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

**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. Recall the level property of a Breadth-First Search (BFS) tree in an undirected graph:** In a BFS tree rooted at node $P$, every vertex $v$ is placed at a level equal to its shortest-path distance $d(P, v)$ from the root $P$: - **Level 0:** $P$ - **Level 1:** $Q, U$ - **Level 2:** $T, R$ - **Level 3:** $S, V$ A fundamental property of BFS on an undirected graph is that every edge $(u, w)$ in the original graph can only connect vertices whose levels differ by **at most 1** ($|d(P, u) - d(P, w)| \le 1$), i.e., vertices in the same level or in adjacent levels. **2. Check each candidate edge:** - **Option A `(P, S)`:** $P$ is at Level 0 and $S$ is at Level 3 ($|0 - 3| = 3 > 1$). If `(P, S)` existed in the original graph, $S$ would have been discovered at Level 1 directly from $P$. (**Impossible**) - **Option B `(T, R)`:** Both $T$ and $R$ are at Level 2 ($|2 - 2| = 0 \le 1$). A non-tree edge between two vertices at the same level is completely valid. (**Possible**) - **Option C `(V, S)`:** Both $V$ and $S$ are at Level 3 ($|3 - 3| = 0 \le 1$). A non-tree edge between two vertices at Level 3 is also valid. (**Possible**) - **Option D `(P, V)`:** $P$ is at Level 0 and $V$ is at Level 3 ($|0 - 3| = 3 > 1$). (**Impossible**) **Answer: Options B, C** A: **Incorrect:** An edge between $P$ (Level 0) and $S$ (Level 3) would force $S$ to be at Level 1 in the BFS tree. B: **Correct:** Both $T$ and $R$ are at Level 2, and BFS allows non-tree edges between vertices in the same level. C: **Correct:** Both $V$ and $S$ are at Level 3, so an edge `(V, S)` can exist in the original graph without changing the BFS tree. D: **Incorrect:** An edge between $P$ (Level 0) and $V$ (Level 3) would force $V$ to be at Level 1.

**1. Complete the square for $g(x) = x^2 - 4x + 5$:** $$g(x) = (x^2 - 4x + 4) + 1 = (x - 2)^2 + 1$$ **2. Evaluate each statement:** - **Options A and B:** On its domain $\mathbb{R}$, $g(x)$ is a parabola symmetric about $x = 2$ (for example, $g(1) = g(3) = 2$), so $g$ is **not one-one** and therefore **does not have an inverse** on its domain. - **Option C:** From the vertex form $g(x) = (x - 2)^2 + 1$, the vertex of the parabola is at **$(2, 1)$**. (**Correct**) - **Option D:** Setting $x = 0$ gives the $y$-intercept $g(0) = 0^2 - 4(0) + 5 = 5$. (**Correct**) **Answer: Options C, D** A: **Incorrect:** Since $g(1) = g(3) = 2$, $g$ is not one-one on $\mathbb{R}$. B: **Incorrect:** Because $g$ is not one-one on its domain, it is not invertible on its domain. C: **Correct:** Completing the square gives $g(x) = (x - 2)^2 + 1$, so the vertex is at $(2, 1)$. D: **Correct:** Evaluating $g(0) = 0 - 0 + 5 = 5$ gives a $y$-intercept of $5$.

