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29 questions · suggested time 62 minutes. Record one letter per question.
For radian x, what is lim as x approaches 0 of sin(5x)/(3x)?
Answer C. Write sin(5x)/(3x)=(5/3)[sin(5x)/(5x)]. The bracket tends to 1 as 5x approaches zero, so the limit is 5/3. The radian convention is essential for this standard limit.
Review focus: Use a standard trigonometric limit with scaling. Unit 1: Limits and continuity.
A function f is continuous on [1,3] and has the table values. Which statement must be true?
| x | 1 | 2 | 3 |
|---|---|---|---|
| f(x) | −2 | 4 | −1 |
Answer A. The intermediate value theorem gives one zero in (1,2) and another in (2,3), since the endpoint signs change on each interval. The intervals are disjoint, so the zeros are distinct. Additional zeros and nondifferentiability are not excluded by the data.
Review focus: Infer distinct zeros from continuity and supplied values. Unit 1: Limits and continuity.
If f(x)=eˣsin x, what is f′(0)?
Answer D. The product rule gives f′(x)=eˣsin x+eˣcos x. At zero, these contributions are 0 and 1, respectively, giving 1.
Review focus: Differentiate a product involving exponential and trigonometric factors. Unit 2: Derivative definitions and rules.
The supplied graph defines f by straight segments. What is f′(1)?
Graph description: The graph joins (−1,0),(0,2),(2,0),(3,0) with straight segments. f (solid): straight segments through (-1, 0), (0, 2), (2, 0), (3, 0). Axes use the labeled scales; no values outside the specified domain are supplied.
Answer B. Input 1 is inside the straight segment joining (0,2) and (2,0), whose slope is (0−2)/(2−0)=−1. Corners at other inputs do not prevent differentiability at 1.
Review focus: Read an exact derivative from a supplied linear segment. Unit 2: Derivative definitions and rules.
For f(x)=exp(2x), what is f″(0)?
Answer A. Each differentiation contributes the derivative 2 of the inner function. Thus f′=2exp(2x), f″=4exp(2x), and f″(0)=4.
Review focus: Combine repeated differentiation with the chain rule. Unit 3: Composite, implicit and inverse derivatives.
The curve xy+ln y=2 has a local differentiable y-branch through (2,1). What is its tangent slope there?
Answer C. Since y>0 near the point, differentiation gives y+xy′+y′/y=0. Thus y′=−y/(x+1/y), which at (2,1) equals −1/(2+1)=−1/3. The nonzero denominator supports the local branch.
Review focus: Differentiate an implicit product and logarithm. Unit 3: Composite, implicit and inverse derivatives.
Water fills a conical container in which the water surface radius is always half the water depth h. Volume increases at π cubic metres per minute. How fast is h increasing when h=2 metres?
Answer D. With r=h/2, the water volume is (1/3)πr²h=πh³/12. Differentiating gives V′=(πh²/4)h′. At h=2, π=πh′, so h′=1 m/min.
Review focus: Use similar geometry in a related-rate model. Unit 4: Contextual differentiation.
What is lim as x approaches 0 of [ln(1+x)−x]/x²?
Answer B. Both numerator and denominator tend to zero. One application of L’Hôpital gives lim [1/(1+x)−1]/(2x). For x≠0 this equals −1/[2(1+x)], whose limit is −1/2. The functions are differentiable on a suitable punctured neighborhood with x>−1.
Review focus: Evaluate an indeterminate limit with justified differentiation. Unit 4: Contextual differentiation.
The supplied graph is f′ for a differentiable function f. At which inputs in (−2,4) does f have local minima?
Graph description: The derivative joins (−2,−1),(0,1),(2,−1),(4,1) by straight segments. Its zeros are −1,1,3. f′ (solid): straight segments through (-2, -1), (0, 1), (2, -1), (4, 1). Axes use the labeled scales; no values outside the specified domain are supplied.
Answer A. At −1 and 3, f′ changes from negative to positive, giving local minima of f. At 1 it changes from positive to negative, giving a local maximum. Turning points of the derivative graph itself answer a different question.
Review focus: Locate extrema from derivative sign changes on a supplied graph. Unit 5: Analytical differentiation.
At which inputs does f(x)=x³−3x attain its absolute maximum on [−2,2]?
Answer D. The critical inputs satisfy 3x²−3=0, so they are −1 and 1. The values at −2,−1,1,2 are −2,2,−2,2. The absolute maximum value 2 occurs at both −1 and 2.
Review focus: Compare endpoints and interior critical points for a global extremum. Unit 5: Analytical differentiation.
For f(x)=|x−2| on [0,4], the endpoint secant has slope zero, but no existing derivative in (0,4) equals zero. Why does this not contradict the mean value theorem?
