Mathematics for Data Science I · Quiz 2 · May 2026

**1. Identify the domain and symmetry:** For \(f(x)=-e^{(\ln x)^2}\), the real domain is x>0. Because \(\ln(1/x)=-\ln x\), squaring gives \(f(1/x)=f(x)\). Distinct reciprocal inputs therefore generally have the same output. **2. Check monotonicity:** \(f\prime(x)=-e^{(\ln x)^2}(2\ln x)/x\), which is positive for 0<x<1. Thus f is strictly increasing on (0,1], where it has an inverse onto its image, but it is not one-to-one on its full domain. **Answer: D — f(x)=f(1/x) for every x>0.** A: **Incorrect:** For example, x=2 and x=1/2 are different but have the same function value. B: **Incorrect:** The restriction to (0,1] is strictly increasing and therefore invertible onto its range. C: **Incorrect:** The derivative is positive on (0,1), so the function is increasing there. D: **Correct:** The squared logarithm is unchanged when x is replaced by its reciprocal.

**1. Compose f with g:** With \(f(x)=x^2\) and \(g(x)=\sqrt{x+1}\), \((f\circ g)(x)=x+1\), but the original square root still requires \(x\ge-1\). **2. Compose g with f and f with itself:** \((g\circ f)(x)=\sqrt{x^2+1}\) has domain all real numbers; \((f\circ f)(x)=x^4\) also has domain all real numbers. **3. Compose g with itself:** \((g\circ g)(x)=\sqrt{\sqrt{x+1}+1}\). Once x>=-1, the outer radicand is at least 1, so its domain is [-1,infinity). Match these results to the table. **Answer: A, B, D — i-a-1, ii-b-2 and iv-d-1.** A: **Correct:** The formula is x+1 and its inherited domain is [-1,infinity), giving i-a-1. B: **Correct:** The formula is sqrt(x squared+1), defined for every real x, giving ii-b-2. C: **Incorrect:** The formula x to the fourth is right, but its domain is all real numbers, not [0,infinity). D: **Correct:** The nested square-root formula has domain [-1,infinity), giving iv-d-1.

**1. Check continuity at zero:** For x nonzero, \(|x^2\sin(1/x)|\le x^2\to0=f(0)\). Thus f is continuous at zero. **2. Compute the derivative at zero:** The difference quotient is \(h\sin(1/h)\), whose absolute value is at most |h|. Therefore \(f\prime(0)=0\). **3. Examine nearby derivatives:** For x nonzero, \(f\prime(x)=2x\sin(1/x)-\cos(1/x)\). Along \(x_n=1/(2\pi n)\), the derivative is -1, while along \(t_n=1/((2n+1)\pi)\), it is 1. So the derivative is not continuous at zero. **4. Check boundedness:** For |x|<1, \(|f\prime(x)|\le2|x|+1<3\); the value at zero is also bounded. Hence the derivative is bounded in a neighborhood despite its oscillation. **Answer: A, B, D — Continuity of f, existence of f prime(0), and local boundedness of f prime.** A: **Correct:** The bound by x squared proves the required limit equals f(0). B: **Correct:** The difference quotient tends to zero, so the derivative exists. C: **Incorrect:** Two sequences approaching zero give derivative values -1 and 1, so it has no limit there. D: **Correct:** The estimate |f prime(x)|<=2|x|+1 bounds the derivative near zero.

**1. Maximize the inner quadratic:** \(g(x)=-2x^2+12x+63=81-2(x-3)^2\). Its maximum is 81, attained at x=3. **2. Check the composition domain:** \(f(t)=\log_2(\log_3t)\) requires t>1. The maximum input 81 satisfies this condition. **3. Use monotonicity of both logarithms:** Both bases exceed 1, so f is increasing on t>1. Hence \(\max h=f(81)=\log_2(4)=2\). **Answer: 2.**

**1. Make an exponential substitution:** Let t=e to the x, so t>0. The equation becomes \(t^2-5t+6=(t-2)(t-3)=0\). **2. Return to x:** Both positive roots are admissible and give \(x=\ln2\) and \(x=\ln3\). Since the exponential is one-to-one, these are exactly two distinct real solutions. **Answer: 2.**

