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AP Precalculus

AP Precalculus Practice Test: Practice Test 3

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

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Question 1 of 25

A sinusoidal function f(x)=asin⁡(bx)+df(x) = a\sin(bx)+df(x)=asin(bx)+d has a maximum value of 10 and a minimum value of -2. What is the amplitude of the function?

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Question 1

A sinusoidal function f(x)=asin⁡(bx)+df(x) = a\sin(bx)+df(x)=asin(bx)+d has a maximum value of 10 and a minimum value of -2. What is the amplitude of the function?

  1. 121212
  2. 888
  3. 666 (correct answer)
  4. 444

Explanation: The amplitude of a sinusoidal function is half the difference between its maximum and minimum values. Amplitude = Maximum Value−Minimum Value2\frac{\text{Maximum Value} - \text{Minimum Value}}{2}2Maximum Value−Minimum Value​. For this function, the amplitude is 10−(−2)2=122=6\frac{10 - (-2)}{2} = \frac{12}{2} = 6210−(−2)​=212​=6.

Question 2

A radioactive isotope has half-life 888 days and initial mass 505050 mg, modeled by A(t)=50(12)t/8A(t)=50\left(\tfrac12\right)^{t/8}A(t)=50(21​)t/8. Determine the time when A(t)=12.5A(t)=12.5A(t)=12.5 mg.

  1. t=16t=16t=16 days (correct answer)
  2. t=8t=8t=8 days
  3. t=24t=24t=24 days
  4. t=4t=4t=4 days

Explanation: This question tests AP Precalculus skills in solving exponential and logarithmic equations and inequalities. Exponential equations involve variables in the exponent, while logarithmic equations involve finding the power to which a number must be raised to obtain another number. Understanding properties of logs and exponentials is crucial. In this problem, the given scenario involves radioactive decay modeled by A(t)=50(1/2)^(t/8), requiring solving the equation 12.5=50(1/2)^(t/8). Choice A is correct because it accurately applies the half-life concept: 12.5/50=0.25=(1/2)^2, so (1/2)^(t/8)=(1/2)^2, giving t/8=2, thus t=16 days. Choice B is incorrect because it results from confusing the half-life period with the answer, thinking one half-life gives the desired amount. To help students: Emphasize that half-life means the time for half the substance to remain. Encourage recognizing when amounts are powers of 1/2 of the initial value for simpler calculations.

Question 3

In an input-output model, A=(0.20.10.30.2)A=\begin{pmatrix}0.2&0.1\\0.3&0.2\end{pmatrix}A=(0.20.3​0.10.2​) and demand d⃗=(5040)\vec d=\begin{pmatrix}50\\40\end{pmatrix}d=(5040​); what is Ad⃗A\vec dAd?

  1. (1423)\begin{pmatrix}14\\23\end{pmatrix}(1423​) (correct answer)
  2. (1323)\begin{pmatrix}13\\23\end{pmatrix}(1323​)
  3. (1422)\begin{pmatrix}14\\22\end{pmatrix}(1422​)
  4. (2314)\begin{pmatrix}23\\14\end{pmatrix}(2314​)

Explanation: This question tests AP Precalculus skills, specifically understanding matrix multiplication in economic input-output models. In these models, the matrix A represents production coefficients showing how much of each sector's output is needed per unit of the other sector's production. The calculation Ad represents intermediate demand when final demand is d. Choice A is correct because Ad = (0.2×50 + 0.1×40, 0.3×50 + 0.2×40) = (10+4, 15+8) = (14, 23), showing how much of each sector's output is consumed internally. Choice B has an arithmetic error in the first component, computing 13 instead of 14. To help students: Break down matrix multiplication into dot products of rows with the vector, and interpret results in the economic context. Practice with decimal coefficients to build computational accuracy.

Question 4

The zeros of the function f(x)=tan⁡(x)f(x) = \tan(x)f(x)=tan(x) occur at which values of xxx?

  1. At x=kπx = k\pix=kπ for any integer kkk. (correct answer)
  2. At x=π2+kπx = \frac{\pi}{2} + k\pix=2π​+kπ for any integer kkk.
  3. At x=π4+kπ2x = \frac{\pi}{4} + \frac{k\pi}{2}x=4π​+2kπ​ for any integer kkk.
  4. At x=2kπx = 2k\pix=2kπ for any integer kkk.