Answer C. The function is continuous on the closed interval but has a corner at 2. Its derivative is −1 to the left and 1 to the right, and does not exist at 2. Differentiability throughout the open interval is a required hypothesis; unequal endpoint values are not required.
Review focus: Diagnose a failed mean value theorem hypothesis. Unit 5: Analytical differentiation.
A rectangle has its lower side on the x-axis and upper vertices on the upper semicircle x²+y²=9, with its sides parallel to the axes. What is the largest possible area?
Answer B. Symmetry gives upper vertices (−x,y),(x,y), where x²+y²=9 and x,y≥0. Area is 2xy. Since (x−y)²≥0, 2xy≤x²+y²=9, with equality at x=y=3/√2. A derivative optimization gives the same global maximum.
Review focus: Optimize a constrained geometric quantity. Unit 5: Analytical differentiation.
What is ∫₀¹x exp(2x)dx?
Answer C. Take u=x and dv=exp(2x)dx, so v=exp(2x)/2. The integral is [x exp(2x)/2]₀¹−(1/2)∫₀¹exp(2x)dx=e²/2−(e²−1)/4=(e²+1)/4.
Review focus: Evaluate a definite integral using integration by parts. Unit 6: Integration and accumulation.
What is ∫₀¹ dx/[(x+1)(x+3)]?
Answer A. The decomposition is (1/2)[1/(x+1)−1/(x+3)]. Endpoint evaluation gives (1/2)[ln(2/4)−ln(1/3)]=(1/2)ln(3/2). Both denominators stay positive on the interval.
Review focus: Use partial fractions with distinct linear factors. Unit 6: Integration and accumulation.
What is the value of the improper integral ∫₀¹x^(−2/3)dx?
Answer D. Use the limit as a approaches zero from above of ∫ₐ¹x^(−2/3)dx. The primitive is 3x^(1/3), giving 3−3a^(1/3), which tends to 3. An unbounded integrand at an endpoint does not automatically make its integral divergent.
Review focus: Evaluate an improper integral using its defining limit. Unit 6: Integration and accumulation.
If G(x)=∫ from sin x to x of exp(t²)dt, which expression equals G′(x)?
Answer B. Subtract primitives at the upper and lower endpoints. FTC gives exp(x²) for the upper endpoint and cos x exp(sin²x) for the lower endpoint, which is subtracted. Differentiating the integrand again is not part of this FTC step.
Review focus: Differentiate an accumulation with two moving endpoints. Unit 6: Integration and accumulation.
Using the table and the composite trapezoid rule on its two indicated intervals, what is the estimate of ∫₀⁵f(x)dx?
| x | 0 | 2 | 5 |
|---|---|---|---|
| f(x) | 1 | 3 | −1 |
Answer A. The widths are 2 and 3. The signed trapezoids give 2(1+3)/2+3(3−1)/2=4+3=7. The negative endpoint is retained with its sign; this estimates a signed integral, not total geometric area.
Review focus: Construct a signed numerical integral from unequal-width data. Unit 6: Integration and accumulation.
For y′=x−y and y(0)=2, what approximation to y(1) results from Euler’s method with two steps of width 1/2?
Answer D. At (0,2) the slope is −2, giving y₁=2+(1/2)(−2)=1 at x=1/2. The next slope is 1/2−1=−1/2, giving y₂=1+(1/2)(−1/2)=3/4 at x=1. Each slope is evaluated at the current approximate point.
Review focus: Perform Euler updates using each updated point. Unit 7: Differential equations.
A population satisfies P′=(1/5)P(1−P/60), with 0≤P≤60. What is the greatest possible value of P′?
Answer C. The rate as a function of P is P/5−P²/300, a downward-opening quadratic with vertex at P=30. Its value there is (1/5)·30·(1/2)=3. The population at fastest growth is 30, but the requested growth rate is 3.
Review focus: Identify the maximum growth rate in a logistic model. Unit 7: Differential equations.
A particle has velocity v(t)=3t²−6t on [0,3]. What is its total distance traveled?
Answer B. Velocity is negative on (0,2) and positive on (2,3). A primitive is t³−3t², with values 0,−4,0 at 0,2,3. Thus distance is 4+4=8, although net displacement is zero.
Review focus: Split at a velocity sign change to find total distance. Unit 8: Applications of integration.
What is the length of y=(1/3)(x²+2)^(3/2) for 0≤x≤1?
Answer A. The derivative is x√(x²+2). Therefore 1+(y′)²=1+x²(x²+2)=(x²+1)². Its positive square root is x²+1, so length is ∫₀¹(x²+1)dx=4/3.
