AP Calculus AB

AP Calculus AB Practice Test: Practice Test 3

Practice Test 3 for AP Calculus AB: real questions and explanations from the Varsity Tutors practice-test pool.

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Question 1 of 25

A particle’s velocity is $v(t)=\sin t$ for $-\pi \le t \le \pi$ . What is its net displacement?

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Question 1

A particle’s velocity is $v(t)=\sin t$ for $-\pi \le t \le \pi$ . What is its net displacement?

$2$
$0$ (correct answer)
$\pi$
$-2$
$1$

Explanation: This problem uses accumulation reasoning to find displacement from sinusoidal velocity over a symmetric interval. The velocity $v(t) = \sin t$ over $[-\pi, \pi]$ creates equal positive and negative contributions due to the symmetry of sine about the origin. We integrate $\int_{-\pi}^{\pi} \sin t \, dt = [-\cos t]_{-\pi}^{\pi} = -\cos(\pi) - (-\cos(-\pi)) = -(-1) - (-(-1)) = 1 - 1 = 0.$ The sine function completes exactly one full period with equal areas above and below the axis. Students might choose 2 by computing the displacement over [0, $\pi$ ] only, but the full interval includes backward motion that cancels the forward motion. For periodic functions over symmetric intervals, positive and negative contributions often cancel completely.

Question 2

A machine’s defect count $D(t)$ rises from 14 to 20 over 3 shifts. What is the average rate of change?

2 defects/shift (correct answer)
6 defects/shift
$\tfrac{1}{2}$ shift/defect
34 defects/shift
2 shifts/defect

Explanation: This question involves average rate calculation for quality control metrics. Defects rise from 14 to 20 over 3 shifts, so the change is 20 - 14 = 6 defects over 3 shifts. The average rate is 6/3 = 2 defects/shift. Students might make arithmetic errors or forget to divide by the number of shifts. The positive result indicates increasing defect rates, which suggests declining quality control. For manufacturing applications, tracking defects per shift helps identify trends and implement corrective measures to maintain product quality.

Question 3

If $S'(x)=s(x)$ and $S(x)=\sqrt{x}+x$ , evaluate $\int_{4}^{9} s(x)\,dx$ .

$S(4)-S(9)$
$S(9)$
$S(9)-S(4)$ (correct answer)
$S(9)+S(4)$
$S(4)$

Explanation: This question applies the Fundamental Theorem of Calculus Part 2 to evaluate a definite integral using a known antiderivative relationship. Since S'(x) = s(x) and S(x) = √x + x, we compute ∫[4 to 9] s(x)dx as S(9) - S(4). Evaluating at the bounds: S(9) = √9 + 9 = 3 + 9 = 12 and S(4) = √4 + 4 = 2 + 4 = 6, so S(9) - S(4) = 12 - 6 = 6. Option A, S(4) - S(9), would give -6, which has the wrong sign due to incorrect ordering of the subtraction. Remember that FTC Part 2 always uses the pattern: antiderivative at upper bound minus antiderivative at lower bound.

Question 4

A function $f$ is continuous on $(0,4)$ but not defined at $x=0$ or $x=4$ ; which statement is true?

$f$ must have an absolute maximum and minimum by EVT.
EVT guarantees absolute extrema only on closed intervals where $f$ is continuous. (correct answer)
If $f$ is continuous, absolute extrema cannot exist.
Having endpoints missing forces a critical point to exist.
Any local extremum is automatically absolute on $(0,4)$ .

Explanation: This question examines the applicability of the Extreme Value Theorem when functions are not defined on closed intervals. The Extreme Value Theorem specifically requires both continuity and a closed, bounded interval to guarantee the existence of absolute extrema. Since $f$ is continuous on the open interval $(0,4)$ but not defined at the endpoints $x=0$ and $x=4$ , the EVT does not apply to this situation. Without the closed interval condition, we cannot guarantee that absolute extrema exist - the function might approach but never attain its supremum or infimum values. Choice A incorrectly applies EVT without the required closed interval condition. The other choices make incorrect claims about critical points or extrema behavior. The crucial EVT requirement: both continuity AND a closed interval are needed to guarantee absolute extrema exist.