**1. Relate derivative and tangent:** Since f is differentiable at -2, \(f\prime(-2)\) is the slope of the tangent through (-2,5). **2. Use the other point on that tangent:** It also passes through (1,-4), so \(f\prime(-2)=(-4-5)/(1-(-2))=-9/3=-3\). **Answer: -3.**

**1. Enforce both logarithm domains:** The inner argument is \(x^2-4x+5=(x-2)^2+1>0\) for every real x. The outer argument requires \(4-\log_3((x-2)^2+1)>0\). **2. Solve the strict inequality:** Since base 3 exceeds 1, this is equivalent to \((x-2)^2+1<3^4=81\), or \(|x-2|<4\sqrt5\). **3. Add the endpoints:** The largest domain interval is \((a,b)=(2-4\sqrt5,2+4\sqrt5)\). Hence a+b=4. **Answer: 4.**

**1. Rationalize the numerator:** For a nonzero sequence term, \(\frac{\sqrt{1+2a_n}-\sqrt{1-3ka_n}}{a_n}=\frac{2+3k}{\sqrt{1+2a_n}+\sqrt{1-3ka_n}}\). **2. Pass to the limit:** As an tends to zero, both square roots tend to 1. The stated limit is therefore \((2+3k)/2=4\). **3. Solve:** \(2+3k=8\), giving k=2. The radicands are positive for all sufficiently small sequence terms. **Answer: 2.** **Convention note:** The quotient must be defined for the sequence terms used in the limit; in particular, those terms are nonzero. This is implicit in the stated existence of the limit.

**1. Set a small variable:** Let \(t=1/a_n\), so t tends to zero from above. The expression becomes \((\sin(5t)-5t\cos(2t))/t^3\). **2. Expand to the first surviving order:** \(\sin(5t)=5t-125t^3/6+O(t^5)\) and \(5t\cos(2t)=5t-10t^3+O(t^5)\). The linear terms cancel. **3. Divide and take the limit:** The cubic coefficient is \(-125/6+10=-65/6\). Thus the limit is \(-65/6=-10.833333\ldots\), inside the supplied range. **Answer: \(-65/6\approx-10.833333\).**

**1. Separate the value from the increment:** Write \(xf(2)-2f(x)=(x-2)f(2)-2(f(x)-f(2))\). **2. Divide and apply the derivative definition:** The limit is \(f(2)-2\lim_{x\to2}(f(x)-f(2))/(x-2)=f(2)-2f\prime(2)\). **3. Substitute:** \(10-2(4)=2\). **Answer: 2.**

**1. Evaluate the right branch at zero:** For x>=0, \(f(x)=\sqrt{1+x}+1\). Thus f(0)=2, and its right-hand limit is 2. **2. Compute the left-hand limit:** For x<0, \(\sin(kx)/x\to k\), using the standard sine limit (also valid with k=0). **3. Match continuity:** The left limit must equal f(0), so k=2. **Answer: 2.**

**1. Form the composition:** With \(f(x)=\log_3x\) and \(g(x)=x^2+8\), \(h(x)=\log_3(x^2+8)\). Since x squared+8 is always positive, its domain is all real numbers. **2. Check the increasing interval:** \(h\prime(x)=2x/((x^2+8)\ln3)>0\) for x>0. Thus h is strictly increasing on (0,infinity). **3. Evaluate the proposed point value:** \(h(1)=\log_3(9)=2\), not 1. **Answer: A, B, C — Increasing on positive inputs, the stated composition, and real domain.** A: **Correct:** Its derivative is positive at every positive x. B: **Correct:** Substituting g(x) into f gives the stated logarithmic formula. C: **Correct:** The logarithm argument is at least 8 for all real x. D: **Incorrect:** h(1)=log base 3 of 9=2, not 1.

**1. Write the equation:** \(h(x)=2\) gives \(\log_3(x^2+8)=2\). **2. Undo the logarithm:** \(x^2+8=3^2=9\), so x squared=1 and x=plus or minus 1. **3. Apply the requested sign:** Only x=1 is positive. **Answer: 1.**