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AP Calculus BC

AP Calculus BC Practice Test: Practice Test 8

Practice Test 8 for AP Calculus BC: real questions and explanations from the Varsity Tutors practice-test pool.

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Question 1 of 25

A temperature correction uses p(x)=x1−xp(x)=\frac{x}{1-x}p(x)=1−xx​ for ∣x∣<1|x|<1∣x∣<1; which power series equals p(x)p(x)p(x)?

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Question 1

A temperature correction uses p(x)=x1−xp(x)=\frac{x}{1-x}p(x)=1−xx​ for ∣x∣<1|x|<1∣x∣<1; which power series equals p(x)p(x)p(x)?

  1. ∑n=0∞xn\displaystyle \sum_{n=0}^{\infty} x^{n}n=0∑∞​xn
  2. ∑n=0∞xn+1\displaystyle \sum_{n=0}^{\infty} x^{n+1}n=0∑∞​xn+1 (correct answer)
  3. ∑n=1∞xn−1\displaystyle \sum_{n=1}^{\infty} x^{n-1}n=1∑∞​xn−1
  4. ∑n=1∞xn+1\displaystyle \sum_{n=1}^{\infty} x^{n+1}n=1∑∞​xn+1
  5. ∑n=0∞(−1)nxn+1\displaystyle \sum_{n=0}^{\infty} (-1)^n x^{n+1}n=0∑∞​(−1)nxn+1

Explanation: This problem asks for the power series of p(x)=x1−xp(x) = \frac{x}{1-x}p(x)=1−xx​. We can factor out x from the geometric series: p(x)=x⋅11−x=x⋅∑n=0∞xn=∑n=0∞xn+1p(x) = x \cdot \frac{1}{1-x} = x \cdot \sum_{n=0}^{\infty} x^n = \sum_{n=0}^{\infty} x^{n+1}p(x)=x⋅1−x1​=x⋅∑n=0∞​xn=∑n=0∞​xn+1. This series is x+x2+x3+…x + x^2 + x^3 + \dotsx+x2+x3+…, which starts with x (not 1). Choice A would give 11−x\frac{1}{1-x}1−x1​ without the x factor, missing the numerator x in the original function. When a function has a factor outside the geometric form, distribute it through the entire series by adjusting the power of x.

Question 2

Determine whether the series ∑n=1∞(−1)n−1n\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n}∑n=1∞​n(−1)n−1​ converges absolutely, conditionally, or diverges.

  1. Converges absolutely
  2. Converges conditionally (correct answer)
  3. Diverges because terms do not approach 000
  4. Diverges by the ratio test
  5. Converges absolutely by the alternating series test

Explanation: This problem tests your ability to classify series convergence as absolute, conditional, or divergent. The series ∑n=1∞(−1)n−1n\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n}∑n=1∞​n(−1)n−1​ is the alternating harmonic series, which converges by the alternating series test since 1n\frac{1}{n}n1​ decreases to 0. To check absolute convergence, we examine ∑n=1∞∣(−1)n−1n∣=∑n=1∞1n\sum_{n=1}^{\infty} \left|\frac{(-1)^{n-1}}{n}\right| = \sum_{n=1}^{\infty} \frac{1}{n}∑n=1∞​​n(−1)n−1​​=∑n=1∞​n1​, which is the harmonic series and diverges. Since the original series converges but the absolute value series diverges, we have conditional convergence. Choice E incorrectly claims the alternating series test proves absolute convergence, but this test only establishes convergence of the alternating series itself. Remember: a series converges conditionally when it converges but its absolute value series diverges.

Question 3

A chemist models concentration by C(t)=5t+1t2−4C(t)=\frac{5t+1}{t^2-4}C(t)=t2−45t+1​. What is ∫5t+1t2−4 dt\int \frac{5t+1}{t^2-4}\,dt∫t2−45t+1​dt?

  1. 52ln⁡∣t−2∣+52ln⁡∣t+2∣+14ln⁡∣t−2t+2∣+C\frac{5}{2}\ln|t-2|+\frac{5}{2}\ln|t+2|+\frac{1}{4}\ln\left|\frac{t-2}{t+2}\right|+C25​ln∣t−2∣+25​ln∣t+2∣+41​ln​t+2t−2​​+C
  2. 114ln⁡∣t−2∣+94ln⁡∣t+2∣+C\frac{11}{4}\ln|t-2|+\frac{9}{4}\ln|t+2|+C411​ln∣t−2∣+49​ln∣t+2∣+C (correct answer)
  3. 94ln⁡∣t−2∣+114ln⁡∣t+2∣+C\frac{9}{4}\ln|t-2|+\frac{11}{4}\ln|t+2|+C49​ln∣t−2∣+411​ln∣t+2∣+C
  4. 52ln⁡∣t2−4∣+14ln⁡∣t2−4∣+C\frac{5}{2}\ln|t^2-4|+\frac{1}{4}\ln|t^2-4|+C25​ln∣t2−4∣+41​ln∣t2−4∣+C
  5. 52ln⁡∣t−2∣+14ln⁡∣t+2∣+C\frac{5}{2}\ln|t-2|+\frac{1}{4}\ln|t+2|+C25​ln∣t−2∣+41​ln∣t+2∣+C

