Mathematics for Data Science I · Quiz 1 · Jan 2026

**1. Identify the entry:** This is a zero-mark instruction acknowledgment, not a mathematical problem. The source asks the candidate to note the rules displayed above. **2. Read the relevant conventions:** Use the stated domain, including \(0\in\mathbb N\) in this paper; give one valid numerical answer where multiple are possible; follow the requested numerical precision. Option A is the keyed acknowledgment. **Answer: A — Instructions have been mentioned above.** **Convention note:** The options contain grammatical errors in the original. This administrative entry is retained as exported and carries zero marks. A: **Correct:** It acknowledges that the exam instructions have been supplied. B: **Incorrect:** It is not the source’s keyed acknowledgment; exam instructions still govern how answers are entered.

**1. Set up inclusion–exclusion:** Let S,Y,A represent the three subscription sets, and let \(t=|S\cap Y\cap A|\). All 250 students are in their union. Pairwise intersection counts include those with all three subscriptions. **2. Substitute the counts:** \[250=100+110+120-30-30-40+t=230+t.\] **3. Solve and check:** \(t=20\). The exactly-two counts are 10,10,20, and the only-S, only-Y, only-A counts are 60,60,70; these seven disjoint regions total 250. **Answer: 20.**

**1. Read the relation:** On \(A=\{a,b,c,d\}\), \(R=\{(a,c),(b,c),(d,c),(b,a)\}\). A function must give exactly one output to every input in A; reflexivity requires every diagonal pair. **2. Test transitivity:** A pair ending at c cannot continue because no pair starts at c. The only composable two-step chain is \(b\to a\to c\), and its required pair \((b,c)\) is present. Thus every two-step chain is closed. **3. Check the remaining properties:** b has two outputs a and c, no diagonal pairs are present, and \((a,c)\) has no reverse \((c,a)\). Therefore only transitivity holds. **Answer: C — Transitive.** A: **Incorrect:** b has two outputs, and c has no output, so this is not a function from A to A. B: **Incorrect:** Pairs such as (a,a) are absent, so it is not reflexive. C: **Correct:** The only nontrivial two-step requirement is (b,c), which is present. D: **Incorrect:** (a,c) is present but (c,a) is absent.

**1. Compute the intercepts:** For \(2x+3y=6\), setting y=0 gives \(m=3\), and setting x=0 gives \(n=2\). Thus \(m+n=5\), not 6. An axis intercept is not generally the shortest distance from the origin. **2. Compare slopes:** The first line has slope \(-2/3\); \(3x-2y=6\) has slope \(3/2\). Their product is −1, so the lines are perpendicular. **3. Compare the proposed parallel lines:** Dividing \(4x+6y=12\) by 2 gives \(2x+3y=6\). They are the same line, so their separation is 0. **Answer: C, D — C and D.** A: **Incorrect:** The intercepts are 3 and 2, so their sum is 5. B: **Incorrect:** The y-intercept is 2 here, whereas perpendicular distance is 6/√13; these are different. C: **Correct:** The slopes −2/3 and 3/2 have product −1. D: **Correct:** The second equation is exactly twice the first, so the lines coincide.

**1. Find the set difference:** With \(A=\{4,5,6\}\) and \(B=\{4,7,8\}\), removing B’s elements from A leaves \(A\setminus B=\{5,6\}\). **2. Find the intersection:** For \(C=\{4,6,7\}\), \(B\cap C=\{4,7\}\). **3. Count ordered pairs:** \[(A\setminus B)\times(B\cap C)=\{(5,4),(5,7),(6,4),(6,7)\}.\] Its cardinality is \(2\times2=4\). **Answer: 4.**

**1. Use equal signed intercepts:** Let both axis intercepts be a. Intercept form gives \(x/a+y/a=1\), or \(x+y=a\). **2. Impose the given point:** Since \((2,3)\) is on the line, \(a=2+3=5\). Thus \[x+y=5\quad\Longleftrightarrow\quad y=-x+5.\] **3. Verify both requirements:** The line passes through (2,3), and its x- and y-intercepts are both 5. **Answer: A — y=−x+5.** **Convention note:** Equal intercepts is interpreted as equal signed intercept values. If only equal absolute distances were intended, y=x+1 would also have equal-magnitude intercepts; the source uses the standard signed convention. A: **Correct:** It passes through (2,3) and has equal intercepts 5 and 5. B: **Incorrect:** Although its intercepts are equal, (2,3) gives x+y=5, not 6. C: **Incorrect:** It passes through (2,3), but its signed intercepts are −1 and 1. D: **Incorrect:** It passes through (2,3), but its intercepts 7/2 and 7 are unequal.