Review focus: Simplify and evaluate a Cartesian arc-length integral. Unit 8: Applications of integration.
For x=t²+1 and y=t³−3t, what is d²y/dx² at t=1?
Answer D. dy/dx=(3t²−3)/(2t)=(3/2)(t−1/t) near t=1. Differentiate with respect to t to get (3/2)(1+1/t²), then divide by dx/dt=2t. At 1 the result is 3/2. Merely differentiating the slope with respect to t gives 3, not the second x-derivative.
Review focus: Compute a second derivative for a parametrized curve. Unit 9: Parametric, polar and vector-valued functions.
The cardioid r=1+sinθ is traced once for 0≤θ≤2π. What is its enclosed area?
Answer B. The area is (1/2)∫₀^(2π)(1+sinθ)²dθ. Over a full period the integrals of 1, 2sinθ and sin²θ are 2π,0,π. Half their sum is 3π/2. The isolated zero radius does not introduce repeated positive-area coverage.
Review focus: Evaluate a polar area with a single tracing interval. Unit 9: Parametric, polar and vector-valued functions.
Which of the following infinite series converges?
Answer C. For the third series, comparison with 1/n² gives limit ratio n³/(n³+1)→1, so it converges. The first is comparable with the divergent harmonic series, the second is a p-series with p=1/2, and the last fails the necessary condition that terms tend to zero.
Review focus: Choose a valid convergence classification by comparison. Unit 10: Infinite sequences and series.
What is Σ from n=1 to ∞ of 5(−1/3)ⁿ?
Answer A. The first term is −5/3 and the common ratio is −1/3, whose absolute value is less than 1. The sum is (−5/3)/(1+1/3)=−5/4. Starting at n=0 instead would add 5 and give 15/4.
Review focus: Sum a geometric series with nonzero starting index. Unit 10: Infinite sequences and series.
For which real x does Σ from n=1 to ∞ of (x−1)ⁿ/(n·2ⁿ) converge?
Answer D. The ratio test gives |x−1|<2 for absolute convergence, or −1<x<3. At x=−1 the series is Σ(−1)ⁿ/n, which converges conditionally. At x=3 it is the divergent harmonic series. Therefore the complete interval is [−1,3).
Review focus: Determine a full interval of convergence including endpoints. Unit 10: Infinite sequences and series.
The table gives exact derivatives at a=2. What is the value at x=5/2 of the degree-three Taylor polynomial centered at 2?
| Quantity | f(2) | f′(2) | f″(2) | f‴(2) |
|---|---|---|---|---|
| Value | 1 | −3 | 4 | −6 |
Answer B. Writing h=x−2, the polynomial is 1−3h+(4/2!)h²−(6/3!)h³=1−3h+2h²−h³. At h=1/2 it equals 1−3/2+1/2−1/8=−1/8. These derivative values determine the polynomial, not an exact value of the original function away from 2.
Review focus: Construct and evaluate a Taylor polynomial from derivative data. Unit 10: Infinite sequences and series.
For Σ from n=1 to ∞ of (−1)^(n+1)/n³, what is the smallest N for which the next-term error bound guarantees that the first N terms have error at most 1/1000?
Answer C. The positive magnitudes 1/n³ decrease to zero, so the remainder is at most 1/(N+1)³. Requiring this bound to be at most 1/1000 gives N+1≥10, hence N≥9. This is the smallest N certified by this bound, not a claim that every smaller approximation actually violates the tolerance.
Review focus: Apply the alternating-series error bound with correct indexing. Unit 10: Infinite sequences and series.
Suppose f has a continuous third derivative on [0,1/5] with |f‴(x)|≤5 throughout. What upper bound does the Lagrange remainder formula give for approximating f(1/5) by its degree-two Taylor polynomial centered at 0?
Answer D. The degree-two remainder bound is M|x−a|³/3!. With M=5 and x−a=1/5, it is 5(1/5)³/6=1/150. The supplied bound applies throughout the interval, not just at its endpoints.
Review focus: Use an interval derivative bound in Taylor’s theorem. Unit 10: Infinite sequences and series.
13 questions · suggested time 38 minutes. Record one letter per question.
Define F(x)=exp(−x²)+k for x<1.4 and F(x)=ln(1+x) for x≥1.4. Which value of k makes F continuous at x=1.4? Round to three decimal places.
Answer B. Both branch formulas are continuous on their respective neighborhoods of 1.4. Matching the left limit to the right branch value requires exp(−1.4²)+k=ln(2.4). Thus k=ln(2.4)−exp(−1.96)≈0.7346103164, rounding to 0.735. Subtract the two calculator values before rounding; matching derivatives is not a requirement for continuity.