Explanation: The skill used here is partial fraction decomposition to integrate the rational function. Factor the denominator as t² - 4 = (t - 2)(t + 2), and express (5t + 1)/((t - 2)(t + 2)) as A/(t - 2) + B/(t + 2). Solving the equation 5t + 1 = A(t + 2) + B(t - 2) yields A = 11/4 and B = 9/4. Integrating term by term gives (11/4) ln|t - 2| + (9/4) ln|t + 2| + C. A tempting distractor like choice D combines the logs with incorrect coefficients, failing to account for the distinct constants from the decomposition. Always solve for the constants in partial fractions by substituting the roots or equating coefficients to ensure accurate integration of rational functions.

Question 4

A sensor’s cumulative drift is ∑n=2∞ln⁡nn2\sum_{n=2}^{\infty}\frac{\ln n}{n^2}∑n=2∞​n2lnn​. Use the integral test to determine convergence.

  1. Diverges, since ∫2∞ln⁡xx2 dx\int_2^\infty \frac{\ln x}{x^2}\,dx∫2∞​x2lnx​dx diverges
  2. Converges, since ∫2∞ln⁡xx2 dx\int_2^\infty \frac{\ln x}{x^2}\,dx∫2∞​x2lnx​dx converges (correct answer)
  3. Diverges, since ∫2∞ln⁡xx2 dx=(ln⁡x)22∣2∞\int_2^\infty \frac{\ln x}{x^2}\,dx=\left.\frac{(\ln x)^2}{2}\right|_2^\infty∫2∞​x2lnx​dx=2(lnx)2​​2∞​
  4. Converges, since ∫2∞ln⁡xx2 dx=ln⁡x∣2∞\int_2^\infty \frac{\ln x}{x^2}\,dx=\left.\ln x\right|_2^\infty∫2∞​x2lnx​dx=lnx∣2∞​
  5. Diverges, because ln⁡nn2\frac{\ln n}{n^2}n2lnn​ is not decreasing for all n≥2n\ge2n≥2

Explanation: The integral test is an important tool in AP Calculus BC for convergence analysis via improper integrals. For ∑_{n=2}^∞ (ln n)/n^2, f(x) = (ln x)/x^2 is positive, continuous, and eventually decreasing for x ≥ 2. Integration by parts yields the antiderivative -(ln x + 1)/x, evaluating from 2 to ∞ to converge to (ln 2 + 1)/2. Thus, the series converges by the integral test. A tempting distractor might claim divergence because the term is not decreasing for all n ≥ 2, but it decreases for n ≥ 3, which suffices. A transferable strategy for the integral test is to always verify the conditions of the integral test: the function must be positive, continuous, and decreasing for sufficiently large x.

Question 5

For ∑k=1∞(−1)k−16k(k+1)\sum_{k=1}^{\infty}(-1)^{k-1}\frac{6}{k(k+1)}∑k=1∞​(−1)k−1k(k+1)6​, what is the maximum error using S9S_9S9​?

  1. ≤69⋅10\le \dfrac{6}{9\cdot 10}≤9⋅106​
  2. ≤610⋅11\le \dfrac{6}{10\cdot 11}≤10⋅116​ (correct answer)
  3. ≤611⋅12\le \dfrac{6}{11\cdot 12}≤11⋅126​
  4. ≤310⋅11\le \dfrac{3}{10\cdot 11}≤10⋅113​
  5. ≤∫9∞6x(x+1) dx\le \displaystyle\int_{9}^{\infty}\frac{6}{x(x+1)}\,dx≤∫9∞​x(x+1)6​dx

Explanation: The skill being tested here is the alternating series error bound, which provides a way to estimate the error when approximating an infinite alternating series with a partial sum. For an alternating series that satisfies the conditions of the alternating series test—terms alternating in sign, decreasing in absolute value, and approaching zero—the error in using the partial sum S_n is less than the absolute value of the next term, a_{n+1}. This bound works because the remainder after n terms is bracketed between zero and the first omitted term, ensuring the actual sum lies within S_n ± |a_{n+1}|. In this case, for the series ∑ (-1)^{k-1} * 6/(k(k+1)) approximated by S_9, the next term is 6/(1011), so the error is at most 6/(1011). A tempting distractor like ≤ 6/(9*10) fails because it uses the last included term instead of the next one, which underestimates the bound since the error is actually smaller than the next term, not the previous. A transferable strategy for error bounds in alternating series is to always identify the first omitted term after verifying the series meets the convergence criteria.

Question 6

A heat-transfer model leads to ∫3(x+1)(x+7) dx\int \frac{3}{(x+1)(x+7)}\,dx∫(x+1)(x+7)3​dx; find an antiderivative.