Question 5

A savings account changes at rate $S'(t)=-2$ dollars/week for $0\le t\le5$ . What is the net change in savings?

$-2$ dollars
$10$ dollars
$-10$ dollars (correct answer)
$-5$ dollars
$2$ dollars

Explanation: This problem requires accumulation reasoning to find the net change in savings from a constant negative rate. The net change equals the integral of S'(t) = -2 from 0 to 5, representing the total accumulation (which is negative) over the time period. Computing ∫(-2)dt = -2t evaluated from 0 to 5 gives -2(5) - (-2(0)) = -10 dollars. The negative rate means money is being withdrawn at a constant rate of 2 dollars per week for 5 weeks. Choice B (10 dollars) might attract students who forget the negative sign, thinking of the magnitude only. When the rate of change is constant and negative, the accumulation is simply the rate times the time interval, keeping the negative sign to indicate a decrease.

Question 6

For $0<x<8$ , the profit is $P(x)= -x^2+8x+3$ ; which $x$ maximizes profit?

$x=8$
$x=0$
$x=2$
$x=4$ (correct answer)
$x=6$

Explanation: This problem requires solving optimization problems using calculus, a key skill in AP Calculus AB. To maximize the profit P(x) = -x² + 8x + 3, compute the derivative P' = -2x + 8 and set it to zero, yielding a critical point at x = 4. Evaluating at x = 4 gives the maximum profit, while at the endpoints x approaching 0 and 8, profits are lower. The second derivative P'' = -2 is negative, confirming a maximum. A tempting distractor is x = 0, but it yields a lower profit and is an endpoint minimum in this context. To solve optimization problems, express the quantity as a function of one variable, find critical points by setting the derivative to zero, and evaluate at critical points and endpoints to find the maximum or minimum.

Question 7

What is the correct volume integral setup if the base is bounded by $y=\ln x$ and $y=0$ on $[1,e]$ with rectangular cross sections perpendicular to the $x$ -axis of height $5$ ?

$\displaystyle \int_{1}^{e} 5\ln x\,dx$ (correct answer)
$\displaystyle \int_{1}^{e} 5(\ln x)^2\,dx$
$\displaystyle \int_{0}^{1} 5e^y\,dy$
$\displaystyle \int_{1}^{e} \ln(5x)\,dx$
$\displaystyle \int_{1}^{e} (5-\ln x)\,dx$

Explanation: This problem requires using the method of cross-sectional volumes to find the volume of a solid with rectangular cross sections. The base of the solid is the region bounded by y = ln x and y = 0 on [1, e]. For cross sections perpendicular to the x-axis, the width of each rectangle is the vertical distance between the curves, which is ln x. The height of each rectangle is given as 5, so the area is 5 ln x. Integrating this area function from 1 to e gives the volume as ∫ from 1 to e of 5 ln x dx. A common mistake is choice B, which squares the width as if for squares. In general, when modeling volumes with known cross sections, identify the axis of integration, determine the varying dimension of the cross section as a function of the integration variable, compute the area function, and integrate over the appropriate limits.

Question 8

A pool’s depth varies as $d(x)=3+\tfrac{x}{2}$ for $0\le x\le 6$ . What is the average depth?

$d(3)$
$\dfrac{1}{6}\int_0^6 d(x)\,dx$
$\int_0^6 d(x)\,dx$
$\dfrac{d(0)+d(6)}{2}$
$\dfrac{9}{2}$ (correct answer)

Explanation: This question asks for the average value of the depth function d(x). The average value of d(x) = 3 + x/2 on [0,6] is (1/6)∫[0 to 6] (3 + x/2)dx = (1/6)[3x + x²/4] from 0 to 6 = (1/6)(18 + 9) = 9/2. Choice B shows the correct formula setup but doesn't evaluate it. Choice A represents the midpoint value, which equals the average for this linear function but isn't the general approach. The correct answer is the numerical result E = 9/2.