Review focus: Determine a continuity parameter at a piecewise junction. Unit 1: Limits and continuity.
Let f(x)=(x²+1)sin x. What is f′(1.3), rounded to three decimal places? Use radians.
Answer D. Apply the product rule with derivatives 2x and cos x: f′(x)=2x sin x+(x²+1)cos x. At x=1.3 this is approximately 3.2248231311, rounding to 3.225. Evaluate both terms before adding; either term by itself omits part of the product rule. The value f(1.3)≈2.5919715188 is the function height, not its derivative. A symmetric numerical difference quotient near 1.3 checks the derivative calculation.
Review focus: Evaluate a product-rule derivative using basic trigonometric derivatives. Unit 2: Derivative definitions and rules.
Let f(x)=x³+eˣ on the real line and let h be its inverse. What is h′(3.2), rounded to three decimal places?
Answer A. Because f′(x)=3x²+eˣ>0, f is strictly increasing; its limits at negative and positive infinity are negative and positive infinity, so the inverse exists on the real line. Solve u³+eᵘ=3.2 to obtain u≈0.9023264876. Then h′(3.2)=1/(3u²+eᵘ)≈0.2037526644, rounding to 0.204. Evaluating f′ at 3.2 instead of at u uses the wrong input.
Review focus: Evaluate an inverse derivative at its correctly recovered input. Unit 3: Composite, implicit and inverse derivatives.
A conical tank has its vertex at the bottom, height 12 meters and top radius 4 meters. Water enters at rate 8+cos t cubic meters per minute, where t is in minutes. At t=2 the water depth is 3 meters. What is the instantaneous rate of increase of depth, in meters per minute, rounded to three decimal places?
Answer C. Similarity gives water-surface radius r=h/3. Thus V=(1/3)πr²h=πh³/27 and dV/dt=(πh²/9)dh/dt. At h=3 the multiplier is π, so dh/dt=(8+cos2)/π≈2.4140154373 m/min. Use radians. The tank dimensions determine the radius-depth relation; substituting the fixed top radius 4 for the current water radius would give the wrong rate.
Review focus: Relate a changing volume to a depth using similar triangles. Unit 4: Contextual differentiation.
What is the absolute minimum value of f(x)=e⁻ˣ cos(2x) on [0,3], rounded to three decimal places?
Answer A. Differentiate: f′(x)=−e⁻ˣ[cos(2x)+2sin(2x)]. Critical points satisfy tan(2x)=−1/2, with cos(2x) nonzero at any such point. In [0,3] the two points are x₁=(π−arctan(1/2))/2≈1.3389725223 and x₂=x₁+π/2≈2.9097688491. Evaluate f at these and at 0 and 3. The first critical value is approximately −0.2344426467; the second and both endpoint values are positive. Therefore the minimum value is −0.234. The input 1.339 is not the requested minimum value.
Review focus: Compare numerical critical values and endpoints for an absolute extremum. Unit 5: Analytical differentiation.
What is the value of the convergent improper integral ∫₁^∞ dx/(x²+4), rounded to three decimal places?
Answer B. For a finite upper bound b, the integral is (1/2)[arctan(b/2)−arctan(1/2)]. Taking b→∞ gives π/4−(1/2)arctan(1/2)≈0.5535743589, rounding to 0.554. The finite limit proves convergence. Merely replacing infinity by a moderately large calculator bound omits a positive tail and does not itself justify the requested precision.
Review focus: Evaluate an improper integral with an explicit limiting argument. Unit 6: Integration and accumulation.
What is ∫₀² x ln(1+x²) dx, rounded to three decimal places?
Answer D. With u=1+x², du=2x dx and bounds u=1 to 5, the integral is (1/2)∫₁⁵ ln u du. Integration by parts gives the primitive u ln u−u. Evaluation yields (5ln5−4)/2≈2.0235947811, rounding to 2.024. Direct numerical integration of the original integrand provides a check. Omitting the substitution factor 1/2 doubles the answer.
Review focus: Combine a substitution and integration by parts with numerical evaluation. Unit 6: Integration and accumulation.
For the initial-value problem y′=t−y², y(0)=0.5, use Euler’s method with four equal steps of size 0.25 to approximate y(1). Which result is obtained, rounded to three decimal places? Do not round the intermediate states.
Answer C. Use yₙ₊₁=yₙ+0.25(tₙ−yₙ²), tₙ=n/4. The successive states are y₁=0.4375, y₂=0.4521484375, y₃≈0.5260388851 and y₄≈0.6443596580. Thus the requested Euler approximation is 0.644. Each slope uses the current updated state and the left-endpoint time; reusing y₀ in every step does not implement Euler’s method. The method’s output is an approximation, not a claim of three-decimal accuracy for the exact solution.