  1. 12ln⁡∣x+1x+7∣+C\frac{1}{2}\ln\left|\frac{x+1}{x+7}\right|+C21​ln​x+7x+1​​+C (correct answer)
  2. 12ln⁡∣x+7x+1∣+C\frac{1}{2}\ln\left|\frac{x+7}{x+1}\right|+C21​ln​x+1x+7​​+C
  3. 12ln⁡∣x+1∣+12ln⁡∣x+7∣+C\frac{1}{2}\ln|x+1|+\frac{1}{2}\ln|x+7|+C21​ln∣x+1∣+21​ln∣x+7∣+C
  4. 3(x+1)(x+7)+C\frac{3}{(x+1)(x+7)}+C(x+1)(x+7)3​+C
  5. ln⁡∣x+1∣−ln⁡∣x+7∣+C\ln|x+1| - \ln|x+7|+Cln∣x+1∣−ln∣x+7∣+C

Explanation: This problem requires the use of partial fraction decomposition to integrate a rational function. To evaluate ∫ 3 / ((x+1)(x+7)) dx, decompose the integrand as 3 / ((x+1)(x+7)) = A/(x+1) + B/(x+7). Solving for the coefficients gives A = 1/2 and B = -1/2. Integrating term by term yields (1/2) ln |x+1| - (1/2) ln |x+7| + C, which simplifies to (1/2) ln |(x+1)/(x+7)| + C. A tempting distractor is choice B, (1/2) ln |(x+7)/(x+1)| + C, but its derivative is -3 / ((x+1)(x+7)), the negative of the integrand. A transferable partial-fraction strategy is to multiply both sides by the common denominator and solve the resulting linear system for the coefficients.

Question 7

A quantity satisfies dydt=ln⁡2 y\frac{dy}{dt}=\ln 2\, ydtdy​=ln2y with y(0)=5y(0)=5y(0)=5. Find y(1)y(1)y(1).

  1. 101010 (correct answer)
  2. 5ln⁡25\ln 25ln2
  3. 5e1/ln⁡25e^{1/\ln 2}5e1/ln2
  4. (ln⁡2)e5(\ln 2)e^{5}(ln2)e5
  5. 5+ln⁡25+\ln 25+ln2

Explanation: This problem involves exponential growth with a logarithmic rate constant. From dy/dt=(ln⁡2)ydy/dt = (\ln 2)ydy/dt=(ln2)y with y(0)=5y(0) = 5y(0)=5, we get y(t)=5e(ln⁡2)ty(t) = 5e^{(\ln 2)t}y(t)=5e(ln2)t. To find y(1)y(1)y(1): y(1)=5e(ln⁡2)×1=5eln⁡2y(1) = 5e^{(\ln 2) \times 1} = 5e^{\ln 2}y(1)=5e(ln2)×1=5eln2. Using the fundamental property eln⁡a=ae^{\ln a} = aelna=a, we have eln⁡2=2e^{\ln 2} = 2eln2=2, so y(1)=5×2=10y(1) = 5 \times 2 = 10y(1)=5×2=10. The logarithmic rate constant ln⁡2\ln 2ln2 creates a specific growth pattern where the quantity doubles over unit time. Choice B gives 5ln⁡25 \ln 25ln2, which incorrectly treats the exponential function as a linear function. When rate constants involve logarithms, use the property eln⁡a=ae^{\ln a} = aelna=a to evaluate the exponential expression.

Question 8

Which slope field corresponds to dydx=sin⁡x\dfrac{dy}{dx}=\sin xdxdy​=sinx for a curve y(x)y(x)y(x) modeling vertical displacement?

  1. Along any vertical line x=cx=cx=c, all segments have the same slope; slopes vary periodically with xxx. (correct answer)
  2. Along any horizontal line y=cy=cy=c, all segments have the same slope; slopes vary periodically with yyy.
  3. Slopes are zero on y=xy=xy=x; slopes are positive below y=xy=xy=x and negative above y=xy=xy=x.
  4. Slopes are zero on x=0x=0x=0 and increase with distance from the yyy-axis in both directions.
  5. Slopes are zero on both axes; slopes are positive in Quadrants I and III and negative in Quadrants II and IV.

Explanation: Sketching slope fields is a key skill in AP Calculus BC for visualizing solutions to differential equations like dy/dx = sin x. To verify, along a vertical line like x = 0, dy/dx = sin 0 = 0, constant zero slope. At x = π/2, dy/dx = sin(π/2) = 1, constant positive, and at x = π, dy/dx = 0 again, showing periodic variation with x, matching choice A. These points confirm slopes are the same for all y at fixed x, independent of y. A tempting distractor like choice B fails because slopes depend on x, not y; along y = c, dy/dx = sin x varies with x, not constant. Always verify slope fields by plugging sample points into the differential equation to check consistency with the described features.

Question 9

A bacteria culture satisfies dNdt=4tN\frac{dN}{dt}=\frac{4}{t}NdtdN​=t4​N. What is the general solution for N(t)N(t)N(t)?

