Mathematics for Data Science I · Quiz 2 · Jan 2026

**1. Apply the domain of the outer function:** The composition f(g(x)) is defined exactly when \(g(x)\le9\), because f has domain \((-\infty,9]\). **2. Solve the quadratic constraint:** \(x^2-6x+5\le9\) gives \((x-3)^2\le13\). Hence the largest closed interval is \([a,b]=[3-\sqrt{13},3+\sqrt{13}]\). **3. Add endpoints:** a+b=6. **Answer: 6.**

**1. Test parity:** If g is odd and f is even, then \((f\circ g)(-x)=f(-g(x))=f(g(x))\). Thus the composition is even. **2. Invert a decreasing bijection:** If y1<y2 and xi=f inverse(yi), strict decrease forces x1>x2. Therefore f inverse is strictly decreasing. **3. Inspect the natural-number example:** The paper includes 0 in the natural numbers. Here h(0)=h(1)=2, and h(2)=h(3)=3, so h is not one-to-one. Its image is {2,3,4,...}, so it is also not onto the stated natural-number codomain. **4. Construct the existence example:** Take A={0}, B={0,1}, C={0}. Define g(0)=0 and f(0)=f(1)=0. Then g is injective but not onto B, f is onto C but not injective, and f composed with g is onto C. **Answer: A, B, D — Even composition, decreasing inverse, and the stated existence claim.** A: **Correct:** Oddness of g and evenness of f yield f(g(-x))=f(g(x)). B: **Correct:** Order is reversed by a decreasing bijection and also by its inverse. C: **Incorrect:** The equal images h(0)=h(1) already disprove injectivity. D: **Correct:** The explicit finite-set construction satisfies all three requested properties.

**1. Complete the square:** \(g(x)=-x^2+10x+100=125-(x-5)^2\). **2. Locate the maximum:** The squared term is nonnegative, so g(x)<=125, with equality at x=5. Hence the maximum is 125. **Answer: 125.**

**1. Find the domain of the outer function:** \(f(t)=\log_2(\log_3(\log_5t))\) requires \(\log_3(\log_5t)>0\), equivalent to \(\log_5t>1\), or t>5. **2. Maximize the inner input:** \(g(x)=125-(x-5)^2\) has maximum 125 at x=5. This point is allowed since 125>5. Every logarithm has base greater than 1, so f is strictly increasing on its domain. **3. Evaluate at the maximum:** \(h(5)=\log_2(\log_3(\log_5 125))=\log_2(\log_3 3)=\log_2 1=0\). **Answer: 0.**

**1. Impose the logarithm domain:** Both x-1 and x-3 must be positive, so x>3. **2. Combine and exponentiate:** \(\log_2((x-1)(x-3))=3\), giving \((x-1)(x-3)=8\). Therefore \(x^2-4x-5=(x-5)(x+1)=0\). **3. Reject the invalid root:** The algebraic roots are 5 and -1, but only x=5 satisfies x>3. **Answer: 5.**

**1. Use first-order limits:** As x tends to zero, \((e^{ax}-1)/x\to a\) and \(\sin(2x)/x\to2\). **2. Match the stated limit:** Thus \(\lim_{x\to0}(e^{ax}-1)/\sin(2x)=a/2=-4\), giving a=-8. **Answer: -8.**

**1. Compute the left limit:** For x<0, \(f(x)=a\sin x/x+\cos x\). Its limit at zero is a+1. **2. Compute the right limit:** For x>0, \(f(x)=x\cos(1/x)+3\). Since \(|x\cos(1/x)|\le|x|\), the right limit is 3. **3. Enforce continuity:** Both limits and f(0)=b must agree. Thus a+1=3=b, giving the requested a=2 and also b=3. **Answer: 2.** **Convention note:** Continuity requires b=3 as well as a=2. If b is fixed to any other value, no choice of a can make f continuous; the question asks only for the necessary a under the continuity assumption.