Review focus: Execute a fixed-step Euler calculation using successive updated states. Unit 7: Differential equations.
What is the length of the graph y=ln(1+x²) from x=0 to x=2, rounded to three decimal places?
Answer A. The derivative is y′=2x/(1+x²). Arc length is ∫₀²√[1+(2x/(1+x²))²]dx≈2.6019041699, so the answer is 2.602. Numerical integration can be refined to check the rounding. The endpoint height ln5 is not length, and √[2²+(ln5)²] is only the endpoint chord length, not the curved path length.
Review focus: Set up and numerically evaluate a Cartesian arc length. Unit 8: Applications of integration.
A particle has position x(t)=t²+sin t and y(t)=exp(t/2), with coordinates in meters and t in seconds. What is its speed at t=1.2, in meters per second, rounded to three decimal places?
Answer B. Velocity components are x′=2t+cos t and y′=(1/2)exp(t/2). Speed is their Euclidean norm, √[(2t+cos t)²+exp(t)/4]. At t=1.2 this is approximately 2.9087195799 m/s, rounding to 2.909. Adding the two component velocities does not give speed; either component alone also omits part of the motion.
Review focus: Evaluate speed from a parametric position. Unit 9: Parametric, polar and vector-valued functions.
In the first quadrant, region A consists of points with 0≤r≤1+θ/2 and 0≤θ≤π/2. Region B consists of points with 0≤r≤2cosθ over the same angle interval. What is the area common to A and B, rounded to three decimal places?
Answer D. The radial boundary changes where 1+α/2=2cosα. The difference 1+θ/2−2cosθ is strictly increasing on [0,π/2], with opposite endpoint signs, so there is one switch α≈0.7965913868. The smaller radius is 1+θ/2 before α and 2cosθ afterward. Area=(1/2)∫₀^α(1+θ/2)²dθ+(1/2)∫_α^(π/2)4cos²θdθ≈0.8523271803, rounding to 0.852. Both regions include every radial point down to zero, and the first-quadrant description covers the overlap once. Store α with extra precision instead of rounding it before integration.
Review focus: Determine a polar boundary switch numerically before integrating an overlap. Unit 9: Parametric, polar and vector-valued functions.
A curve is given by x=t+ln(1+t) and y=t²+sin t for t>−1. What is d²y/dx² at t=1.4, rounded to three decimal places?
Answer A. Use x′=1+1/(1+t), x″=−1/(1+t)², y′=2t+cos t and y″=2−sin t. Since x′>0 here, d²y/dx²=(y″x′−y′x″)/(x′)³. At t=1.4 this is approximately 0.6868734386, rounding to 0.687. Differentiating dy/dx with respect to t alone is incomplete: divide by x′ once more to convert that derivative to one with respect to x.
Review focus: Compute a second derivative of a parametric curve. Unit 9: Parametric, polar and vector-valued functions.
Let Pₙ(x)=Σₖ₌₀ⁿ xᵏ/k! be the degree-n Maclaurin polynomial for eˣ. What is the smallest n for which the alternating-series error bound guarantees |e^(−1.7)−Pₙ(−1.7)|<10⁻⁶?
Answer C. At x=−1.7 the terms alternate. From k=1 onward their magnitudes decrease, because the next/current ratio is 1.7/(k+1)<1; they tend to zero. For n≥1 the omitted tail therefore has error at most 1.7^(n+1)/(n+1)!. For n=11 this bound is approximately 1.216326286×10⁻⁶, too large; for n=12 it is approximately 1.590580528×10⁻⁷, which is below 10⁻⁶. Earlier eligible bounds are larger, so the smallest guaranteed degree is 12. This uses the bound requested, not a comparison of rounded polynomial and exponential displays.
Review focus: Choose a polynomial degree from a numerical alternating-series bound. Unit 10: Infinite sequences and series.
2 questions · suggested time 30 minutes. Each question has 9 practice-rubric points.
CALCULATOR PERMITTED. Questions 1–2 share a 30-minute section. A surveying drone follows a synthetic horizontal-plane motion model for 0≤t≤3 seconds. Its velocity is v(t)=(2+cos(t²), 1+sin t) meters/second, and its initial position is (−1,2) meters. Angles are in radians. Treat the velocity components as exact model functions.
Find the drone’s position at t=3. Show how the initial point enters the calculation.
Find the total distance traveled. Explain why integrating the sum of the two velocity components, or taking the distance between endpoints, would generally give a different quantity.
Find the slope dy/dx of the drone’s path at t=1. Explain why the slope formula is valid at this time.
Is the drone’s speed increasing or decreasing at t=1? Justify using velocity and acceleration.