  1. ln⁡∣N∣=4t+C\ln|N|=4t+Cln∣N∣=4t+C
  2. N=4ln⁡∣t∣+CN=4\ln|t|+CN=4ln∣t∣+C
  3. ln⁡∣N∣=4ln⁡∣t∣+C\ln|N|=4\ln|t|+Cln∣N∣=4ln∣t∣+C (correct answer)
  4. ln⁡∣t∣=4ln⁡∣N∣+C\ln|t|=4\ln|N|+Cln∣t∣=4ln∣N∣+C
  5. ln⁡∣N∣=4t+C\ln|N|=\frac{4}{t}+Cln∣N∣=t4​+C

Explanation: This bacteria growth model requires separation of variables for dNdt=4tN\frac{dN}{dt}=\frac{4}{t}NdtdN​=t4​N. Separating variables yields dNN=4tdt\frac{dN}{N}=\frac{4}{t}dtNdN​=t4​dt. Both sides integrate to natural logarithms: ln⁡∣N∣=4ln⁡∣t∣+C\ln|N|=4\ln|t|+Cln∣N∣=4ln∣t∣+C. This is the general solution, which could also be written as ln⁡∣N∣=ln⁡∣t4∣+C\ln|N|=\ln|t^4|+Cln∣N∣=ln∣t4∣+C using logarithm properties. Choice A incorrectly integrates 4t\frac{4}{t}t4​ as 4t4t4t instead of 4ln⁡∣t∣4\ln|t|4ln∣t∣, confusing it with the integral of a constant. When separating variables: (1) move all terms with one variable to each side, (2) remember that ∫1xdx=ln⁡∣x∣+C\int\frac{1}{x}dx=\ln|x|+C∫x1​dx=ln∣x∣+C, and (3) simplify using logarithm properties when possible.

Question 10

Let p(t)=(tan⁡t)(t2−1)p(t)=(\tan t)(t^2-1)p(t)=(tant)(t2−1). What is p′(t)p'(t)p′(t)?

  1. p′(t)=sec⁡2t+(2t)p'(t)=\sec^2 t+(2t)p′(t)=sec2t+(2t)
  2. p′(t)=sec⁡2t (t2−1)+tan⁡t (2t)p'(t)=\sec^2 t\,(t^2-1)+\tan t\,(2t)p′(t)=sec2t(t2−1)+tant(2t) (correct answer)
  3. p′(t)=sec⁡2t (2t)p'(t)=\sec^2 t\,(2t)p′(t)=sec2t(2t)
  4. p′(t)=tan⁡t (t2−1)p'(t)=\tan t\,(t^2-1)p′(t)=tant(t2−1)
  5. p′(t)=tan⁡t (2t)p'(t)=\tan t\,(2t)p′(t)=tant(2t)

Explanation: This problem requires the product rule to find p'(t) for a product of tangent and a quadratic. Let u(t) = tan(t) with u'(t) = sec²(t), and v(t) = t² - 1 with v'(t) = 2t. Then, p'(t) = u'(t)v(t) + u(t)v'(t) = sec²(t)(t² - 1) + tan(t)(2t), which is choice B. This incorporates the trigonometric derivative properly. Choice C fails as a distractor by neglecting the second term and the (t² - 1) factor. When dealing with trig functions in products, recall their derivatives accurately and apply the rule to both parts consistently.

Question 11

For f(x)=x3−6x2+9xf(x)=x^3-6x^2+9xf(x)=x3−6x2+9x, on which interval(s) is fff increasing?

  1. (−∞,1)∪(3,∞)(-\infty,1)\cup(3,\infty)(−∞,1)∪(3,∞) (correct answer)
  2. (1,3)(1,3)(1,3)
  3. (−∞,3)(-\infty,3)(−∞,3)
  4. (−∞,1)(-\infty,1)(−∞,1)
  5. (0,3)(0,3)(0,3)

Explanation: This problem tests your ability to determine where a function is increasing by analyzing its derivative. First, find f'(x) = 3x² - 12x + 9 = 3(x² - 4x + 3) = 3(x - 1)(x - 3). The derivative equals zero when x = 1 and x = 3, creating test intervals (-∞, 1), (1, 3), and (3, ∞). Testing x = 0 in the first interval: f'(0) = 3(-1)(-3) = 9 > 0, so f is increasing on (-∞, 1). Testing x = 2 in the middle interval: f'(2) = 3(1)(-1) = -3 < 0, so f is decreasing on (1, 3). Testing x = 4 in the last interval: f'(4) = 3(3)(1) = 9 > 0, so f is increasing on (3, ∞). A common error is to think the function increases only where both factors are positive, missing that two negative factors also yield a positive product. The key strategy is to create a sign chart showing where each factor changes sign, then multiply to find where f'(x) > 0.

Question 12

The region bounded by x=4−y2x=4-y^2x=4−y2 and the y-axis on −2≤y≤2-2\le y\le 2−2≤y≤2 is revolved about the y-axis; which integral gives volume?

