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

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.

A quotient f1(x)/f2(x) is defined where both functions are defined and the denominator is nonzero. Both original domains are D, and the denominator's zeros form Z2. Thus its natural domain is D\Z2. A zero numerator is allowed if the denominator is nonzero; cancelling symbolic factors does not automatically restore points excluded from the original quotient. **Answer: C.** A: **Incorrect:** Can include denominator zeros, where division is undefined. B: **Incorrect:** Removes numerator zeros instead of denominator zeros. C: **Correct:** Removes exactly the denominator's zero set. D: **Incorrect:** Removes only common zeros and can leave other denominator zeros in the domain.

The composition g∘f is sqrt(1-e^x). A real square root needs 1-e^x>=0, or e^x<=1. Since the exponential function is increasing and e^0=1, this requires x<=0. Therefore the natural real domain is (-infinity,0], including zero because sqrt(0) is defined. **Answer: B.** A: **Incorrect:** Contains positive x, for which the radicand is negative. B: **Correct:** Exactly satisfies e^x<=1. C: **Incorrect:** Includes positive values up to 1, which are invalid. D: **Incorrect:** Every positive point in this interval is invalid.

Continuity at x=1 requires the left value a+2 to equal the right value 1+b, so b=a+1. Differentiability requires matching derivatives: the left derivative is a and the right derivative at 1 is 2. Thus a=2 and b=3. The requested combination is 2a+b=4+3=7. Differentiability must be checked along with continuity because matching values alone would leave many pairs. **Answer: 7.**

Use the reciprocal-log identity 1/log_b(e)=ln(b), with b>0 and b≠1. The equation becomes ln(x-1)+ln(x-2)=ln6, so (x-1)(x-2)=6. Expanding gives x²-3x-4=0, with roots 4 and -1. Logarithm bases require x>2, excluding -1; x=4 also avoids base 1 for both logs. Hence x=4. **Answer: 4.**

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 degree is 3+2+4=9 and the leading coefficient is negative. Therefore f(x) tends to +infinity as x→-infinity and to -infinity as x→+infinity. Differentiate and factor: \[f'(x)=-(x+1)^2(x-1)(x-3)^3(9x^2-16x-1).\] The roots of the final factor are (8±sqrt73)/9, both simple and distinct from -1,1,3. At -1 the derivative has even multiplicity and no sign change. At the other four derivative roots it changes sign, giving four turning points, not eight. **Answer: A, C.** A: **Correct:** The four odd-multiplicity derivative zeros are turning points. B: **Incorrect:** Degree nine permits at most eight, but this polynomial has only four actual turns. C: **Correct:** Odd degree with negative leading coefficient rises on the far left. D: **Incorrect:** The far-right end falls to negative infinity.

DFS can backtrack when a vertex has no unvisited outgoing neighbour; consecutive discovery labels need not all be one continuous path. A explores 0→5, backtracks, then 0→1→4→2→3→6→7. B explores 0→1→4→5, backtracks to 4, then 3→6→7 before returning to visit 2. BFS from 0 can discover 1 and 5, then 4, then 2 and 3, then 6, then 7, matching C. In D, after discovering 3, DFS must explore its still-unvisited outgoing 6 before backtracking to visit 2. **Answer: A, B, C.** A: **Correct:** A valid DFS with an early dead-end at 5 and backtracking to 0. B: **Correct:** A valid DFS with backtracking after 5 and after the 3–6–7 branch. C: **Correct:** Respects BFS levels {0},{1,5},{4},{2,3},{6},{7}. D: **Incorrect:** Attempts to return to 2 before exploring the available outgoing branch from 3 to 6.

Divide numerator and denominator by 3^n: \[a_n=\frac{9-7n^3/3^n}{1+6n^3/3^n}.\] An exponential with base greater than one grows faster than any fixed polynomial, so n³/3^n→0. The limit is therefore (9-0)/(1+0)=9. Dividing only by n³ would obscure this exponential dominance. **Answer: 9.**

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.**

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.**

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.**

An entry (A²)uv is the sum over intermediate k of A(u,k)A(k,v), counting directed two-step walks. The displayed alpha is the (i,j) entry. Vertex i goes only to m, and m has no outgoing edges, so alpha=0. Beta is the (k,m) entry; k→l→m is its sole two-step walk, so beta=1. Therefore 2alpha+beta=1. **Answer: 1.**

Apply Kruskal's algorithm, taking the cheapest edges that do not create cycles: C–D(1), C–F(2), D–G(2), A–B(4), D–E(5). The components are then {C,D,F,G,E} and {A,B}; the cheapest connecting edge has weight 8, either A–C or B–F. Six edges connect all seven vertices, with total 1+2+2+4+5+8=22. Edges such as E–G(6) are skipped because their endpoints are already connected. **Answer: 22.**

Dijkstra solves single-source shortest paths for nonnegative edge weights, rather than the minimum spanning-tree problem. Bellman–Ford handles negative edges and detects a negative cycle reachable from its source; using an added zero-edge super-source extends detection to any component. Floyd–Warshall can detect negative cycles via negative diagonal entries, but ordinary finite shortest-path answers are undefined for pairs that can traverse such a cycle. The source's D is interpreted as claiming a valid finite shortest-path solution in that situation, not merely cycle detection. **Answer: B, C.** A: **Incorrect:** A shortest-path tree need not minimise total spanning-tree weight. B: **Correct:** Bellman–Ford detects reachable negative cycles, or all cycles with a super-source. C: **Correct:** This is Bellman–Ford's primary shortest-path task when no reachable negative cycle makes distances unbounded. D: **Incorrect:** Negative cycles invalidate finite distances for affected pairs; Floyd–Warshall can still detect them, which is distinct from returning ordinary shortest paths.