**1. Find the domain of the composite function $(g \circ f)(x) = g(f(x))$:** - Given $f(x) = \sqrt{x + 2}$ and $g(x) = \log(1 + x^2)$: - The inner function $f(x) = \sqrt{x + 2}$ is defined for all real $x$ such that $x + 2 \ge 0 \iff x \ge -2$, so the domain of $f$ is $[-2, \infty)$. - Since $1 + (f(x))^2 > 0$ is always positive for every $f(x) \in \mathbb{R}$, the domain of $(g \circ f)(x)$ is **$[-2, \infty)$** (**Option B is correct**, Option D is incorrect). **2. Simplify the expression for $(g \circ f)(x)$ on $[-2, \infty)$:** $$ (g \circ f)(x) = g\left(\sqrt{x + 2}\right) = \log\left(1 + \left(\sqrt{x + 2}\right)^2\right) = \log(1 + x + 2) = \log(x + 3)$$ (**Option C is correct**, Option A is incorrect). **Answer: Options B, C** A: **Incorrect:** $1 + (\sqrt{x + 2})^2 = x + 3$, not $2x + 1$. B: **Correct:** $\sqrt{x + 2}$ requires $x + 2 \ge 0$, so the domain of $(g \circ f)(x)$ is $[-2, \infty)$. C: **Correct:** Substituting $f(x) = \sqrt{x + 2}$ into $g(x) = \log(1 + x^2)$ gives $\log(1 + x + 2) = \log(x + 3)$. D: **Incorrect:** At $x = -2$, $f(-2) = 0$ and $g(0) = \log(1) = 0$ is well-defined, so $-2$ is included in the domain.

**1. Analyze $R_1 = \{(x, y) \in \mathbb{Z}^2 \mid |x - y| \le 5\}$:** - **Reflexive:** For every $x \in \mathbb{Z}$, $|x - x| = 0 \le 5$, so $(x, x) \in R_1$. Thus $R_1$ **is** a reflexive relation (Option C is incorrect). - **Relation vs. Function:** For a single input $x = 0$, multiple outputs satisfy $|0 - y| \le 5$ (e.g., $(0, 0) \in R_1$ and $(0, 1) \in R_1$). Thus, $R_1$ **represents a relation but not a function** (**Option B is correct**). **2. Analyze $R_2 = \{(x, y) \in \mathbb{Z}^2 \mid 5 \text{ divides } (x - y)\}$:** - **Reflexive:** For every $x \in \mathbb{Z}$, $x - x = 0 = 5 \times 0$ is divisible by $5$, so $R_2$ **represents a Reflexive Relation** (**Option A is correct**). - **Symmetric:** If $5$ divides $(x - y)$, then $x - y = 5k$ for some $k \in \mathbb{Z}$, so $y - x = 5(-k)$ is also divisible by $5$. Thus, $R_2$ **represents a Symmetric relation** (**Option D is correct**). **Answer: Options A, B, D** A: **Correct:** Since $5$ divides $x - x = 0$ for all $x \in \mathbb{Z}$, $R_2$ is reflexive. B: **Correct:** Each integer $x$ is related to 11 different integers $y$ under $R_1$, so $R_1$ is a relation and not a function. C: **Incorrect:** Since $|x - x| = 0 \le 5$ for all $x \in \mathbb{Z}$, $R_1$ is reflexive. D: **Correct:** If $5 \mid (x - y)$, then $5 \mid (y - x)$, so $R_2$ is symmetric.

**1. List the directed edges of $G$ from the adjacency matrix:** Reading rows $0, 1, 2, 3, 4, 5$: - From $0$: edges to $2, 5$ ($0 \to 2,\; 0 \to 5$) - From $1$: edges to $0, 2, 3$ ($1 \to 0,\; 1 \to 2,\; 1 \to 3$) - From $2$: edges to $3, 4$ ($2 \to 3,\; 2 \to 4$) - From $3$: edge to $4$ ($3 \to 4$) - From $4$: no outgoing edges (sink) - From $5$: edge to $2$ ($5 \to 2$) **2. Evaluate each statement:** - **Options A and B:** Consider the vertex ordering **$1, 0, 5, 2, 3, 4$**. Every directed edge listed above goes strictly from left to right in this sequence ($1 \to 0, 2, 3$; $0 \to 5, 2$; $5 \to 2$; $2 \to 3, 4$; $3 \to 4$). Therefore, **$1, 0, 5, 2, 3, 4$ is a valid topological sequence** (**Option B is correct**), which also proves that **$G$ is a directed acyclic graph (DAG)** (**Option A is correct**). - **Option C:** Column $1$ of the adjacency matrix is all zeros, meaning vertex $1$ has in-degree $0$ and cannot be reached from vertex $0$. (**Incorrect**) - **Option D:** The directed path $1 \to 0 \to 5 \to 2 \to 3 \to 4$ visits all 6 vertices of the 6-vertex DAG and has length **$5$ edges**, which is the maximum possible length. (**Correct**) **Answer: Options A, B, D** A: **Correct:** Since $G$ admits a topological ordering, it has no directed cycles and is a DAG. B: **Correct:** Every edge $(u, v)$ in $G$ has $u$ appearing before $v$ in the sequence $1, 0, 5, 2, 3, 4$. C: **Incorrect:** Vertex 1 has in-degree 0, so it is not reachable from vertex 0. D: **Correct:** The Hamiltonian directed path $1 \to 0 \to 5 \to 2 \to 3 \to 4$ contains 5 edges.

