Mathematics for Data Science I · Quiz 2 · Sep 2025

**1. Identify the grammatical form:** “Children” is already the irregular plural of “child”; it does not end in s. The sentence means the laughter belonging to the children. **2. Form the possessive:** Add apostrophe+s to the whole plural word: “children’s laughter.” The corrected sentence is “The children’s laughter echoed through the playground.” **Answer: A — Children’s.** **Convention note:** This valid English grammar question is a spillover from the preceding section of the source PDF; it is included because it is present in the supplied export. A: **Correct:** Children’s is the correct possessive of the irregular plural children. B: **Incorrect:** The apostrophe must follow the complete word children, not split it after child. C: **Incorrect:** This incorrectly inserts an apostrophe inside the word. D: **Incorrect:** An apostrophe alone follows an ordinary plural ending in s; children is not such a plural.

**1. Separate the two positive factors:** \(a_n=\frac{n}{n+1}\frac{n+2}{n+3}\). For nonnegative integer n, the first factor is \(1-1/(n+1)\) and the second is \(1-1/(n+3)\). Each strictly increases with n. **2. Compare consecutive products:** For n>=1 both factors are positive and both increase, so their product strictly increases. At n=0, a0=0<a1=3/8. Hence the sequence is strictly increasing from its first index, whether indexing starts at 0 or 1. **Answer: A — The sequence is strictly increasing.** A: **Correct:** Both nonnegative factors increase, and the product increases strictly at every step. B: **Incorrect:** Consecutive terms increase rather than decrease. C: **Incorrect:** There is no later decreasing phase because both factors keep increasing. D: **Incorrect:** There is no initial decreasing phase.

**1. Enforce the outer logarithm:** For \(g(x)=\log_a(\log_a((x-1)/(x+2)))\), the inner logarithm must be strictly positive. Since a>1, this requires \((x-1)/(x+2)>1\). **2. Solve the rational inequality:** Subtracting 1 gives \(-3/(x+2)>0\), so x+2<0, or x<-2. This automatically makes the inner logarithm’s own argument positive. **3. Exclude the endpoint:** At x=-2 the fraction is undefined. Thus the exact domain is \((-\infty,-2)\). **Answer: A — \((-\infty,-2)\).** A: **Correct:** For x<-2 the fraction exceeds 1, so both logarithms are defined. B: **Incorrect:** The endpoint -2 makes the denominator zero and cannot be included. C: **Incorrect:** For x>1 the fraction lies between 0 and 1, making the inner logarithm negative and the outer one undefined. D: **Incorrect:** The fraction is less than 1 on x>1, so this interval is invalid for the outer logarithm.

**1. Respect the order of operations:** The bonus is applied first, giving g(x)=x+5, and then GPA conversion gives \((f\circ g)(x)=(x+5)/25\). **2. Transform the endpoints:** This affine function is increasing. For raw scores 80 through 90, GPA runs from \(85/25=3.4\) to \(95/25=3.8\). Thus the range is [3.4,3.8]. **Answer: C — [3.4, 3.8].** A: **Incorrect:** These endpoints do not result from dividing the bonus-adjusted scores by 25. B: **Incorrect:** Its lower and upper bounds are below the correctly transformed endpoints. C: **Correct:** It equals [(80+5)/25,(90+5)/25]. D: **Incorrect:** These values are on the wrong GPA scale.

**1. Read the graph near x=2:** The left-hand curve approaches the hollow point (2,10), so \(\lim_{t\to2^-}f(t)=10\). The filled point is (2,5), and the line on the right approaches 5. Thus f(2)=5 and f is not continuous at 2. **2. Read the graph near x=6:** The line from the left approaches the hollow point (6,10), but the right-hand curve starts at the filled point (6,5). Therefore \(\lim_{t\to6^+}f(t)=5\), not 10. **3. Check differentiability at x=3:** The graph there lies inside the straight segment joining (2,5) and (6,10), with slope \((10-5)/(6-2)=5/4\). Hence f is differentiable at 3. **Answer: A, D — Left limit 10 at x=2 and differentiability at x=3.** A: **Correct:** The left-hand curve approaches the hollow point at height 10. B: **Incorrect:** The right-hand limit at 6 is 5; the limit from the left is 10. C: **Incorrect:** The left-hand limit at 2 is 10, while the function value and right-hand limit are 5. D: **Correct:** An interior point of the shown straight segment has the well-defined slope 5/4.