Question 9

A car’s speed $v(t)$ increases at a constant rate of $3$ m/s $^2$ . Which differential equation models this?

$\dfrac{dv}{dt}=3$ (correct answer)
$\dfrac{dv}{dt}=3v$
$\dfrac{dv}{dt}=\dfrac{3}{v}$
$\dfrac{dt}{dv}=3$
$\dfrac{dv}{dt}=3t$

Explanation: This problem models constant acceleration in physics. A "constant rate of 3 m/s²" means the velocity's rate of change is constant, so dv/dt = 3. This represents uniform acceleration where velocity increases by 3 m/s every second. Choice B (dv/dt = 3v) would model exponential growth of velocity, not constant acceleration. When the rate of change is described as constant, the differential equation has a constant on the right side, not a variable-dependent term.

Question 10

Let $f(x)=\dfrac{2}{x(x-4)}$ . Which limit notation matches the behavior as $x\to 4^+$ ?

$\displaystyle \lim_{x\to 4^+} f(x)=+\infty$ (correct answer)
$\displaystyle \lim_{x\to 4^-} f(x)=+\infty$
$\displaystyle f(4)=+\infty$
$\displaystyle \lim_{x\to 4^+} f(x)=-\infty$
$\displaystyle \lim_{x\to 0^+} f(x)=+\infty$

Explanation: The function f(x) = 2/(x(x-4)), as x→4^+, x positive, (x-4) positive small, denominator positive small, 2/positive = +∞. The notation lim_{x→4^+} f(x) = +∞ is correct. This is valid for the right-hand sign. A common error is assuming -∞ or writing f(4) = +∞. Overlooking the positive product is typical. Transferable notation checklist: Identify if the limit is one-sided or two-sided; determine the sign of the expression near the asymptote; use ±∞ appropriately; avoid equating to function value; specify direction with ^+ or ^- when necessary.

Question 11

The graph of a function $f(x)$ has a vertical asymptote at the line $x=3$ . As $x$ approaches 3 from both the left and the right, the graph of $f(x)$ increases without bound.

Based on the description of the graph of $f(x)$ , what is $\lim_{x \to 3} f(x)$ ?

$0$
$3$
$\infty$ (correct answer)
$-\infty$

Explanation: The statement that the graph increases without bound as $x$ approaches 3 means that the function values are approaching positive infinity. Since this is the behavior from both sides, the limit is $\infty$ . While an infinite limit means the limit does not exist as a finite number, $\infty$ is the most precise description of the behavior.

Question 12

A particle’s acceleration is positive at $t=3$ and its velocity is zero at $t=3$ ; which is correct at $t=3$ ?

At rest and speeding up (correct answer)
Moving right and slowing down
Moving left and slowing down
Moving left and speeding up
Moving right and speeding up

Explanation: This straight-line motion problem involves a particle at rest with positive acceleration. Since the velocity is zero at $t = 3$ , the particle is momentarily at rest. Given that acceleration is positive at $t = 3$ , the particle will begin moving in the positive direction (right). When at rest with non-zero acceleration, the particle is effectively "speeding up" from zero velocity. Students might think about current motion, but the particle is transitioning from rest to motion. The key motion analysis strategy is that being at rest with non-zero acceleration means accelerating away from rest in the direction of the acceleration.

Question 13

For $f$ with $f'(x)=(x+2)(x-1)^2$ , on which interval(s) is $f$ increasing?