**1. List the directed edges in the given 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 against the directed edges:** - **Option A (`ABCDEFGH`):** - $A$ and $B$ appear before $C$; $B$ and $C$ appear before $D$; $C$ and $D$ appear before $E$; $E$ appears before $F, G, H$; and both $F$ and $G$ appear before $H$. Every directed edge goes from left to right! (**Valid**) - **Option B (`ABCDEFHG`):** Places $H$ before $G$, violating the directed edge $G \to H$. - **Option C (`BACEDFGH`):** Places $E$ before $D$, violating the directed edge $D \to E$. - **Option D (`BACEDFHG`):** Violates both $D \to E$ and $G \to H$. **Answer: Option A** A: **Correct:** Every directed edge in the graph points from an earlier vertex to a later vertex in `ABCDEFGH`. B: **Incorrect:** Violates the edge $G \to H$ because $H$ appears before $G$. C: **Incorrect:** Violates the edge $D \to E$ because $E$ appears before $D$. D: **Incorrect:** Violates both $D \to E$ and $G \to H$.

**1. Find the equation of the control response line $R(x)$:** - First, evaluate $F(1)$ using $F(x) = x^2 + x - 4$: $$F(1) = 1^2 + 1 - 4 = -2$$ - The line $R(x)$ passes through $(1, F(1)) = (1, -2)$ with slope $m = -2$: $$R(x) - (-2) = -2(x - 1) \implies R(x) = -2x + 2 - 2 = -2x$$ **2. Compute the total response $D(x) + R(x) \cdot F(x)$:** - Multiply $R(x) \cdot F(x)$: $$R(x) \cdot F(x) = (-2x)(x^2 + x - 4) = -2x^3 - 2x^2 + 8x$$ - Add $D(x) = x^4 - 3x^3 + 3x + 3$: $$D(x) + R(x) \cdot F(x) = (x^4 - 3x^3 + 3x + 3) + (-2x^3 - 2x^2 + 8x) = x^4 - 5x^3 - 2x^2 + 11x + 3$$ **Answer: Option D** A: **Incorrect:** Uses $+2x$ instead of $-2x$ for $R(x)$ on the $x^3$ term. B: **Incorrect:** Both the $x^3$ and $x^2$ coefficients have the wrong sign. C: **Incorrect:** $-2x(-4) = +8x$, so the coefficient of $x$ is $3 + 8 = +11$, not $-11$. D: **Correct:** With $R(x) = -2x$, $D(x) + R(x)F(x) = x^4 - 5x^3 - 2x^2 + 11x + 3$.

