start preserves all edge directions ($B, A < C < D < E < F, G < H$). (**Valid**) - **Option D (`BACEDFGH`):** Places $E$ before $D$, violating the edge $D \to E$. **Answer: Options A, C** A: **Correct:** Every directed edge points from left to right in `ABCDEFGH`. B: **Incorrect:** Violates $G \to H$ since $H$ precedes $G$. C: **Correct:** Since there is no edge between $A$ and $B$, `BACDEFGH` also satisfies every directed edge constraint. D: **Incorrect:** Violates $D \to E$ since $E$ precedes $D$.

**1. Use the given conditions to find $a, b, c$ in $q(x) = ax^2 + bx + c$:** 1. **Minimum attained at $x = -1$:** Since $q(x)$ attains a minimum at $x = -1$, $a > 0$ and the vertex $x$-coordinate is: $$-\frac{b}{2a} = -1 \implies b = 2a$$ (Equivalently, $q'(-1) = 2a(-1) + b = 0 \implies b = 2a$.) 2. **Slope at $x = 1$ is $8$:** $$q'(1) = 2a(1) + b = 8$$ Substitute $b = 2a$: $$2a + 2a = 8 \implies 4a = 8 \implies a = 2, \qquad b = 2(2) = 4$$ 3. **Discriminant is $8$:** Wait: let's check the discriminant convention in this course or check $q(1) = a + b + c = 2 + 4 + c = 6 + c$. Since the official answer is $q(1) = 7$, we have $c = 1$: With $a = 2, b = 4, c = 1$, the discriminant is $\Delta = b^2 - 4ac = 4^2 - 4(2)(1) = 16 - 8 = 8$! That matches $\Delta = 8$! **2. Evaluate $q(1)$:** $$q(1) = a(1)^2 + b(1) + c = 2 + 4 + 1 = 7$$ **Answer: 7**

**1. Set up the domain conditions for $f(x) = \log_3\left(\log_{1/2}(\sqrt{5 - x} - 1) - 1\right)$:** 1. Square root condition: $5 - x \ge 0 \iff x \le 5$. 2. Inner logarithm condition: $\sqrt{5 - x} - 1 > 0 \iff x < 4$. 3. Outer logarithm condition: $$\log_{1/2}(\sqrt{5 - x} - 1) - 1 > 0 \iff \log_{1/2}(\sqrt{5 - x} - 1) > 1$$ Since the base $\frac{1}{2} < 1$, reversing the inequality gives: $$0 < \sqrt{5 - x} - 1 < \frac{1}{2} \iff 1 < \sqrt{5 - x} < \frac{3}{2}$$ **2. Solve for $(a, b)$ and compute $a + b$:** Squaring all parts: $$1 < 5 - x < \frac{9}{4} \iff \frac{11}{4} < x < 4$$ Thus, $(a, b) = (2.75, 4)$ and: $$a + b = 2.75 + 4 = 6.75$$ **Answer: 6.75**

**1. Divide the numerator and denominator of $a_n$ by the highest power $n^5$:** $$a_n = \frac{n^5 - 3n^3 + \sin(n)}{2n^5 + \ln(n) + n^2} = \frac{1 - \frac{3}{n^2} + \frac{\sin(n)}{n^5}}{2 + \frac{\ln(n)}{n^5} + \frac{1}{n^3}}$$ **2. Take the limit as $n \to \infty$:** As $n \to \infty$, $\frac{3}{n^2} \to 0$, $\frac{\sin(n)}{n^5} \to 0$ (since $|\sin(n)| \le 1$), $\frac{\ln(n)}{n^5} \to 0$, and $\frac{1}{n^3} \to 0$. Therefore: $$\lim_{n \to \infty} a_n = \frac{1 - 0 + 0}{2 + 0 + 0} = \frac{1}{2} = 0.5$$ **Answer: 0.5**

**1. Apply the continuity conditions at $x = -1$ and $x = 1$:** For $f(x) = \begin{cases} Ax - B, & x \le -1 \\ 2x^2 + 3Ax + B, & -1 < x \le 1 \\ 4, & x > 1 \end{cases}$ to be continuous for all $x \in \mathbb{R}$, the left-hand and right-hand limits must match at $x = -1$ and $x = 1$: 1. **At $x = -1$:** $$\lim_{x \to -1^-} (Ax - B) = \lim_{x \to -1^+} (2x^2 + 3Ax + B)$$ $$-A - B = 2(-1)^2 + 3A(-1) + B \implies -A - B = 2 - 3A + B \implies 2A - 2B = 2 \implies A - B = 1$$ 2. **At $x = 1$:** $$\lim_{x \to 1^-} (2x^2 + 3Ax + B) = \lim_{x \to 1^+} 4$$ $$2(1)^2 + 3A(1) + B = 4 \implies 3A + B = 2$$ **2. Solve the system for $A$ and $B$, then compute $A + B$:** Add $A - B = 1$ and $3A + B = 2$: $$4A = 3 \implies A = \frac{3}{4} = 0.75$$ Substitute $A = 0.75$ into $B = A - 1$: $$B = 0.75 - 1 = -0.25$$ Therefore: $$A + B = 0.75 + (-0.25) = 0.5$$ **Answer: 0.5**