After your attempt: selected-criterion repair, changed case and coverage boundaries. This optional worksheet adds no form points.
Position is (−1+∫₀³[2+cos(t²)]dt, 2+∫₀³[1+sin t]dt). This is approximately (5.702863558,6.989992497) meters. The second coordinate also equals 6−cos3 exactly.
Speed is √[(2+cos(t²))²+(1+sin t)²]. Its integral from 0 to 3 is approximately 8.557009953 meters. Adding coordinate velocities does not give their Euclidean norm. Endpoint distance measures the norm of integrated velocity, which omits the extra length of a path whose direction changes.
dy/dx=y′/x′=(1+sin1)/(2+cos1). The denominator is nonzero, since 2+cos1≥1. This exact expression is an acceptable answer; it is a Cartesian path slope, not the speed.
Acceleration is a(t)=(−2t sin(t²),cos t). At t=1 the dot product is (2+cos1)(−2sin1)+(1+sin1)cos1<0 (approximately −3.280230347). Speed is positive, so its derivative (v·a)/|v| is negative. The drone’s speed is decreasing.
Review focus: Planar motion, numerical length and changing speed. Unit 9: Parametric, polar and vector-valued functions.
CALCULATOR PERMITTED. Questions 1–2 share a 30-minute section. A laboratory vessel contains a dissolved tracer. This is a synthetic mass-balance model. For 0≤t≤6 minutes, tracer enters at R(t)=4+2exp(−t²/5) milligrams/minute. The amount in the vessel is stipulated to be Q(t)=exp(−t/4)[10+∫₀ᵗexp(u/4)R(u)du] milligrams. Thus Q(0)=10. The model has no capacity or mixing limitations during this interval.
Differentiate the supplied formula to show that Q′(t)=R(t)−Q(t)/4. Interpret the term Q(t)/4.
Find the amount of tracer at t=6, using the supplied weighted-integral formula.
Find the first time Q(t)=15. Give a bracket and justify why there cannot be a second crossing of this level on [0,6].
Use one Euler step of width 1 minute from t=0 to estimate Q(1). Compare it with a direct numerical evaluation of the supplied formula at t=1.
After your attempt: selected-criterion repair, changed case and coverage boundaries. This optional worksheet adds no form points.
The product rule gives Q′=−(1/4)exp(−t/4)[10+∫₀ᵗexp(u/4)R(u)du]+exp(−t/4)exp(t/4)R(t), by FTC. These terms are −Q/4 and R. The term Q/4 is the removal rate, in milligrams/minute, proportional to the amount currently present; the coefficient has units 1/minute.
Q(6)=exp(−6/4)[10+∫₀⁶exp(u/4)(4+2exp(−u²/5))du]≈15.910674281 milligrams. Integrating R alone would ignore removal of previously introduced tracer.
Numerically, Q(2)≈14.759532935 and Q(3)≈15.542882664, so continuity gives a crossing in (2,3). Solving gives t≈2.218381604 minutes. Whenever Q=15, the differential equation gives Q′=R−15/4≥4−15/4=1/4>0, so every crossing is upward. More explicitly, W=exp(t/4)(Q−15) has derivative exp(t/4)(R−15/4)>0; hence W is strictly increasing and has at most one zero. This proves uniqueness and therefore that the computed crossing is the first.
Q′(0)=R(0)−10/4=6−2.5=3.5 mg/min. Euler gives 10+1·3.5=13.5 mg. Direct evaluation gives Q(1)≈12.978711960 mg, so this particular Euler estimate is an overestimate by approximately 0.521288040 mg. This comparison does not assert that Euler always overestimates.
Review focus: Accumulation with exponential weighting, FTC and numerical differential equations. Unit 7: Differential equations.
4 questions · suggested time 60 minutes. Each question has 9 practice-rubric points.
NO CALCULATOR. Questions 3–6 share a 60-minute section. Consider the polar curve r=2+sinθ for 0≤θ≤2π, with θ in radians. Its radius is always positive. Use Cartesian coordinates x=r cosθ and y=r sinθ when finding tangents.
Find the point and tangent slope at θ=π/6. Show both coordinate derivatives.
Find the area lying outside the circle r=2 and inside the given curve in the upper half-plane. Justify the angular interval and radial order.
Use the supplied Cartesian parameterization x(θ)=(2+sinθ)cosθ and y(θ)=(2+sinθ)sinθ to set up, but do not evaluate, the arc length for 0≤θ≤π. Start with the parametric arc-length formula and simplify the integrand.
Find all points with horizontal tangents during 0≤θ<2π, checking regularity.