**1. Set up the domain conditions for $f(x) = \log_3\left(\log_{1/2}(\sqrt{5 - x} - 1) - 1\right)$:** 1. For the square root $\sqrt{5 - x}$ to be defined: $5 - x \ge 0 \iff x \le 5$. 2. For the inner logarithm $\log_{1/2}(\sqrt{5 - x} - 1)$ to be defined, its argument must be strictly positive: $$\sqrt{5 - x} - 1 > 0 \iff \sqrt{5 - x} > 1 \iff 5 - x > 1 \iff x < 4$$ 3. For the outer logarithm $\log_3(\cdot)$ to be defined, its argument must be strictly positive: $$\log_{1/2}(\sqrt{5 - x} - 1) - 1 > 0 \iff \log_{1/2}(\sqrt{5 - x} - 1) > 1$$ Since the base $\frac{1}{2} \in (0, 1)$, $\log_{1/2}(u)$ is strictly decreasing, so reversing the inequality gives: $$0 < \sqrt{5 - x} - 1 < \left(\frac{1}{2}\right)^1 = \frac{1}{2}$$ **2. Solve the inequality for $x$:** $$1 < \sqrt{5 - x} < \frac{3}{2} \implies 1 < 5 - x < \frac{9}{4} \implies 5 - \frac{9}{4} < x < 5 - 1 \implies \frac{11}{4} < x < 4$$ Thus, the domain is $(a, b) = \left(\frac{11}{4},\; 4\right) = (2.75,\; 4)$, and: $$a + b = 2.75 + 4 = 6.75$$ **Answer: 6.75**

**1. Divide the numerator and denominator by $n^3$:** $$a_n = \frac{11n^3 + 2n^2 - 1}{(n^3 + 3n)\left(\frac{1}{8} + \frac{(-1)^n}{n}\right)} = \frac{11 + \frac{2}{n} - \frac{1}{n^3}}{\left(1 + \frac{3}{n^2}\right)\left(\frac{1}{8} + \frac{(-1)^n}{n}\right)}$$ **2. Take the limit as $n \to \infty$:** As $n \to \infty$, $\frac{2}{n} \to 0$, $\frac{1}{n^3} \to 0$, $\frac{3}{n^2} \to 0$, and $\frac{(-1)^n}{n} \to 0$ (by the Squeeze Theorem). Therefore: $$\lim_{n \to \infty} a_n = \frac{11 + 0 - 0}{(1 + 0)\left(\frac{1}{8} + 0\right)} = \frac{11}{1/8} = 88$$ **Answer: 88**

**1. Check the behavior of $f(x)$ at the boundary point $x = 1$ and other points:** Given: $$f(x) = \begin{cases} \frac{x}{(x + 3)(x + 4)}, & x \ge 1, \\ \frac{1}{x - 5}, & x < 1. \end{cases}$$ - **At $x = 1$:** - Left-hand limit: $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} \frac{1}{x - 5} = -\frac{1}{4}$. - Right-hand limit (and value $f(1)$): $\lim_{x \to 1^+} f(x) = \frac{1}{(1 + 3)(1 + 4)} = \frac{1}{20}$. Since $-\frac{1}{4} \ne \frac{1}{20}$, **$f$ is discontinuous at $x = 1$** (**Option D is correct**, and Option B is false). - **Option A:** At $x = -2 < 1$, we use the branch $f(x) = \frac{1}{x - 5}$, so $\lim_{x \to -2^+} f(x) = -\frac{1}{7} \ne \infty$. (**Incorrect**) - **Option C:** At $x = 5 > 1$, we use the branch $f(x) = \frac{x}{(x + 3)(x + 4)}$, so $\lim_{x \to 5} f(x) = \frac{5}{8 \times 9} = \frac{5}{72} \ne \frac{5}{42}$. (**Incorrect**) **Answer: Option D** A: **Incorrect:** For $x = -2 < 1$, $f(x) = \frac{1}{x - 5}$, which approaches $-1/7$ as $x \to -2^+$. B: **Incorrect:** The left-hand limit ($-1/4$) and right-hand limit ($1/20$) at $x = 1$ are unequal, so $f$ is not continuous. C: **Incorrect:** At $x = 5 > 1$, $\frac{5}{(5+3)(5+4)} = \frac{5}{72}$, not $\frac{5}{42}$. D: **Correct:** At $x = 1$, $\lim_{x \to 1^-} f(x) = -1/4 \ne \lim_{x \to 1^+} f(x) = 1/20$, so $f$ is discontinuous at $x = 1$.