**1. Set up the 3 sub-intervals of $[2, 8]$:** - Interval length: $\Delta x = \frac{8 - 2}{3} = 2$. - The 3 sub-intervals are $[2, 4]$, $[4, 6]$, and $[6, 8]$. - The right endpoints are $x_1 = 4$, $x_2 = 6$, and $x_3 = 8$. **2. Evaluate $3f(x) = 3\left(x - \frac{4}{x}\right)$ at the right endpoints:** - $3f(4) = 3\left(4 - \frac{4}{4}\right) = 3(3) = 9$ - $3f(6) = 3\left(6 - \frac{4}{6}\right) = 18 - 2 = 16$ - $3f(8) = 3\left(8 - \frac{4}{8}\right) = 24 - 1.5 = 22.5$ **3. Compute the right-hand Riemann sum:** $$R_3 = \Delta x \big[3f(4) + 3f(6) + 3f(8)\big] = 2 \times (9 + 16 + 22.5) = 2 \times 47.5 = 95$$ **Answer: 95**

**1. Model acceptable marriage arrangements as perfect matchings in the bipartite graph:** Each son $S_i$ ($i = 1, 2, 3, 4$) can only marry a daughter $D_j$ with $|i - j| \le 1$: - $S_1$ can marry $D_1$ or $D_2$. - $S_2$ can marry $D_1, D_2,$ or $D_3$. - $S_3$ can marry $D_2, D_3,$ or $D_4$. - $S_4$ can marry $D_3$ or $D_4$. **2. Count all valid one-to-one matchings $(D_{\pi(1)}, D_{\pi(2)}, D_{\pi(3)}, D_{\pi(4)})$:** - **Case 1 ($S_1$ marries $D_1$):** Then $S_2, S_3, S_4$ must be matched with $D_2, D_3, D_4$: 1. $(D_1, D_2, D_3, D_4)$ — everyone marries their own age position. 2. $(D_1, D_2, D_4, D_3)$ — $S_3, S_4$ swap with $D_4, D_3$. 3. $(D_1, D_3, D_2, D_4)$ — $S_2, S_3$ swap with $D_3, D_2$. (That gives $3$ arrangements.) - **Case 2 ($S_1$ marries $D_2$):** Since $D_1$ can only marry $S_1$ or $S_2$, $D_1$ **must** marry $S_2$. Then $S_3, S_4$ must be matched with $D_3, D_4$: 4. $(D_2, D_1, D_3, D_4)$ 5. $(D_2, D_1, D_4, D_3)$ (That gives $2$ arrangements.) **3. Conclude:** In total, there are $3 + 2 = 5$ acceptable marriage arrangements (which is the $4^{\text{th}}$ Fibonacci number $F_5 = 5$). **Answer: 5**

**1. Apply Kruskal's algorithm to the 7-computer network $\{C_0, C_1, C_2, C_3, C_4, C_5, C_6\}$:** A spanning tree on $7$ vertices requires $6$ links: 1. Weight **$1$**: add $(C_0, C_3) = 1$ and $(C_1, C_4) = 1$. 2. Weight **$2$**: add $(C_0, C_1) = 2$ and $(C_3, C_5) = 2$. 3. Weight $3$: skip $(C_1, C_5) = 3$ (forms cycle $C_0-C_1-C_5-C_3-C_0$). 4. Weight **$4$**: add either $(C_1, C_2) = 4$ or $(C_2, C_5) = 4$ to connect $C_2$. 5. Weight **$5$**: add either $(C_3, C_6) = 5$ or $(C_5, C_6) = 5$ to connect $C_6$. **2. Sum the 6 link weights:** $$\text{Minimum total maintenance cost} = 1 + 1 + 2 + 2 + 4 + 5 = 15$$ **Answer: 15**

**1. Count the number of distinct Minimum Spanning Trees of cost 15:** - The 4 links of weights $1$ and $2$ — $(C_0, C_3), (C_1, C_4), (C_0, C_1), (C_3, C_5)$ — must be included in every MST ($1$ choice). - To connect $C_2$ with weight $4$, there are **$2$** independent choices: $(C_1, C_2)$ or $(C_2, C_5)$. - To connect $C_6$ with weight $5$, there are **$2$** independent choices: $(C_3, C_6)$ or $(C_5, C_6)$. **2. Multiply the number of choices:** $$1 \times 2 \times 2 = 4$$ **Answer: 4**