Explanation: The zeros of tan⁡(x)=sin⁡(x)cos⁡(x)\tan(x) = \frac{\sin(x)}{\cos(x)}tan(x)=cos(x)sin(x)​ occur when the numerator, sin⁡(x)\sin(x)sin(x), is equal to 0, provided the denominator is not also 0. The function sin⁡(x)\sin(x)sin(x) is zero at all integer multiples of π\piπ. At these values, cos⁡(x)\cos(x)cos(x) is either 1 or -1, so the denominator is not zero.

Question 5

Which of the following values is the greatest?

  1. arcsin⁡(0.5)\arcsin(0.5)arcsin(0.5)
  2. arccos⁡(0.5)\arccos(0.5)arccos(0.5) (correct answer)
  3. arctan⁡(1)\arctan(1)arctan(1)
  4. arcsin⁡(−1)\arcsin(-1)arcsin(−1)

Explanation: We evaluate each expression: A) arcsin⁡(0.5)=π6\arcsin(0.5) = \frac{\pi}{6}arcsin(0.5)=6π​. B) arccos⁡(0.5)=π3\arccos(0.5) = \frac{\pi}{3}arccos(0.5)=3π​. C) arctan⁡(1)=π4\arctan(1) = \frac{\pi}{4}arctan(1)=4π​. D) arcsin⁡(−1)=−π2\arcsin(-1) = -\frac{\pi}{2}arcsin(−1)=−2π​. Comparing the values, π6≈0.524\frac{\pi}{6} \approx 0.5246π​≈0.524, π3≈1.047\frac{\pi}{3} \approx 1.0473π​≈1.047, π4≈0.785\frac{\pi}{4} \approx 0.7854π​≈0.785, and −π2≈−1.571-\frac{\pi}{2} \approx -1.571−2π​≈−1.571. The greatest value is π3\frac{\pi}{3}3π​.

Question 6

A sound sensor models intensity by g(x)=10xg(x)=10^xg(x)=10x and converts to decibels with f(I)=10log⁡10(I)f(I)=10\log_{10}(I)f(I)=10log10​(I). Based on the scenario, what is f(g(3))f(g(3))f(g(3))?

  1. 303030 (correct answer)
  2. 300300300
  3. 10log⁡10(3)10\log_{10}(3)10log10​(3)
  4. 10310^3103

Explanation: This question tests AP Precalculus skills: Composition of Functions, specifically involving exponential functions and logarithmic decibel conversion. Composition requires applying one function to the output of another, where g(x) = 10^x models intensity and f(I) = 10log₁₀(I) converts to decibels. In this scenario, we need to find f(g(3)), which means first calculating g(3) = 10³ = 1000, then applying f to get the decibel level. Choice A is correct because f(g(3)) = f(1000) = 10log₁₀(1000) = 10·3 = 30 decibels. Choice D is incorrect because it gives g(3) = 1000 instead of f(g(3)), confusing the intensity value with the decibel measurement. To help students: Draw diagrams showing the flow from input through each function, practice with the decibel formula, and emphasize that log₁₀(10^n) = n is a key logarithm property.

Question 7

A startup’s user base grows according to f(t)=150⋅4tf(t)=150\cdot 4^tf(t)=150⋅4t, where ttt is weeks. The product team uses the inverse to find when users reach a benchmark. Here, f(t)f(t)f(t) is the number of users after ttt weeks and 150 is the initial number. Assume the weekly growth factor remains 4. Using the inverse, solve for ttt when f(t)=Uf(t)=Uf(t)=U.