**1. Set up the 3 sub-intervals of $[1, 7]$:** - Interval length: $\Delta x = \frac{7 - 1}{3} = 2$. - The 3 sub-intervals are $[1, 3]$, $[3, 5]$, and $[5, 7]$. - The right endpoints are $x_1 = 3$, $x_2 = 5$, and $x_3 = 7$. **2. Evaluate the integrand $g(x) = 3f(x) = 3\left(x - \frac{4}{x}\right)$ at the right endpoints:** - $3f(3) = 3\left(3 - \frac{4}{3}\right) = 9 - 4 = 5$ - $3f(5) = 3\left(5 - \frac{4}{5}\right) = 15 - 2.4 = 12.6$ - $3f(7) = 3\left(7 - \frac{4}{7}\right) = 21 - \frac{12}{7} \approx 19.2857$ **3. Compute the right-hand Riemann sum:** $$R_3 = \Delta x \big[3f(3) + 3f(5) + 3f(7)\big] = 2 \times \left(5 + 12.6 + 19.2857\right) = 2 \times 36.8857 = 73.7714 \approx 73.77$$ **Answer: 73.77**

**1. Apply Kruskal's algorithm to find a Minimum Spanning Tree (MST) of the 8-vertex graph:** Sort the edges in non-decreasing order of weight and add each edge that does not form a cycle: 1. $(A, B)$ with weight **$6$** — add. 2. $(A, C)$ with weight **$7$** — add. 3. $(B, C)$ with weight $8$ — skip (forms cycle $A-B-C$). 4. $(C, D)$ with weight **$9$** — add. 5. $(B, E)$ with weight **$10$** — add. 6. $(D, E)$ with weight $11$ — skip (forms cycle $A-B-E-D-C-A$). 7. $(E, F)$ with weight **$12$** — add. 8. $(D, F)$ with weight $13$ — skip (forms cycle $E-F-D-C-A-B-E$). 9. $(F, G)$ with weight **$14$** — add. 10. $(F, H)$ with weight **$15$** — add. (Edge $(G, H)$ of weight $16$ is skipped.) **2. Sum the weights of the $8 - 1 = 7$ MST edges:** $$\text{Total MST weight} = 6 + 7 + 9 + 10 + 12 + 14 + 15 = 73$$ **Answer: 73**

**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 $7 - 1 = 6$ 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 $(C_3, C_5) = 2$. (Now $\{C_0, C_1, C_3, C_4, C_5\}$ are all connected in one component.) 3. Weight $3$: skip $(C_1, C_5) = 3$ because $C_1$ and $C_5$ are already connected via $C_1 - C_0 - C_3 - C_5$. 4. Weight **$4$**: to connect $C_2$ to the component, add either $(C_1, C_2) = 4$ or $(C_2, C_5) = 4$ (one edge of weight $4$). 5. Weight **$5$**: to connect $C_6$ to the component, add either $(C_3, C_6) = 5$ or $(C_5, C_6) = 5$ (one edge of weight $5$). **2. Sum the 6 link weights:** $$\text{Minimum total maintenance cost} = 1 + 1 + 2 + 2 + 4 + 5 = 15$$ **Answer: 15**

**1. Count the independent choices when constructing a Minimum Spanning Tree of weight 15:** From Question 15: - The four edges of weights $1$ and $2$ — namely $(C_0, C_3)=1$, $(C_1, C_4)=1$, $(C_0, C_1)=2$, and $(C_3, C_5)=2$ — are uniquely forced into every MST, connecting $\{C_0, C_1, C_3, C_4, C_5\}$. - To connect vertex $C_2$ at minimum cost (weight $4$), we can choose **either** $(C_1, C_2) = 4$ **or** $(C_2, C_5) = 4$ ($2$ choices). - To connect vertex $C_6$ at minimum cost (weight $5$), we can choose **either** $(C_3, C_6) = 5$ **or** $(C_5, C_6) = 5$ ($2$ choices). **2. Multiply the independent choices:** $$\text{Number of optimum subsets of links} = 1 \times 2 \times 2 = 4$$ **Answer: 4**