$(-2,1)$
$(-\infty,-2)$
$(-\infty,-2)\cup(-2,\infty)$
$(1,\infty)$
$(-2,\infty)$ (correct answer)

Explanation: This problem tests your ability to determine where a function is increasing by analyzing the sign of its derivative. A function f is increasing on intervals where f'(x) > 0, so we need to find where (x+2)(x-1)² > 0. The critical points occur where f'(x) = 0, which gives us x = -2 and x = 1. Testing intervals: for x < -2, both factors are negative so f'(x) > 0; for -2 < x < 1, (x+2) > 0 and (x-1)² > 0 so f'(x) > 0; for x > 1, both factors are positive so f'(x) > 0. Many students incorrectly choose option A, thinking the function changes behavior at x = 1, but (x-1)² is always non-negative and equals zero only at x = 1. The key insight is that squared factors don't change the sign of the derivative when crossing their zeros, so f is increasing on (-2, ∞).

Question 14

A bacterial culture increases by $1210^6$ cells in $3$ hours; what is the average growth rate?

$410^6$ cells per hour (correct answer)
$3610^6$ cells per hour
$1210^6$ hours per cell
$3$ cells per million hours
$1210^6$ cells after $1$ hour

Explanation: This problem tests your understanding of average rates of change in biological contexts. To find the average growth rate, divide the total change by the time interval: (12×10^6 cells)/(3 hours) = 4×10^6 cells per hour. A common error is to multiply instead of divide, getting 36×10^6, or to confuse the units by inverting the fraction. The key is recognizing that rate means change per unit time, requiring division of the total change by the time elapsed. When calculating average rates, always structure your calculation as (change in quantity)/(change in time) to ensure correct units and interpretation.

Question 15

A function satisfies $\dfrac{dy}{dx}=x y$ . What is the general solution for $y$ ?

$y=\dfrac{x^{2}}{2}+C$
$\ln|y|=\dfrac{x^{2}}{2}$
$y=Ce^{x^{2}/2}$ (correct answer)
$y=e^{x^{2}/2}+C$
$y=C\dfrac{x^{2}}{2}$

Explanation: This problem requires solving a differential equation using separation of variables. For dy/dx = x y, separate as dy/y = x dx, assuming y ≠ 0. Integrate: ln|y| = (x^2)/2 + C. Exponentiate to y = C e^{x^2/2}. The distractor y = e^{x^2/2} + C fails by adding C outside, not as a multiplier. To recognize separable equations, look for the form dy/dx = f(x) g(y), which allows rewriting as (1/g(y)) dy = f(x) dx for integration.

Question 16

For $f(x)=\cos(x^2)$ , what is the second derivative $f''(x)$ ?

$f''(x)=-2 sin(x^2)-4x^2 cos(x^2)$ (correct answer)
$f''(x)=2 sin(x^2)-4x^2 cos(x^2)$
$f''(x)=-2x sin(x^2)$
$f''(x)=4x^2 sin(x^2)$
$f''(x)=-4x^2 sin(x^2)$

Explanation: Higher-order derivatives of ( f(x) = cos(x^2) ) require repeated chain rule use, similar to trigonometric composites. First derivative: ( f'(x) = -sin(x^2) cdot 2x ), second: ( f''(x) = -cos(x^2) cdot 2x cdot 2x - sin(x^2) cdot 2 = -4x^2 cos(x^2) - 2 sin(x^2) ), which is ( -2sin(x^2) - 4x^2 cos(x^2) ), matching A. Each step builds on the previous by differentiating products. A typical mistake is reversing signs in the chain rule, leading to positive terms instead. Verify by computing numerically at a point if possible. Practice by breaking down composites into inner functions for reliable higher-order results.

Question 17

What is the correct volume integral setup if the base is bounded by $y=4-x^2$ and $y=0$ on $[-2,2]$ with rectangular cross sections perpendicular to the $x$ -axis of height $3$ ?