  1. f−1(U)=log⁡4 ⁣(U150)f^{-1}(U)=\log_4\!\left(\dfrac{U}{150}\right)f−1(U)=log4​(150U​) (correct answer)
  2. f−1(U)=log⁡150(4U)f^{-1}(U)=\log_{150}(4U)f−1(U)=log150​(4U)
  3. f−1(U)=log⁡4(U)+150f^{-1}(U)=\log_4(U)+150f−1(U)=log4​(U)+150
  4. f−1(U)=U150f^{-1}(U)=\dfrac{U}{150}f−1(U)=150U​

Explanation: This question tests understanding of inverses of exponential functions in AP Precalculus, focusing on converting exponential forms to their inverses. Finding the inverse of an exponential function involves expressing the independent variable in terms of the dependent variable, using logarithms to solve for the original input variable. In this problem, the function f(t) = 150·4^t models user growth, and students must find f^(-1)(U) to determine time t given user count U. Choice A is correct because starting with U = 150·4^t, dividing by 150 gives U/150 = 4^t, then applying log base 4 yields t = log₄(U/150). Choice C is incorrect because it adds 150 outside the logarithm instead of dividing inside, misunderstanding how to properly isolate the exponential expression. To help students: Emphasize the systematic approach to finding inverses of exponential functions. Practice recognizing that the logarithm base must match the base of the original exponential function.

Question 8

A rational function is a quotient of polynomials, f(x)=p(x)q(x)f(x)=\frac{p(x)}{q(x)}f(x)=q(x)p(x)​. Zeros are the xxx-values where f(x)=0f(x)=0f(x)=0, so p(x)=0p(x)=0p(x)=0 and q(x)≠0q(x)\neq 0q(x)=0. Vertical asymptotes occur at real solutions of q(x)=0q(x)=0q(x)=0 that do not cancel with the numerator. Consider the rational function f(x)=x2+1x2−1.f(x)=\frac{x^2+1}{x^2-1}.f(x)=x2−1x2+1​. Here, x2−1=(x−1)(x+1)x^2-1=(x-1)(x+1)x2−1=(x−1)(x+1), so the function is undefined at x=±1x=\pm 1x=±1, producing vertical asymptotes at x=1x=1x=1 and x=−1x=-1x=−1 because nothing cancels. The numerator x2+1=0x^2+1=0x2+1=0 has solutions x=±ix=\pm ix=±i, which are complex zeros; this means there are no real xxx-intercepts.

Using the function provided, determine the vertical asymptotes of f(x)=x2+1x2−1f(x)=\frac{x^2+1}{x^2-1}f(x)=x2−1x2+1​.

  1. x=0x=0x=0 only
  2. x=1x=1x=1 only
  3. x=±1x=\pm 1x=±1 (correct answer)
  4. x=±ix=\pm ix=±i

Explanation: This question tests understanding of vertical asymptotes in rational functions where no factor cancellation occurs. Vertical asymptotes happen where the denominator equals zero and the factor doesn't cancel with the numerator. For f(x) = (x²+1)/(x²-1), the denominator x²-1 factors as (x-1)(x+1), equaling zero at x=±1. Since the numerator x²+1 has no real factors (its zeros are x=±i), nothing cancels with the denominator factors. Choice C is correct because vertical asymptotes occur at both x=1 and x=-1 where the denominator equals zero. Choice D is incorrect because x=±i are the complex zeros of the function, not vertical asymptotes. Students should practice factoring both numerator and denominator to identify which factors cancel and which create vertical asymptotes.

Question 9

A curve is parameterized by x(t)=2sin⁡(t)+3x(t) = 2\sin(t) + 3x(t)=2sin(t)+3 and y(t)=cos⁡2(t)y(t) = \cos^2(t)y(t)=cos2(t) for all real ttt. What is the range of possible y-values for this curve?

  1. [0,1][0, 1][0,1] (correct answer)
  2. [−1,1][-1, 1][−1,1]
  3. [1,5][1, 5][1,5]
  4. [0,4][0, 4][0,4]

Explanation: The y-values of the curve are given by the function y(t)=cos⁡2(t)y(t) = \cos^2(t)y(t)=cos2(t). The range of the function f(t)=cos⁡(t)f(t)=\cos(t)f(t)=cos(t) is [−1,1][-1, 1][−1,1]. When we square the values in this range, the outputs are always non-negative. The smallest possible value is 02=00^2 = 002=0, which occurs when cos⁡(t)=0\cos(t)=0cos(t)=0. The largest possible value is (−1)2=12=1(-1)^2 = 1^2 = 1(−1)2=12=1. Therefore, the range of y(t)=cos⁡2(t)y(t) = \cos^2(t)y(t)=cos2(t) is the closed interval [0,1][0, 1][0,1].