  1. π∫−22(4−y2)2 dy\pi\int_{-2}^{2} (4-y^2)^2\,dyπ∫−22​(4−y2)2dy (correct answer)
  2. π∫−22(4−y2) dy\pi\int_{-2}^{2} (4-y^2)\,dyπ∫−22​(4−y2)dy
  3. π∫02(4−y2)2 dy\pi\int_{0}^{2} (4-y^2)^2\,dyπ∫02​(4−y2)2dy
  4. π∫−22(y2)2 dy\pi\int_{-2}^{2} (y^2)^2\,dyπ∫−22​(y2)2dy
  5. π∫04x2 dx\pi\int_{0}^{4} x^2\,dxπ∫04​x2dx

Explanation: This problem uses the disc method since the region bounded by x = 4 - y² and the y-axis on -2 ≤ y ≤ 2 is revolved about the y-axis, creating solid discs. The radius of each disc at height y is R(y) = 4 - y², extending from the y-axis to the curve. The volume integral is π∫₋₂² [R(y)]² dy = π∫₋₂² (4 - y²)² dy. Choice B incorrectly omits the squaring of the radius function, giving area instead of volume. The essential principle: disc method about the y-axis always involves π times the square of the x-function for proper volume calculation.

Question 13

A series is ∑n=1∞(−1)n+11n(n+1)\sum_{n=1}^{\infty}(-1)^{n+1}\frac{1}{n(n+1)}∑n=1∞​(−1)n+1n(n+1)1​. Does it converge?

  1. Diverges because 1n(n+1)\frac{1}{n(n+1)}n(n+1)1​ does not approach 000.
  2. Converges by the Alternating Series Test because 1n(n+1)\frac{1}{n(n+1)}n(n+1)1​ decreases and approaches 000. (correct answer)
  3. Diverges because ∑1n(n+1)\sum \frac{1}{n(n+1)}∑n(n+1)1​ diverges.
  4. Converges because 1n(n+1)\frac{1}{n(n+1)}n(n+1)1​ is increasing.
  5. Diverges because alternating series require bnb_nbn​ to be increasing.

Explanation: The skill being tested here is the Alternating Series Test for convergence. b_n positive, decreasing to 0; telescoping via partial fractions. Confirms decrease. For ∑ (-1)^{n+1} rac{1}{n(n+1)}, b_n = 1/n - 1/(n+1) decreases to 0, converges. A tempting distractor is choice C, saying diverges because absolute diverges, but AST applies regardless. A transferable strategy for alternating series is to use telescoping to verify conditions explicitly.

Question 14

Which integral represents the volume: base bounded by y=2xy=2xy=2x and y=0y=0y=0 for 0≤x≤30\le x\le 30≤x≤3; cross sections ⟂ xxx-axis are equilateral triangles?

  1. V=∫0334 (2x) dxV=\displaystyle\int_{0}^{3}\frac{\sqrt{3}}{4}\,(2x)\,dxV=∫03​43​​(2x)dx
  2. V=∫0334 (2x)2 dxV=\displaystyle\int_{0}^{3}\frac{\sqrt{3}}{4}\,(2x)^2\,dxV=∫03​43​​(2x)2dx (correct answer)
  3. V=∫0634 y2 dyV=\displaystyle\int_{0}^{6}\frac{\sqrt{3}}{4}\,y^2\,dyV=∫06​43​​y2dy
  4. V=∫0312 (2x)2 dxV=\displaystyle\int_{0}^{3}\frac{1}{2}\,(2x)^2\,dxV=∫03​21​(2x)2dx
  5. V=∫03π8 (2x)2 dxV=\displaystyle\int_{0}^{3}\frac{\pi}{8}\,(2x)^2\,dxV=∫03​8π​(2x)2dx

Explanation: This problem involves calculating the volume of a solid using cross-sectional areas, specifically equilateral triangles perpendicular to the x-axis. The base region is bounded by y = 2x and y = 0 from x = 0 to 3, where the height at each x is 2x, serving as the side length of the equilateral triangle. The area is (√3/4) (2x)² = √3 x². Integrating this from 0 to 3 gives the volume as in choice B. A tempting distractor like choice A uses the side without squaring, mistaking for a non-area dimension. For transferable strategy, always express the cross-section's area formula in terms of the integration variable, using squared terms for areas like in (√3/4) s².

Question 15

Using P3P_3P3​ about x=0x=0x=0 to approximate s(0.6)s(0.6)s(0.6), given ∣s(4)(x)∣≤2|s^{(4)}(x)|\le 2∣s(4)(x)∣≤2 on [0,0.6][0,0.6][0,0.6], what is the maximum error?