**1. Check the inverse statement:** A decreasing invertible function reverses order, and solving for its input reverses order again in the sense that its inverse is also decreasing. Thus A is true. **2. Check the quadratic:** \(x^2-2x+3=(x-1)^2+2\) decreases to x=1 and then increases. It is not increasing on the entire real line. **3. Solve the exponential equation:** Real logarithms require x>0. Taking natural logs of \(e^{\log_2x}=2^{\ln x}\) gives \((\ln x)/\ln2=(\ln2)(\ln x)\). Since \(1/\ln2-\ln2\ne0\), ln x=0, so x=1 is the only solution. **4. Check the proposed inverse:** The unrestricted real function x squared+1 is not one-to-one: f(1)=f(-1). It therefore has no global inverse. The negative square-root branch would be an inverse only after restricting the original domain to nonpositive inputs. **Answer: A, C — Decreasing inverse and the unique solution x=1.** A: **Correct:** The inverse of an invertible decreasing map is decreasing on its range. B: **Incorrect:** For instance f(0)=3>f(1)=2 despite 0<1, so it is not globally increasing. C: **Correct:** Taking logarithms proves that x=1 is the sole admissible solution. D: **Incorrect:** The original real quadratic is not globally injective; a domain restriction is required for this branch to be its inverse.

**1. Rationalize the square-root difference:** Put \(s=\sqrt{1+x}\). Then \(s-1=x/(s+1)\), so \(s-1-x/2=-x^2/[2(s+1)^2]\). **2. Cancel the quadratic factor:** For x nonzero, the required quotient is \(-1/[2(\sqrt{1+x}+1)^2]\). Its limit at zero is \(-1/(2\cdot4)=-1/8=-0.125\). **Answer: -0.125.**

**1. Impose the logarithm domain:** The arguments x-1 and x+3 must both be positive, so x>1. **2. Use a common base:** Since \(\log_4(x+3)=\frac12\log_2(x+3)\), the equation becomes \(\log_2((x-1)(x+3))=5\). Thus \((x-1)(x+3)=32\). **3. Solve and check:** \(x^2+2x-35=(x+7)(x-5)=0\), giving -7 or 5. Only x=5 satisfies x>1. **Answer: 5.**

**1. Obtain the tangent slope:** In y=ax+b, the slope is a. Since \(f\prime(4)=1\), a=1. **2. Use the point of tangency:** The line passes through (4,-3), so \(-3=1(4)+b\). Therefore b=-7, and the tangent is y=x-7. **Answer: -7.**

**1. Apply the outer domain:** The composition f(g(x)) exists exactly when \(-5\le|2x+5|\le7\). The lower bound is automatic because an absolute value is nonnegative. **2. Solve the remaining constraint:** \(|2x+5|\le7\) means \(-7\le2x+5\le7\), hence \(-6\le x\le1\). **3. Add the endpoints:** The largest interval is [a,b]=[-6,1], giving a+b=-5. **Answer: -5.**

**1. Divide by the highest power of n:** \(a_n=(8n^2+10n)/(2n^2+6n-7)=(8+10/n)/(2+6/n-7/n^2)\). **2. Take the limit:** The reciprocal terms vanish as n tends to infinity. Thus the limit is 8/2=4. **Answer: 4.**

**1. Compute the one-sided limits:** The left branch is \(8\sin x+p\cos x\), so its limit at zero is p. The right branch is \(4e^x-rx+3\), so its limit is 7, independent of r. **2. Equate the one-sided limits:** Existence of the limit requires p=7. The value assigned at zero, q-1, does not affect existence of the limit. **Answer: 7.**

**1. Compute the one-sided limits:** The left branch is \(8\sin x+p\cos x\), so its limit at zero is p. The right branch is \(4e^x-rx+3\), so its limit is 7, independent of r. **2. Match the limit and point value when needed:** A two-sided limit exists exactly when p=7. Continuity additionally requires \(f(0)=q-1=7\), so q=8. **3. Evaluate the requested expression:** q=8 implies q/2=4. Continuity also imposes p=7, while r remains unrestricted by continuity alone. **Answer: 4.**