Question 10

As an angle θ\thetaθ increases from π\piπ to 3π2\frac{3\pi}{2}23π​, what is the behavior of sin⁡θ\sin\thetasinθ and cos⁡θ\cos\thetacosθ?

  1. sin⁡θ\sin\thetasinθ decreases from 0 to -1, and cos⁡θ\cos\thetacosθ increases from -1 to 0. (correct answer)
  2. sin⁡θ\sin\thetasinθ increases from -1 to 0, and cos⁡θ\cos\thetacosθ decreases from 0 to -1.
  3. Both sin⁡θ\sin\thetasinθ and cos⁡θ\cos\thetacosθ decrease over the interval.
  4. Both sin⁡θ\sin\thetasinθ and cos⁡θ\cos\thetacosθ increase over the interval.

Explanation: This interval represents Quadrant III. At θ=π\theta=\piθ=π, the point on the unit circle is (−1,0)(-1, 0)(−1,0), so cos⁡π=−1\cos\pi=-1cosπ=−1 and sin⁡π=0\sin\pi=0sinπ=0. At θ=3π2\theta=\frac{3\pi}{2}θ=23π​, the point is (0,−1)(0, -1)(0,−1), so cos⁡(3π2)=0\cos(\frac{3\pi}{2})=0cos(23π​)=0 and sin⁡(3π2)=−1\sin(\frac{3\pi}{2})=-1sin(23π​)=−1. As θ\thetaθ goes from π\piπ to 3π2\frac{3\pi}{2}23π​, the y-coordinate (sin⁡θ\sin\thetasinθ) goes from 0 to -1 (a decrease), and the x-coordinate (cos⁡θ\cos\thetacosθ) goes from -1 to 0 (an increase).

Question 11

A lighthouse beacon is modeled by P(θ)=sin⁡θcsc⁡θP(\theta)=\dfrac{\sin\theta}{\csc\theta}P(θ)=cscθsinθ​, where θ\thetaθ is measured in radians. The maintenance log asks for a simplified expression to reduce computational error. Use exactly one reciprocal identity and keep the result in terms of sin⁡θ\sin\thetasinθ only. Do not introduce additional trigonometric functions. Using the identity, simplify the expression for P(θ)P(\theta)P(θ).​

  1. 111
  2. sin⁡2θ\sin^2\thetasin2θ (correct answer)
  3. csc⁡2θ\csc^2\thetacsc2θ
  4. sin⁡θ\sin\thetasinθ

Explanation: This question tests AP Precalculus skills in trigonometric and polar functions, specifically focusing on simplifying expressions using reciprocal identities. The reciprocal identity csc(θ) = 1/sin(θ) allows complex fractions to be simplified by converting between reciprocal pairs. In this scenario, P(θ) = sin(θ)/csc(θ) needs simplification using the cosecant reciprocal identity. Choice B is correct because it accurately applies the identity: P(θ) = sin(θ)/csc(θ) = sin(θ)/(1/sin(θ)) = sin(θ) × sin(θ) = sin²(θ). Choice A is incorrect because it would result from incorrectly canceling sin(θ) with csc(θ) as if they were reciprocals that multiply to 1, rather than understanding that division by a reciprocal means multiplication. To help students: Remember that dividing by a fraction means multiplying by its reciprocal. Practice rewriting all six trigonometric functions in terms of sine and cosine to build fluency with identities.

Question 12

A company's daily production cost is C(x)=0.01x3−0.6x2+15x+500C(x) = 0.01x^3 - 0.6x^2 + 15x + 500C(x)=0.01x3−0.6x2+15x+500 dollars for producing xxx units. If they want to minimize cost per unit, what function represents cost per unit?

