Mathematics for Data Science I · End Term · Jan 2026 · AN

This is an instruction card. It specifies that some problems have discrete domains, that NAT questions require one admissible answer, and that this paper includes zero in the natural numbers. These conventions should be applied to subsequent questions rather than treated as a mathematical computation. **Answer: A.** A: **Correct:** Acknowledges the instructions as selected by the source. B: **Incorrect:** This also describes their reference purpose, but the first acknowledgement is the marked administrative choice.

Set t=log_2(x), which is positive because x>1. Since log_4(x)=t/2, the equation becomes log_2(t)+log_2(t/2)=1. Combining logs gives log_2(t²/2)=1, hence t²/2=2 and t²=4. Positivity selects t=2, so x=2²=4. The negative t root would violate the outer logarithms' domains. **Answer: 4.**

Use the four rows actually supplied in the table: (1,3),(2,5),(3,6),(4,8). The model yhat=2x+2 predicts 4,6,8,10. Observed-minus-predicted residuals are -1,-1,-2,-2; their squared sum is 1+1+4+4=10. The prose mentions five students, but only four data rows are shown, and the source answer uses these four. No missing fifth observation should be invented. **Answer: 10.**

For x≠0, the oscillating term satisfies |2x²sin(1/x)|<=2x², which tends to zero as x→0. By the squeeze theorem, the limit of f(x) is b. Continuity requires that limit equal f(0)=5, so b=5. The sine factor itself has no limit, but the vanishing x² factor makes their product converge. **Answer: 5.**

The interval [0,6] split into three equal pieces has width 2. Left endpoints are 0,2,4. For f(x)=28x²+1/2, the heights are 1/2,112+1/2,448+1/2. The left rectangle sum is \[2(0.5+112.5+448.5)=2(561.5)=1123.\] This is the specified rectangular estimate, not the exact integral or a right-endpoint sum. **Answer: 1123.**

Rationalise the numerator: \[\frac{\sqrt{1+2x}-\sqrt{1-x}}x=\frac{(1+2x)-(1-x)}{x(\sqrt{1+2x}+\sqrt{1-x})}=\frac3{\sqrt{1+2x}+\sqrt{1-x}}\] for x≠0 near zero. The denominator tends to 2, so the limit is 3/2=1.5. Cancelling x is valid for the nearby nonzero values used to compute a limit. **Answer: 1.5.**

Set the given quadratics equal: x²-4x+1=x²-6x-3 gives 2x=-4, so x=-2 and y=13. This is the required vertex. A vertical-axis parabola therefore has form y=a(x+2)²+13. Its y-intercept is 2, so 2=4a+13, giving a=-11/4. Hence y=-(11/4)(x+2)²+13. **Answer: A.** A: **Correct:** Has vertex (-2,13) and y-intercept 2. B: **Incorrect:** Places the vertex at x=2 instead of -2. C: **Incorrect:** Opens upward and gives y-intercept 24 rather than 2. D: **Incorrect:** Places the vertex height at -13 instead of 13.

The polynomials are p(x)=-x^4+5x³-7x-2 and q(x)=-x^4+5x³-x²-2. Their difference is p-q=x²-7x=x(x-7), so equality occurs at exactly x=0 and x=7: two intersection points. Both have degree four with negative leading coefficient, so both ends tend to negative infinity. A quartic has at most three turning points, because its derivative has degree three. **Answer: A.** A: **Correct:** p-q has exactly the two distinct roots 0 and 7. B: **Incorrect:** The negative quartic leading term sends p to negative infinity on the right. C: **Incorrect:** A degree-four polynomial cannot have five turning points. D: **Incorrect:** The negative quartic leading term sends q to negative infinity on the left.

a+b is even precisely when a and b have the same parity. The relation is reflexive because 2a is even, symmetric because a+b=b+a, and transitive because if a and b have the same parity and b and c have the same parity, then a and c do too. Its equivalence classes are {1,3} and {2,4}. **Answer: B, D.** A: **Incorrect:** The same-parity condition is transitive. B: **Correct:** Reversing an ordered pair preserves its even sum. C: **Incorrect:** Every (a,a) belongs to the relation. D: **Correct:** Two same-parity links imply a same-parity endpoint pair.

Set t=3^x, which is strictly positive. Then 9^x=(3^x)²=t² and the equation becomes t²+t-6=(t+3)(t-2)=0. Only t=2 is positive, so 3^x=2 has exactly one real solution, x=log_3(2). The question asks how many solutions, hence enter 1 rather than the solution value. **Answer: 1.**

Kruskal's algorithm chooses A–F(1), A–B(2), then skips B–F(3) because it would form a cycle. Choose B–C(4), C–D(5), A–E(6), and D–G(8). These six edges connect all seven vertices and have total 1+2+4+5+6+8=26. The alternative F–G(9) is more expensive than D–G, and F–E(11) is more expensive than A–E. **Answer: 26.**

A root vertex must reach every vertex by following arrow directions. Vertex 3 reaches 1 and 0. From 1 it reaches 2, from 2 it reaches 4, and from 4 it reaches 5, so all six vertices are reachable from 3. No other vertex can reach 3 because the diagram has no incoming edge to 3. Thus the unique root vertex is 3. **Answer: 3.**

For a connected undirected weighted graph, both Prim and Kruskal return a minimum spanning tree, so the minimum total weight is the same. With tied edge weights, several different trees can attain that minimum, and the algorithms may choose different edge sets. Every spanning tree on n vertices has n-1 edges, while the full graph can have more edges. **Answer: A.** A: **Correct:** Both algorithms attain the same optimum total weight. B: **Incorrect:** Ties can allow different optimal edge sets. C: **Incorrect:** A spanning tree is, by definition, a tree. D: **Incorrect:** Equal edge weights give cost w(n-1) for a spanning tree, but the full graph has cost wm; these need not agree.

For BFS from 0, the distance levels are {0}, {1,2,6}, {4,5}, {3,7}. Thus 0,1,2,6,4,5,3,7 is achievable with suitable neighbour ordering. The DFS sequence 7,4,3,5,6,0,1,2 follows successive edges and visits all vertices, so it is valid. A BFS cannot visit a level-two vertex before an unvisited level-one vertex, and a traversal starting at 0 must list 0 first. **Answer: B, C.** A: **Incorrect:** Places 4 before 6 even though 6 is a direct neighbour of the start and 4 is two steps away. B: **Correct:** The chain 7–4–3–5–6–0–1–2 follows existing edges and visits every vertex once. C: **Correct:** Respects all BFS levels and can be realised by a suitable neighbour order. D: **Incorrect:** Claims a start at 0 but begins its discovery sequence with 7.

A 5×5 adjacency matrix describes five vertices. The row sums are 2,4,3,3,2, giving total degree 14. The matrix is symmetric with zero diagonal, so each undirected edge is counted twice. Thus the graph has 14/2=7 edges. Degrees are not all 3; in particular the second vertex has degree 4. **Answer: B, C.** A: **Incorrect:** Would require total degree 16, but the row sums total 14. B: **Correct:** The matrix has five rows/columns, one per vertex. C: **Correct:** The handshake lemma gives seven edges. D: **Incorrect:** The row sums include 2 and 4 as well as 3.