  1. 2(0.6)45!\dfrac{2(0.6)^4}{5!}5!2(0.6)4​
  2. 2(0.6)34!\dfrac{2(0.6)^3}{4!}4!2(0.6)3​
  3. 2(0.6)44!\dfrac{2(0.6)^4}{4!}4!2(0.6)4​ (correct answer)
  4. 2(0.6)43!\dfrac{2(0.6)^4}{3!}3!2(0.6)4​
  5. 2(0.6)44\dfrac{2(0.6)^4}{4}42(0.6)4​

Explanation: The Lagrange error bound is a key concept in AP Calculus BC for estimating the maximum error when using a Taylor polynomial to approximate a function. For P_n(x) centered at a, error |R_n(x)| ≤ (M / (n+1)!) * |x - a|^{n+1}, M as max |s^{(n+1)}(t)|. Using P_3 about 0 for s(0.6) with |s^{(4)}(x)| ≤ 2 on [0,0.6], n=3, M=2, |x-a|=0.6, bound 2*(0.6)^4 / 4!. Apply by ensuring the bound uses the fourth derivative, matching power and factorial to 4. Choice B uses ^3 instead of ^4, which mismatches the n+1 exponent and underestimates. Always remember to match the factorial to n+1 and the exponent to n+1 for a reliable error estimate in Taylor approximations.

Question 16

Looking at the graph of z(x)z(x)z(x), there's a gap in the curve at x=−3x = -3x=−3. The left piece of the curve ends at (−3,5)(-3, 5)(−3,5) with an open circle, and the right piece begins at (−3,5)(-3, 5)(−3,5) with an open circle. What is lim⁡x→−3z(x)\lim_{x \to -3} z(x)limx→−3​z(x)?

  1. 5, since both pieces approach the same y-value (correct answer)
  2. -3, representing the x-coordinate of the gap location
  3. The limit does not exist because there's a gap
  4. Undefined since the function has no value at x = -3

Explanation: The correct answer is A. Despite the gap at x = -3, both sides of the function approach the same y-value (5). The open circles indicate that the function isn't defined at x = -3, but this doesn't prevent the limit from existing. Since both one-sided limits equal 5, the two-sided limit is 5.

Question 17

A function ggg has g′(−1)=0g'(-1)=0g′(−1)=0 and g′′(−1)=7g''(-1)=7g′′(−1)=7; what is the classification of x=−1x=-1x=−1?

  1. A point of inflection at x=−1x=-1x=−1
  2. A relative minimum at x=−1x=-1x=−1 (correct answer)
  3. A relative maximum at x=−1x=-1x=−1
  4. Neither a maximum nor minimum at x=−1x=-1x=−1 because g′′(−1)>0g''(-1)>0g′′(−1)>0
  5. Cannot be determined because g′(−1)=0g'(-1)=0g′(−1)=0

Explanation: This problem requires the Second Derivative Test to classify critical points of a function. The Second Derivative Test indicates that if g'(c) = 0 and g''(c) > 0, then there is a relative minimum at x = c due to the graph being concave up. For g, g'(-1) = 0 marks a critical point, and g''(-1) = 7, which is positive, showing upward concavity. This upward concavity implies the function curves toward the tangent line from below, forming a local valley. A tempting distractor is choice C, a relative maximum, but this is incorrect as a negative second derivative, not positive, signals a maximum. To use the test reliably, ensure the function is twice differentiable and the first derivative vanishes at the point in question.

Question 18

If f′(x)=(x−2)(x+1)f'(x)=(x-2)(x+1)f′(x)=(x−2)(x+1) for all real xxx, on which interval(s) is fff increasing?

  1. (−∞,−1)∪(2,∞)(-\infty,-1)\cup(2,\infty)(−∞,−1)∪(2,∞) (correct answer)
  2. (−1,2)(-1,2)(−1,2)
  3. (2,∞)(2,\infty)(2,∞) only
  4. (−∞,2)(-\infty,2)(−∞,2)
  5. (−∞,−1)∪(−1,2)(-\infty,-1)\cup(-1,2)(−∞,−1)∪(−1,2)

Explanation: Determining the intervals where a function is increasing or decreasing is a key skill in AP Calculus BC, relying on the first derivative test. The function f is increasing where its derivative f'(x) is positive, which occurs when (x-2)(x+1) > 0. By finding the critical points x = -1 and x = 2, we divide the real line into intervals and test the sign of f'(x) in each: positive in (-∞, -1) and (2, ∞), and negative in (-1, 2). Thus, f is increasing on (-∞, -1) ∪ (2, ∞). A tempting distractor like (-1, 2) fails because that's where f'(x) is negative, indicating the function is actually decreasing there. Always create a sign chart using the roots of the derivative to systematically determine where it is positive or negative.

Question 19

A bacteria culture satisfies dPdt=0.5P(1−P)\frac{dP}{dt}=0.5P(1-P)dtdP​=0.5P(1−P). Which expression gives the general solution for P(t)P(t)P(t)?