  1. C(x)x=0.01x2−0.6x+15+500x\frac{C(x)}{x} = 0.01x^2 - 0.6x + 15 + \frac{500}{x}xC(x)​=0.01x2−0.6x+15+x500​ (correct answer)
  2. C(x)x=0.01x3−0.6x2+15x+500\frac{C(x)}{x} = 0.01x^3 - 0.6x^2 + 15x + 500xC(x)​=0.01x3−0.6x2+15x+500
  3. C(x)x=0.01x2−0.6x+15+500\frac{C(x)}{x} = 0.01x^2 - 0.6x + 15 + 500xC(x)​=0.01x2−0.6x+15+500
  4. C(x)x=0.01x−0.6+15x+500x\frac{C(x)}{x} = 0.01x - 0.6 + \frac{15}{x} + \frac{500}{x}xC(x)​=0.01x−0.6+x15​+x500​

Explanation: Cost per unit is C(x)x=0.01x3−0.6x2+15x+500x=0.01x2−0.6x+15+500x\frac{C(x)}{x} = \frac{0.01x^3 - 0.6x^2 + 15x + 500}{x} = 0.01x^2 - 0.6x + 15 + \frac{500}{x}xC(x)​=x0.01x3−0.6x2+15x+500​=0.01x2−0.6x+15+x500​. Choice B doesn't divide by xxx. Choice C doesn't properly handle the constant term division. Choice D incorrectly divides each coefficient by xxx.

Question 13

A factory tests a catalyst: output rises sharply at low doses, then extra catalyst adds less improvement. Model 1: Exponential, y=10(1.4)dy=10(1.4)^dy=10(1.4)d (base 1.41.41.4). Model 2: Logarithmic, y=25+12log⁡(1.2d+1)y=25+12\log(1.2d+1)y=25+12log(1.2d+1) (coefficient 121212). Exponential growth seems plausible for strong early effects, while logarithmic growth reflects saturation of available reaction sites. Based on the models described in the passage, why might the logarithmic model be more appropriate than the exponential model in this scenario?

  1. Because it models a strong initial rise that gradually tapers as saturation occurs. (correct answer)
  2. Because an exponential model with base 1.41.41.4 must eventually decrease as dose increases.
  3. Because the coefficient 12 is the exponential base, so both models are identical.
  4. Because logarithmic models always fit better whenever data are measured in factories.
  5. Because both models predict the same output once ddd is large enough.

Explanation: This question tests AP Precalculus skills: Competing Function Model Validation, specifically understanding when to apply exponential versus logarithmic models. Exponential models describe rapid growth that continues unchecked, whereas logarithmic models account for factors that limit growth, such as saturation of reaction sites in chemical processes. In the passage, the exponential model predicts 40% growth per dose unit, while the logarithmic model reflects saturation of available reaction sites as catalyst dose increases. Choice A is correct because it accurately identifies that the logarithmic model captures a strong initial rise that gradually tapers as saturation occurs, matching the scenario where extra catalyst adds less improvement. Choice B is incorrect because exponential models with base > 1 never decrease - they continue increasing indefinitely. To help students: Emphasize understanding chemical saturation concepts and diminishing returns in catalysis, practice identifying when processes face physical limits versus unlimited scaling. Encourage students to consider whether the underlying mechanism can sustain constant percentage improvements.

Question 14

Which of the following is a parametrization of the curve defined by the equation y=x3−2x+1y = x^3 - 2x + 1y=x3−2x+1?

  1. x(t)=t,y(t)=t3−2t+1x(t) = t, y(t) = t^3 - 2t + 1x(t)=t,y(t)=t3−2t+1 (correct answer)
  2. x(t)=t3−2t+1,y(t)=tx(t) = t^3 - 2t + 1, y(t) = tx(t)=t3−2t+1,y(t)=t
  3. x(t)=t+1,y(t)=t3−2tx(t) = t+1, y(t) = t^3 - 2tx(t)=t+1,y(t)=t3−2t
  4. x(t)=t,y(t)=x3−2x+1x(t) = t, y(t) = x^3 - 2x + 1x(t)=t,y(t)=x3−2x+1

Explanation: To parametrize a function of the form y=f(x)y = f(x)y=f(x), the simplest method is to set the independent variable xxx equal to a parameter ttt. So, let x(t)=tx(t) = tx(t)=t. Then, we substitute ttt for xxx in the original equation to find y(t)y(t)y(t). This gives y(t)=t3−2t+1y(t) = t^3 - 2t + 1y(t)=t3−2t+1.