**1. Check logarithm bases and inputs:** a>0, a!=1. The base x in log base x of a also requires x>0 and x!=1. Set \(t=\log_a x\), so t!=0 and \(\log_x a=1/t\). **2. Reduce to a quadratic:** Because x>0, \(\log_a(x^2)=2t\). The equation becomes \(1/t+4t=5\), or \((4t-1)(t-1)=0\). **3. Recover the possible values:** t=1/4 or t=1 gives \(x=a^{1/4}\) or x=a. Among the displayed choices, only the former is listed, so select C. **Answer: C — \(x=a^{1/4}\).** **Convention note:** The full equation also has x=a, which is not among the choices. Both solutions are admissible for every stated a>0, a!=1. A: **Incorrect:** This has t=-1/4 and gives 1/t+4t=-5 rather than 5. B: **Incorrect:** This has t=1/2 and gives 4 rather than 5. C: **Correct:** This has t=1/4 and gives 4+1=5. D: **Incorrect:** This has t=-1/2 and gives -4 rather than 5.

**1. Compare highest powers:** The numerator and denominator are cubic polynomials, with leading coefficients 2 and -1. **2. Divide by n cubed:** \(\frac{2n^3+3n^2-4}{-n^3+4n+1}=\frac{2+3/n-4/n^3}{-1+4/n^2+1/n^3}\). As n tends to infinity all reciprocal powers vanish, giving -2. **Answer: -2.**

**1. Compare values to the right of x0:** For x>x0, strict increase gives \(f(x)>f(x_0)\), while strict decrease gives \(g(x)<g(x_0)\). Since f(x0)=g(x0), it follows that f(x)>g(x). At x0 they are equal. **2. Compare values to the left:** For x<x0, \(f(x)<f(x_0)=g(x_0)<g(x)\). Thus g(x)>=f(x) for all x<=x0. **3. Rule out another crossing:** The strict inequalities on either side mean that x0 is the unique intersection. Hence only A and D are correct. **Answer: A, D — f is at least g on the right, and g is at least f on the left.** A: **Correct:** The inequality holds strictly for x>x0 and as equality at x0. B: **Incorrect:** For every x1>x0 the inequality is strict, so another intersection is impossible. C: **Incorrect:** For x>x0, f is above g, not below it. D: **Correct:** For x<x0, g is above f, with equality at x0. E: **Incorrect:** The inequality is reversed on every x<x0.

**1. Write the tangent equation:** Differentiability at x0 gives \(y-f(x_0)=f\prime(x_0)(x-x_0)\). The tangent passes through \((x_0,f(x_0))\) and has slope f prime(x0). **2. Check possible directions:** The derivative may equal zero, so the tangent may be horizontal, parallel to the x-axis. An ordinary real derivative is finite, so the tangent cannot be vertical, parallel to the y-axis. **Answer: A, D — Passes through the point of tangency and is not vertical.** A: **Correct:** The tangent equation includes the point of tangency. B: **Incorrect:** The slope is the derivative f prime(x0), not the function value f(x0). C: **Incorrect:** A zero derivative gives a horizontal tangent, so the claimed impossibility is false. D: **Correct:** A vertical tangent would require an infinite/undefined ordinary derivative, contrary to differentiability.

**1. Express the function piecewise:** \(f(x)=x|x|\) equals x squared for x>=0 and negative x squared for x<0. Both pieces approach zero at the origin, equal to f(0), so f is continuous there and elsewhere. **2. Differentiate away from zero:** The derivative is 2x for x>0 and -2x for x<0. **3. Differentiate at zero:** \(f\prime(0)=\lim_{h\to0}h|h|/h=\lim_{h\to0}|h|=0\). Thus f is differentiable at zero as well as at every other real point. **Answer: D — f is continuous and differentiable on the real line.** A: **Incorrect:** Both continuity and differentiability hold at every real input. B: **Incorrect:** The absolute-value factor alone has a corner, but multiplication by x makes the derivative exist at zero. C: **Incorrect:** Differentiability implies continuity, and both properties hold here. D: **Correct:** The two smooth branches join continuously with common derivative zero.