**1. Predict each response from the supplied line:** For \(\hat y=12x-10\) at x=2,3,4,5,6, the predictions are 14,26,38,50,62. **2. Subtract predictions from observations:** Actual charges are 14,25,39,52,63, giving residuals 0,−1,1,2,1. **3. Square and sum:** \[SSE=0^2+(-1)^2+1^2+2^2+1^2=7.\] SSE is a sum, not a mean, and squaring prevents positive and negative errors from canceling. **Answer: 7.**

**1. Name the consecutive integers:** Let the smaller positive integer be n; the larger is \(n+1\). Then \(n(n+1)=182\). **2. Solve the quadratic:** \[n^2+n-182=0\quad\Longleftrightarrow\quad(n-13)(n+14)=0.\] **3. Apply positivity and verify:** Reject \(n=-14\); take \(n=13\). The next integer is 14, and \(13\times14=182\). **Answer: 13.**

**1. Differentiate the function:** For \(f(x)=ax^2+2x+3\), the tangent slope is \(f^{\prime}(x)=2ax+2\). **2. Use the slope at x=2:** \[f^{\prime}(2)=4a+2=-2.\] **3. Solve:** \(4a=-4\), so \(a=-1\). Substituting back gives \(f^{\prime}(2)=-4+2=-2\), as required. **Answer: -1.**

**1. Apply the discriminant test:** For \(f(x)=x^2-4x+c\), two distinct real zeros require \(16-4c>0\), hence \(c<4\). For \(g(x)=-x^2+3x+c\), they require \(9+4c>0\), hence \(c>-9/4\). **2. Intersect the conditions:** Both have two distinct real zeros exactly when \[-\frac94<c<4.\] The endpoints are excluded because one discriminant becomes zero there. **3. Determine vertex type:** f has positive leading coefficient and opens upward, so its vertex is a minimum. g has negative leading coefficient and opens downward, so its vertex is a maximum. **Answer: A, D — A and D.** A: **Correct:** The simultaneous discriminant condition is −9/4<c<4, with strict endpoints. B: **Incorrect:** The proposed interval −4<c<9/4 is not the intersection of the two discriminant conditions. C: **Incorrect:** f opens upward, so its vertex gives a minimum, not a maximum. D: **Correct:** g opens downward, so its vertex gives its maximum.

**1. Factor and identify zeros:** \[p(x)=-(x-4)(x+4)(x-3)^2(x-2)^2(x+9).\] It has degree 7 and distinct real zeros −9,−4,2,3,4; the zeros 2 and 3 are double. **2. Count critical points exactly:** Rolle’s theorem gives at least one zero of \(p^{\prime}\) in each of the four gaps between consecutive distinct zeros. The double zeros also give \(p^{\prime}(2)=p^{\prime}(3)=0\). These are six distinct derivative zeros. Since \(p^{\prime}\) has degree 6, there are exactly six, all simple; the derivative changes sign at each, so all six are turning points. **3. Determine the right-tail direction:** For x>4, all displayed factors are positive before the leading minus sign, so p<0. Moreover \[\frac{p^{\prime}(x)}{p(x)}=\frac1{x-4}+\frac1{x+4}+\frac2{x-3}+\frac2{x-2}+\frac1{x+9}>0.\] Therefore \(p^{\prime}(x)<0\), and p is strictly decreasing there. **4. Reject the remaining claims:** It cannot be increasing throughout (−∞,9), which includes the decreasing interval (4,9). The six derivative zeros also rule out seven horizontal tangents. **Answer: A, B — A and B.** A: **Correct:** The logarithmic-derivative calculation proves p′<0 for every x>4. B: **Correct:** There are exactly six distinct simple critical points, each a turn. C: **Incorrect:** The claim includes (4,9), where the function is strictly decreasing. D: **Incorrect:** The derivative has degree 6 and exactly six zeros, not seven.