$\displaystyle \int_{-2}^{2} 3(4-x^2)\,dx$ (correct answer)
$\displaystyle \int_{-2}^{2} 3(4-x^2)^2\,dx$
$\displaystyle \int_{0}^{4} 3(4-y)\,dy$
$\displaystyle \int_{-2}^{2} (4-x^2)\,dx$
$\displaystyle \int_{-2}^{2} (3-x^2)\,dx$

Explanation: This problem requires using the method of cross-sectional volumes to find the volume of a solid with rectangular cross sections. The base of the solid is the region bounded by y = 4 - x² and y = 0 on [-2, 2]. For cross sections perpendicular to the x-axis, the width of each rectangle is the vertical distance between the curves, which is 4 - x². The height of each rectangle is given as 3, so the area is 3(4 - x²). Integrating this area function from -2 to 2 gives the volume as ∫ from -2 to 2 of 3(4 - x²) dx. A tempting distractor is choice B, which squares the width incorrectly for rectangles. In general, when modeling volumes with known cross sections, identify the axis of integration, determine the varying dimension of the cross section as a function of the integration variable, compute the area function, and integrate over the appropriate limits.

Question 18

Let $f(x) = \frac{x^4}{4} - \frac{x^2}{2} + 5$ . For which value of $x$ does $f'(x) = 0$ ?

$x = 1$ (correct answer)
$x = 2$
$x = 4$
$x = 5$

Explanation: First, find the derivative of $f(x)$ : $f'(x) = \frac{1}{4}(4x^3) - \frac{1}{2}(2x) + 0 = x^3 - x$ . Now, set the derivative equal to zero: $x^3 - x = 0$ . Factor out an $x$ : $x(x^2 - 1) = 0$ . This gives solutions $x=0$ and $x^2=1$ , so $x = -1, 0, 1$ . From the given choices, $x=1$ is a correct answer.

Question 19

If $f(x) = 4\sin(x) - e^x + 1$ , what is the value of $f'(0)$ ?

$3$ (correct answer)
$4$
$-1$
$0$

Explanation: First, find the derivative of the function: $f'(x) = rac{d}{dx}(4\sin(x) - e^x + 1) = 4\cos(x) - e^x$ . Next, evaluate the derivative at $x=0$ : $f'(0) = 4\cos(0) - e^0 = 4(1) - 1 = 3$ .

Question 20

For $\lambda(x)=\begin{cases}1,&x\le2\\3,&x>2\end{cases}$ , what type of discontinuity does $\lambda$ have at $x=2$ ?

Removable discontinuity
Jump discontinuity (correct answer)
Infinite discontinuity
No discontinuity (continuous)
Corner discontinuity

Explanation: This function has a jump discontinuity at x = 2. The left-hand limit is $\lim_{x \to 2^{-}} 1 = 1$ , while the right-hand limit is $\lim_{x \to 2^{+}} 3 = 3$ , and $\lambda(2) = 1$ . Since both one-sided limits exist but are unequal ( $1 \neq 3$ ), this creates a jump discontinuity. The function jumps from 1 on the left to 3 on the right at x = 2. Students might think this is continuous because $\lambda(2)$ equals the left-hand limit, but continuity requires both one-sided limits to be equal. To identify jump discontinuities: verify that both one-sided limits exist but have different values.

Question 21

A company’s revenue is $R(t)=2000+300t$ dollars after $t$ weeks; what is the weekly revenue rate?

$2000$ dollars per week
$300$ dollars per week (correct answer)
$2300$ dollars total after $1$ week
$500$ dollars per week
$300$ weeks per dollar

Explanation: This question requires interpreting the rate of change from a linear function in an applied business context. In the revenue function R(t) = 2000 + 300t, the coefficient of t (which is 300) represents the rate of change of revenue with respect to time in weeks. This means revenue increases by $300 per week. Students often confuse the constant term (2000) with the rate or mistakenly add both terms together. The constant 2000 represents the initial revenue at t = 0, not the rate of change. To find rates in linear functions of the form f(t) = a + bt, always identify the coefficient b of the variable as the rate of change.

Question 22

A continuous function $W$ satisfies $W(-4)=\!-2$ and $W(-1)=6$ . Which must exist?