Question 15

A rational function is a quotient of two polynomials. For f(x)=x2−4x−2f(x)=\frac{x^2-4}{x-2}f(x)=x−2x2−4​, zeros occur where the function equals zero, typically where the numerator is zero and the denominator is not. Vertical asymptotes occur at xxx-values that make the denominator zero and do not cancel with the numerator. Since x2−4=(x−2)(x+2)x^2-4=(x-2)(x+2)x2−4=(x−2)(x+2), the factor (x−2)(x-2)(x−2) cancels, so f(x)=x+2f(x)=x+2f(x)=x+2 for x≠2x\ne2x=2. The point x=2x=2x=2 is still excluded from the domain, creating a hole instead of a vertical asymptote. Using the function provided, determine the vertical asymptotes of f(x)=x2−4x−2f(x)=\frac{x^2-4}{x-2}f(x)=x−2x2−4​.​

  1. No vertical asymptotes (correct answer)
  2. x=2x=2x=2
  3. x=−2x=-2x=−2
  4. x=±2x=\pm2x=±2

Explanation: This question tests understanding of vertical asymptotes in rational functions and how they differ from removable discontinuities. Vertical asymptotes occur at x-values where the denominator equals zero and this zero does not cancel with the numerator. In this case, f(x) = (x²-4)/(x-2) = (x-2)(x+2)/(x-2), where the (x-2) factor cancels completely. Choice A is correct because after cancellation, there are no remaining zeros in the denominator, meaning no vertical asymptotes exist. Choice B is incorrect because while x = 2 makes the original denominator zero, this factor cancels with the numerator, creating a removable discontinuity (hole) rather than a vertical asymptote. Students should distinguish between vertical asymptotes and holes by checking whether denominator factors cancel. Practice factoring both numerator and denominator completely before identifying asymptotes.

Question 16

In a projectile model, h(t)=2t5−10t3+8th(t)=2t^5-10t^3+8th(t)=2t5−10t3+8t, which statement best describes end behavior?

  1. As t→∞t\to\inftyt→∞, h(t)→−∞h(t)\to-\inftyh(t)→−∞; as t→−∞t\to-\inftyt→−∞, h(t)→∞h(t)\to\inftyh(t)→∞.
  2. As t→∞t\to\inftyt→∞, h(t)→∞h(t)\to\inftyh(t)→∞; as t→−∞t\to-\inftyt→−∞, h(t)→−∞h(t)\to-\inftyh(t)→−∞. (correct answer)
  3. As t→±∞t\to\pm\inftyt→±∞, h(t)→∞h(t)\to\inftyh(t)→∞ because the degree is odd.
  4. End behavior depends on the zeros, so h(t)h(t)h(t) approaches 000 for large ∣t∣|t|∣t∣.

Explanation: This question tests AP Precalculus skills: understanding polynomial functions and their end behavior. Polynomial end behavior is determined by the degree and leading coefficient; higher degree means steeper growth or decline, while the sign of the leading coefficient indicates the direction. Given h(t) = 2t^5 - 10t^3 + 8t, the leading term 2t^5 dictates that as t approaches infinity, the function behaves like a positive odd-degree polynomial. Choice B is correct because with a positive leading coefficient (2) and odd degree (5), as t→∞, h(t)→∞ and as t→-∞, h(t)→-∞. Choice C is incorrect because it misinterprets odd degree behavior - odd degree polynomials have opposite end behaviors, not both ends going to infinity. To help students: Focus on identifying leading terms, use graphing tools to visualize end behavior, and practice distinguishing polynomial characteristics from other functions. Watch for misconceptions about degree and leading coefficient impacts.

Question 17

A point P(x,y)P(x,y)P(x,y) on a circle of radius 8 corresponds to an angle θ\thetaθ in standard position. If sin⁡θ=34\sin\theta = \frac{3}{4}sinθ=43​ and the terminal ray of θ\thetaθ is in Quadrant II, what are the coordinates of point PPP?