After your attempt: selected-criterion repair, changed case and coverage boundaries. This optional worksheet adds no form points.
r=5/2 and r′=√3/2, so the point is (5√3/4,5/4). x′=r′cosθ−r sinθ=3/4−5/4=−1/2; y′=r′sinθ+r cosθ=√3/4+5√3/4=3√3/2. The slope is −3√3.
In the upper half-plane, 0≤θ≤π and sinθ≥0, so 2≤r≤2+sinθ describes the requested region once. Area=(1/2)∫₀π[(2+sinθ)²−4]dθ=(1/2)∫₀π(4sinθ+sin²θ)dθ=4+π/4.
Parametric length is ∫₀π√[(dx/dθ)²+(dy/dθ)²]dθ. With r=2+sinθ and r′=cosθ, the squared speed is (r′cosθ−r sinθ)²+(r′sinθ+r cosθ)². The cross terms cancel, leaving (r′)²+r²=cos²θ+(2+sinθ)²=5+4sinθ. Thus length is ∫₀π√(5+4sinθ)dθ. This follows from the Cartesian derivatives; a separate memorized polar-length formula is unnecessary.
y′=r′sinθ+r cosθ=2cosθ(1+sinθ), which vanishes at θ=π/2 and 3π/2. The points are (0,3) and (0,−1). At these angles x′ is respectively −3 and 1, so both are regular horizontal tangents. The factor 1+sinθ=0 adds no new angle beyond 3π/2.
Review focus: Polar derivatives, geometric region selection and arc length. Unit 9: Parametric, polar and vector-valued functions.
NO CALCULATOR. Let A(x)=Σ from n=1 to infinity of (−1)^(n+1)(x−2)^n/(n·3^n) wherever this power series converges. Inside its open interval of convergence, termwise differentiation is valid.
Find the radius and interval of convergence. Classify convergence at each endpoint.
Find A′(x) and an elementary expression for A(x) on (−1,5).
Use the first three nonzero terms to approximate A(5/2), and give an alternating-series error bound.
Independently obtain a Lagrange remainder bound for the same degree-three Taylor approximation using the formula from part (b). Explain why a derivative bound must hold over an interval.
After your attempt: selected-criterion repair, changed case and coverage boundaries. This optional worksheet adds no form points.
The absolute-term ratio tends to |x−2|/3, so the radius is 3 and the open interval is (−1,5). At x=−1, every term is −1/n, so the series diverges. At x=5 it is the alternating harmonic series, which converges conditionally. The interval is (−1,5].
Differentiation gives A′=(1/3)Σ from k=0 to infinity of [−(x−2)/3]^k=1/(x+1). Since A(2)=0, integration yields A(x)=ln(x+1)−ln3=ln((x+1)/3) on (−1,5). No conclusion about endpoints is needed for this interior identification.
At x=5/2, the ratio input is (x−2)/3=1/6. The approximation is 1/6−1/72+1/648=25/162. The positive term magnitudes (1/6)^n/n decrease to zero, so the error is at most the next magnitude (1/6)^4/4=1/5184.
A⁽⁴⁾(x)=−6/(x+1)^4. On [2,5/2], its absolute value is at most 6/3^4=2/27. The bound is (2/27)(1/2)^4/4!=1/5184. A single value of the fourth derivative would not generally control the unknown intermediate point in Taylor’s theorem; here monotonicity of 6/(x+1)^4 gives an interval-wide bound.
Review focus: Power-series endpoints, Taylor construction and two remainder arguments. Unit 10: Infinite sequences and series.
NO CALCULATOR. A population model satisfies N′=N(6−N)/3 with N(0)=2 and t≥0. Population is measured in thousands and time in days. Treat this as a stipulated mathematical model.
Use Euler’s method with two steps of width 1/2 to approximate N(1). Show the slope at each numerical point.
Solve the initial-value problem by separation, including the integration constant.
Find the time when N changes concavity, and justify the change.
Evaluate the improper integral ∫₀∞[6−N(t)]dt, showing its limiting interpretation.
After your attempt: selected-criterion repair, changed case and coverage boundaries. This optional worksheet adds no form points.
Starting slope is 2·4/3=8/3, so N₁=2+(1/2)(8/3)=10/3. The next slope is (10/3)(8/3)/3=80/27, so N₂=10/3+(1/2)(80/27)=130/27.
For the branch 0<N<6, separation gives 3dN/[N(6−N)]=dt. Since 3/[N(6−N)]=(1/2)[1/N+1/(6−N)], integration gives (1/2)ln[N/(6−N)]=t+C. Initially N/(6−N)=1/2, so N/(6−N)=(1/2)e^(2t). Therefore N(t)=6/(1+2e^(−2t)), which remains between 0 and 6 and satisfies the initial condition.