A value $c\in(-4,-1)$ such that $W(c)=0$ . (correct answer)
A value $c\in(-4,-1)$ such that $W(c)=7$ .
A value $c\in(-4,-1)$ such that $W(c)=-3$ .
A value $c\in(-1,2)$ such that $W(c)=0$ .
No conclusion can be drawn without knowing $W$ is differentiable.

Explanation: The Intermediate Value Theorem (IVT) guarantees W continuous with W(-4) = -2 and W(-1) = 6 attains all values between -2 and 6 in (-4,-1). 0 is between them, making choice A correct. Choice B is wrong as 7 > 6, and choice C because -3 < -2. A common error is requiring differentiability for IVT, but continuity is enough, refuting choice E. Wrong intervals invalidate choice D. Transferable IVT checklist: 1. Confirm continuity on [a,b]. 2. Verify k is between f(a) and f(b). 3. Ensure c is in (a,b).

Question 23

What is the fundamental algebraic obstacle that makes it impossible to directly compute an instantaneous rate of change without using the concept of a limit?

The function itself may be undefined at the point of interest, preventing any meaningful calculation.
The change in the independent variable for an instant is zero, leading to division by zero in the rate formula. (correct answer)
The rates of change are often irrational numbers, which algebra can only approximate but not find exactly.
The change in the dependent variable might be zero, which would incorrectly imply a constant rate.

Explanation: The expression for an average rate of change, $\frac{\Delta y}{\Delta x}$ , becomes undefined when applied to a single instant because the interval length, $\Delta x$ , is zero. This division-by-zero problem is the central algebraic hurdle that the limiting process of calculus is designed to overcome.

Question 24

A control system uses $h(x)=\arctan(5x+4)$ . What is $h'(x)$ ?

$\dfrac{5}{1+(5x+4)}$
$\dfrac{1}{1+(5x+4)^2}$
$5\sec^2(5x+4)$
$\dfrac{5}{1+(5x+4)^2}$ (correct answer)
$-\dfrac{5}{1+(5x+4)^2}$

Explanation: This problem tests the skill of differentiating inverse trigonometric functions, specifically the arctan function. The formula for the derivative of arctan(u) is 1 over 1 plus u squared, multiplied by u'. Here, u is 5x+4, so u' is 5, yielding 5 over 1 plus (5x+4) squared after chain rule application. This shows how the derivative measures sensitivity to changes in the linear argument. Choice A errs by placing the 5 in the numerator without squaring the denominator term correctly. Recognize arctan patterns by ensuring the chain rule multiplier matches the coefficient of x in the argument.

Question 25

Define $F(x)=\int_{-2}^{x}\sqrt{1+t^4}\,dt$ . What is $F'(x)$ ?

$\int_{-2}^{x}\sqrt{1+t^4}\,dt$
$\sqrt{1+x^4}$ (correct answer)
$\dfrac{2x^3}{\sqrt{1+x^4}}$
$\sqrt{1+t^4}$
$\left[\int \sqrt{1+t^4}\,dt\right]_{-2}^{x}$

Explanation: This problem requires applying the Fundamental Theorem of Calculus Part 1 to find the derivative of an accumulation function. The function F(x) is defined as the integral from -2 to x of √(1 + t⁴) dt, which accumulates the area under the curve of f(t) = √(1 + t⁴) from a fixed lower limit to the variable upper limit x. According to FTC1, the derivative F'(x) equals the integrand evaluated at x, which is √(1 + x⁴). This result holds because the instantaneous rate of change of the accumulated integral at x is precisely the value of the function being integrated at that point. A tempting distractor is choice E, which represents the integral of √(1 + t⁴) dt evaluated from -2 to x, but that actually computes F(x) itself, not its derivative. To recognize FTC Part 1 problems, look for functions defined as definite integrals with a variable upper limit and a fixed lower limit, and questions asking for the derivative of that function.

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