  1. (−27,6)(-2\sqrt{7}, 6)(−27​,6) (correct answer)
  2. (27,6)(2\sqrt{7}, 6)(27​,6)
  3. (6,−27)(6, -2\sqrt{7})(6,−27​)
  4. (−27,34)(-2\sqrt{7}, \frac{3}{4})(−27​,43​)

Explanation: The coordinates are (rcos⁡θ,rsin⁡θ)(r\cos\theta, r\sin\theta)(rcosθ,rsinθ). We are given r=8r=8r=8 and sin⁡θ=34\sin\theta = \frac{3}{4}sinθ=43​. The y-coordinate is y=rsin⁡θ=8(34)=6y = r\sin\theta = 8(\frac{3}{4}) = 6y=rsinθ=8(43​)=6. To find the x-coordinate, we use the identity sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1sin2θ+cos2θ=1. This gives (34)2+cos⁡2θ=1(\frac{3}{4})^2 + \cos^2\theta = 1(43​)2+cos2θ=1, so cos⁡2θ=1−916=716\cos^2\theta = 1 - \frac{9}{16} = \frac{7}{16}cos2θ=1−169​=167​. Since θ\thetaθ is in Quadrant II, cos⁡θ\cos\thetacosθ is negative, so cos⁡θ=−74\cos\theta = -\frac{\sqrt{7}}{4}cosθ=−47​​. The x-coordinate is x=rcos⁡θ=8(−74)=−27x = r\cos\theta = 8(-\frac{\sqrt{7}}{4}) = -2\sqrt{7}x=rcosθ=8(−47​​)=−27​. The coordinates of P are (−27,6)(-2\sqrt{7}, 6)(−27​,6).

Question 18

Let p(x)=x2−6xp(x) = x^2 - 6xp(x)=x2−6x. The graph of the function q(x)q(x)q(x) is obtained by vertically stretching the graph of p(x)p(x)p(x) by a factor of 3 and then reflecting it across the x-axis. Which of the following is the equation for q(x)q(x)q(x)?

  1. q(x)=−3x2+18xq(x) = -3x^2 + 18xq(x)=−3x2+18x (correct answer)
  2. q(x)=3x2−18xq(x) = 3x^2 - 18xq(x)=3x2−18x
  3. q(x)=−13x2+2xq(x) = -\frac{1}{3}x^2 + 2xq(x)=−31​x2+2x
  4. q(x)=(−3x)2−6(−3x)q(x) = (-3x)^2 - 6(-3x)q(x)=(−3x)2−6(−3x)

Explanation: A vertical stretch by a factor of 3 means multiplying the function by 3, resulting in 3p(x)=3(x2−6x)=3x2−18x3p(x) = 3(x^2 - 6x) = 3x^2 - 18x3p(x)=3(x2−6x)=3x2−18x. A reflection across the x-axis means multiplying the result by -1. Therefore, q(x)=−1(3p(x))=−(3x2−18x)=−3x2+18xq(x) = -1(3p(x)) = -(3x^2 - 18x) = -3x^2 + 18xq(x)=−1(3p(x))=−(3x2−18x)=−3x2+18x.

Question 19

Based on the scenario above, find ΔpH\Delta\text{pH}ΔpH if [H+]A=10−3[H^+]_A=10^{-3}[H+]A​=10−3 and [H+]B=10−5[H^+]_B=10^{-5}[H+]B​=10−5.

  1. ΔpH=−2\Delta\text{pH}=-2ΔpH=−2
  2. ΔpH=log⁡10(10−3)log⁡10(10−5)\Delta\text{pH}=\dfrac{\log_{10}(10^{-3})}{\log_{10}(10^{-5})}ΔpH=log10​(10−5)log10​(10−3)​
  3. ΔpH=2\Delta\text{pH}=2ΔpH=2 (correct answer)
  4. ΔpH=10−2\Delta\text{pH}=10^{-2}ΔpH=10−2

Explanation: This question tests AP Precalculus skills: understanding and applying logarithmic expressions and their properties. Logarithms are the inverse of exponential functions and have properties such as the product, quotient, and power rules which simplify expressions. In this scenario, you are required to find the change in pH given hydrogen ion concentrations, where pH = -log₁₀[H⁺]. Choice C is correct because pH_A = -log₁₀(10⁻³) = 3 and pH_B = -log₁₀(10⁻⁵) = 5, so ΔpH = pH_B - pH_A = 5 - 3 = 2. Choice A is incorrect because it calculates pH_A - pH_B = 3 - 5 = -2, reversing the order of subtraction and getting a negative change when the pH actually increases. To help students: Emphasize that lower [H⁺] means higher pH (they are inversely related due to the negative sign). Practice calculating pH from [H⁺] and understanding that a decrease in [H⁺] leads to an increase in pH.