  1. P(t)=11+Ce−0.5tP(t)=\dfrac{1}{1+Ce^{-0.5t}}P(t)=1+Ce−0.5t1​ (correct answer)
  2. P(t)=11+Ce0.5tP(t)=\dfrac{1}{1+Ce^{0.5t}}P(t)=1+Ce0.5t1​
  3. P(t)=11+Ce−tP(t)=\dfrac{1}{1+Ce^{-t}}P(t)=1+Ce−t1​
  4. P(t)=1+Ce−0.5tP(t)=1+Ce^{-0.5t}P(t)=1+Ce−0.5t
  5. P(t)=Ce0.5t1+Ce0.5tP(t)=\dfrac{Ce^{0.5t}}{1+Ce^{0.5t}}P(t)=1+Ce0.5tCe0.5t​

Explanation: This problem requires using separation of variables to solve the differential equation for the bacteria population. Start by rewriting the equation as dP / [P(1 - P)] = 0.5 dt, separating the variables P and t. Use partial fractions to express 1/[P(1 - P)] as 1/P + 1/(1 - P), then integrate both sides to get ln|P| - ln|1 - P| = 0.5 t + K, which simplifies to ln|P / (1 - P)| = 0.5 t + K. Exponentiating yields |P / (1 - P)| = e^K e^{0.5 t}, and solving for P gives the logistic form P(t) = 1 / (1 + C e^{-0.5 t}), where C accounts for the constant and sign. A tempting distractor like choice E presents an equivalent form but with a different parameterization of the constant, which might confuse students but represents the same family of solutions. To solve separable equations, always isolate variables, integrate each side, solve for the dependent variable, and incorporate the constant appropriately.

Question 20

A population satisfies dNdt=kN\frac{dN}{dt}=kNdtdN​=kN, N(0)=3N(0)=3N(0)=3, and N(4)=3e2N(4)=3e^{2}N(4)=3e2. Find kkk.

  1. k=12k=\frac{1}{2}k=21​ (correct answer)
  2. k=2k=2k=2
  3. k=4k=4k=4
  4. k=3k=3k=3
  5. k=e2k=e^{2}k=e2

Explanation: This problem requires finding the growth constant using population data over time. From dN/dt = kN with N(0) = 3 and N(4) = 3e^2, we know N(t) = 3e^(kt). Substituting the condition at t = 4: 3e^(k×4) = 3e^2, which simplifies to e^(4k) = e^2. Therefore, 4k = 2, giving k = 1/2. The time interval of 4 units with exponential factor e^2 determines the growth rate per unit time. Choice A gives k = 1/2 correctly, while choice B would produce N(4) = 3e^8 instead of the required 3e^2. When finding growth constants, divide the logarithmic change by the time interval to get the rate per unit time.

Question 21

The area is modeled by A(x)=sin⁡x (x3−1)A(x)=\sin x\,(x^3-1)A(x)=sinx(x3−1). What is A′(x)A'(x)A′(x)?

  1. cos⁡x (x3−1)+sin⁡x (3x2)\cos x\,(x^3-1)+\sin x\,(3x^2)cosx(x3−1)+sinx(3x2) (correct answer)
  2. cos⁡x (3x2)\cos x\,(3x^2)cosx(3x2)
  3. cos⁡x (x3−1)\cos x\,(x^3-1)cosx(x3−1)
  4. sin⁡x (3x2)\sin x\,(3x^2)sinx(3x2)
  5. cos⁡x (x3−1)sin⁡x (3x2)\cos x\,(x^3-1)\sin x\,(3x^2)cosx(x3−1)sinx(3x2)

Explanation: This problem requires the product rule to differentiate A(x) = sin x · (x³ - 1). The product rule states that if f(x) = u(x)v(x), then f'(x) = u'(x)v(x) + u(x)v'(x). Here, u(x) = sin x with u'(x) = cos x, and v(x) = x³ - 1 with v'(x) = 3x². Applying the product rule: A'(x) = (cos x)(x³ - 1) + (sin x)(3x²) = cos x(x³ - 1) + sin x(3x²). A common mistake is to compute only cos x(3x²) (choice B), which incorrectly pairs the derivative of sin x with the derivative of (x³ - 1). The product rule pairs each function with the derivative of the other—not derivative with derivative.

Question 22

For p(x)=1xp(x)=\frac{1}{x}p(x)=x1​ on [1,∞)[1,\infty)[1,∞), does ∫1∞p(x) dx\int_{1}^{\infty} p(x)\,dx∫1∞​p(x)dx converge, and to what value?​

  1. Converges to 111
  2. Converges to 000
  3. Converges to ln⁡(∞)−ln⁡(1)\ln(\infty)-\ln(1)ln(∞)−ln(1)
  4. Diverges (correct answer)
  5. Converges to ln⁡2\ln 2ln2