N″=(2−2N/3)N′. Along this solution N′>0 and N increases from 2 toward 6. Thus concavity changes from up to down at N=3. Solving 3=6/(1+2e^(−2t)) gives t=(ln2)/2.
6−N=12e^(−2t)/(1+2e^(−2t)). An antiderivative is −3ln(1+2e^(−2t)). Therefore ∫₀ᵀ(6−N)dt=3ln3−3ln(1+2e^(−2T)); as T→∞ this tends to 3ln3 thousand-population-days.
Review focus: Logistic separation, Euler steps, inflection and improper accumulation. Unit 7: Differential equations.
NO CALCULATOR. Investigate the following integrals using exact methods. For parts (b)–(d), let f(x)=1/[(x+1)(x+2)].
Evaluate ∫₀¹x ln(1+x)dx by integration by parts. Show the algebra needed for the remaining rational integral.
Evaluate ∫₀¹f(x)dx by partial fractions.
Determine whether ∫₁∞f(x)dx converges, and find its value if it does.
Explain why there is c in (0,1) with f(c)=ln(4/3). Is that c unique? You do not need to solve for it.
After your attempt: selected-criterion repair, changed case and coverage boundaries. This optional worksheet adds no form points.
Take u=ln(1+x), dv=x dx. The integral becomes (1/2)ln2−(1/2)∫₀¹x²/(1+x)dx. Division gives x²/(1+x)=x−1+1/(1+x). Its integral is 1/2−1+ln2=ln2−1/2. Therefore the original integral is 1/4.
f(x)=1/(x+1)−1/(x+2). The integral is [ln(x+1)−ln(x+2)]₀¹=ln(4/3). All logarithm arguments are positive on this interval.
For finite T>1, ∫₁ᵀf=ln((T+1)/(T+2))−ln(2/3). The first term tends to zero as T→∞, so the integral converges to ln(3/2). The individual divergent logarithm terms must be combined before taking their difference at infinity.
The continuous function f has average value ∫₀¹f/(1−0)=ln(4/3). The integral mean value theorem gives an interior point with this value; alternatively apply the ordinary mean value theorem to an antiderivative. Also f′=−(2x+3)/[(x+1)²(x+2)²]<0 on [0,1], so f is strictly decreasing and the point is unique. The average lies strictly between f(1) and f(0), so it cannot occur only at an endpoint.
Review focus: Integration by parts, partial fractions and integral mean value. Unit 6: Integration and accumulation.
Each free-response rubric entry is worth one point. Accept equivalent exact expressions. For calculator results, accept values within 0.001 of the stated reference unless an exact value is requested; retain unrounded intermediate values. This tolerance is a convention for this practice form, not a universal official scoring rule.
Award one point for each correct multiple-choice answer and zero for an incorrect or omitted answer. For free response, use the stated criteria rather than rewarding a matching final number alone. Accept equivalent valid mathematics. Unless a criterion explicitly depends on a correct earlier value, award later method points for mathematically consistent carry-forward work; do not award a correctness point for a wrong value. Record the specific error rather than subtracting the same point twice.
Keep the two raw totals separate: multiple choice out of 42 and free response out of 54. An optional practice-only weighted percentage is 50×(MC/42)+50×(FR/54). This is not an AP 1–5 conversion or a passing threshold. An independently reviewed explanation is stronger evidence than self-marking after reading a solution.
For each missed or uncertain item, record its ID, your original reasoning, the error type, a corrected argument and a later unfamiliar retry. Distinguish a method-selection gap from algebra, graph reading, missing justification, calculator operation or premature rounding. Repeating this exposed form does not create a fresh independent test.
Each multiple-choice item has one primary unit label. Questions can use several ideas; this table is not proof that every learning objective is tested.
| Primary unit | Questions |
|---|---|
| 1. Limits and continuity | 3 |
| 2. Derivative definitions and rules | 3 |
| 3. Composite, implicit and inverse derivatives | 3 |
| 4. Contextual differentiation | 3 |
| 5. Analytical differentiation | 5 |
| 6. Integration and accumulation | 7 |
| 7. Differential equations | 3 |
| 8. Applications of integration | 3 |
| 9. Parametric, polar and vector-valued functions | 5 |
| 10. Infinite sequences and series | 7 |
Return to graphs, tables and accumulation, derivative reasoning, integration, calculus synthesis or differential equations. The full mathematics curriculum and additional examination forms remain in development.
For BC-specific review, revisit sequences, series and Taylor approximation and parametric and polar synthesis.
Also available: the original Calculus AB practice form.
Format checked 25 October 2026 against the official BC exam overview; unit organization checked against the official BC course overview. For the actual digital environment and device rules, use the official Bluebook guidance and calculator policy.