Explanation: This problem asks whether the improper integral of 1x\frac{1}{x}x1​ converges on [1,∞)[1,\infty)[1,∞), a fundamental example in calculus. To evaluate ∫1∞1x dx\int_{1}^{\infty} \frac{1}{x}\,dx∫1∞​x1​dx, we write it as lim⁡b→∞∫1b1x dx=lim⁡b→∞[ln⁡∣x∣]1b=lim⁡b→∞(ln⁡b−ln⁡1)=lim⁡b→∞ln⁡b\lim_{b \to \infty} \int_{1}^{b} \frac{1}{x}\,dx = \lim_{b \to \infty} [\ln|x|]_{1}^{b} = \lim_{b \to \infty} (\ln b - \ln 1) = \lim_{b \to \infty} \ln blimb→∞​∫1b​x1​dx=limb→∞​[ln∣x∣]1b​=limb→∞​(lnb−ln1)=limb→∞​lnb. Since ln⁡b→∞\ln b \to \inftylnb→∞ as b→∞b \to \inftyb→∞, the integral diverges. A tempting error is to think that since 1x→0\frac{1}{x} \to 0x1​→0 as x→∞x \to \inftyx→∞, the integral must converge, but the decay is too slow. The key insight is that ∫1∞1xp dx\int_{1}^{\infty} \frac{1}{x^p}\,dx∫1∞​xp1​dx converges if and only if p>1p > 1p>1; when p=1p = 1p=1, we get logarithmic divergence.

Question 23

Let p(x)=ln⁡xxp(x)=\dfrac{\ln x}{x}p(x)=xlnx​. What is p′(x)p'(x)p′(x)?​

  1. x(1x)−ln⁡xx2\dfrac{x\left(\dfrac{1}{x}\right)-\ln x}{x^2}x2x(x1​)−lnx​ (correct answer)
  2. x(1x)+ln⁡xx2\dfrac{x\left(\dfrac{1}{x}\right)+\ln x}{x^2}x2x(x1​)+lnx​
  3. ln⁡x−x(1x)x2\dfrac{\ln x-x\left(\dfrac{1}{x}\right)}{x^2}x2lnx−x(x1​)​
  4. x(1x)−ln⁡xx\dfrac{x\left(\dfrac{1}{x}\right)-\ln x}{x}xx(x1​)−lnx​
  5. x(1x)−ln⁡xx3\dfrac{x\left(\dfrac{1}{x}\right)-\ln x}{x^3}x3x(x1​)−lnx​

Explanation: For p(x) = ln x/x, we apply the quotient rule to find p'(x). The quotient rule states that (f/g)' = [g·f' - f·g']/g². The numerator ln x has derivative 1/x, and the denominator x has derivative 1. Substituting: p'(x) = [x·(1/x) - ln x·1]/x² = [1 - ln x]/x². Choice C incorrectly reverses the subtraction, giving [ln x - 1]/x² instead, which would make the derivative positive for x > e rather than x < e. When applying the quotient rule, maintain the exact order of operations: denominator times numerator's derivative, then subtract numerator times denominator's derivative.

Question 24

Let r(x)=x2+1x2−9r(x)=\dfrac{x^2+1}{x^2-9}r(x)=x2−9x2+1​. Which statement correctly identifies all vertical asymptotes?

  1. Vertical asymptotes at x=−3x=-3x=−3 and x=3x=3x=3 (correct answer)
  2. Horizontal asymptotes at x=−3x=-3x=−3 and x=3x=3x=3
  3. Vertical asymptote at x=0x=0x=0 only
  4. Finite limits at x=−3x=-3x=−3 and x=3x=3x=3
  5. Vertical asymptote at y=1y=1y=1 only

Explanation: This problem tests identifying all vertical asymptotes by analyzing where infinite limits occur in rational functions. The function r(x) = (x²+1)/(x²-9) has its denominator factor as (x+3)(x-3), giving zeros at x = -3 and x = 3. At both values, the numerator x²+1 is non-zero ((-3)²+1 = 10 and 3²+1 = 10), while the denominator equals zero. This causes the function to approach ±∞ at both x = -3 and x = 3, creating vertical asymptotes at these locations. Choice C incorrectly suggests a vertical asymptote at x = 0, where actually r(0) = 1/(-9) = -1/9, a finite value. When finding vertical asymptotes, always factor the denominator completely and check that the numerator doesn't simultaneously equal zero at those points.

Question 25

The cost function is C(x)=12x4−5x3+2x+11C(x)=12x^4-5x^3+2x+11C(x)=12x4−5x3+2x+11. What is C′(x)C'(x)C′(x)?

  1. 48x3−15x2+248x^3-15x^2+248x3−15x2+2 (correct answer)
  2. 48x4−15x3+2x48x^4-15x^3+2x48x4−15x3+2x
  3. 12x3−5x2+212x^3-5x^2+212x3−5x2+2
  4. 48x3−15x2+2x+1148x^3-15x^2+2x+1148x3−15x2+2x+11
  5. 36x3−10x2+236x^3-10x^2+236x3−10x2+2

Explanation: To find C'(x), we apply the power rule to each term of the cost function C(x) = 12x^4 - 5x^3 + 2x + 11. Using the power rule d/dx(x^n) = nx^(n-1), we get: the derivative of 12x^4 is 4·12x^3 = 48x^3, the derivative of -5x^3 is 3·(-5)x^2 = -15x^2, the derivative of 2x is 2·1x^0 = 2, and the derivative of the constant 11 is 0. Choice C (12x^3 - 5x^2 + 2) makes the common error of only bringing down the exponent without multiplying by the original coefficient. The key to the power rule is to multiply the coefficient by the exponent before reducing the exponent by 1.