Question 20

A ship travels 7 km east, then 9 km north, forming a right triangle with legs 7 and 9. The angle θ\thetaθ between the ship’s final displacement and the east direction satisfies tan⁡(θ)=97\tan(\theta)=\frac{9}{7}tan(θ)=79​. Find θ\thetaθ in degrees (nearest tenth).

  1. 37.9∘37.9^\circ37.9∘
  2. 52.1∘52.1^\circ52.1∘ (correct answer)
  3. 0.910 rad0.910\text{ rad}0.910 rad
  4. 127.9∘127.9^\circ127.9∘

Explanation: This question tests understanding of inverse trigonometric functions, focusing on arctan to find a direction angle from a tangent ratio. Inverse trigonometric functions find the angle whose trigonometric function gives a specific value. For example, arctan(x) gives the angle θ such that tan(θ) = x. In this question, the problem provides a navigation scenario where tan(θ) = 9/7 ≈ 1.286, requiring calculation of the angle from east using arctan. Choice B is correct because θ = arctan(9/7) ≈ 52.1°, which accurately calculates the angle between the displacement vector and the east direction. Choice A (37.9°) is incorrect because it represents the complementary angle (90° - 52.1°), a common error when students confuse the angle from east with the angle from north. Encourage students to draw displacement vectors and clearly label reference directions. Practice interpreting navigation problems where angles are measured from different cardinal directions. Watch for: confusion about which side is opposite vs adjacent to the desired angle.

Question 21

Which of the following is equivalent to sin⁡(2arcsin⁡(x))\sin(2\arcsin(x))sin(2arcsin(x)) for −1≤x≤1-1 \le x \le 1−1≤x≤1?

  1. 2x2x2x
  2. 2x22x^22x2
  3. 2x1−x22x\sqrt{1-x^2}2x1−x2​ (correct answer)
  4. 1−x2\sqrt{1-x^2}1−x2​

Explanation: Let θ=arcsin⁡(x)\theta = \arcsin(x)θ=arcsin(x), which means sin⁡(θ)=x\sin(\theta) = xsin(θ)=x. The expression becomes sin⁡(2θ)\sin(2\theta)sin(2θ). Using the double-angle identity, sin⁡(2θ)=2sin⁡(θ)cos⁡(θ)\sin(2\theta) = 2\sin(\theta)\cos(\theta)sin(2θ)=2sin(θ)cos(θ). We know sin⁡(θ)=x\sin(\theta) = xsin(θ)=x. To find cos⁡(θ)\cos(\theta)cos(θ), we use the Pythagorean identity: cos⁡(θ)=1−sin⁡2(θ)=1−x2\cos(\theta) = \sqrt{1-\sin^2(\theta)} = \sqrt{1-x^2}cos(θ)=1−sin2(θ)​=1−x2​ (cosine is positive because the range of arcsin⁡\arcsinarcsin is [−π/2,π/2][-\pi/2, \pi/2][−π/2,π/2], where cosine is non-negative). Substituting back, we get 2sin⁡(θ)cos⁡(θ)=2x1−x22\sin(\theta)\cos(\theta) = 2x\sqrt{1-x^2}2sin(θ)cos(θ)=2x1−x2​.

Question 22

The terminal ray of an angle θ=2π3\theta = \frac{2\pi}{3}θ=32π​ intersects the unit circle at a point PPP. What are the coordinates of point PPP?

  1. (−12,32)(-\frac{1}{2}, \frac{\sqrt{3}}{2})(−21​,23​​) (correct answer)
  2. (12,−32)(\frac{1}{2}, -\frac{\sqrt{3}}{2})(21​,−23​​)
  3. (−32,12)(-\frac{\sqrt{3}}{2}, \frac{1}{2})(−23​​,21​)
  4. (32,−12)(\frac{\sqrt{3}}{2}, -\frac{1}{2})(23​​,−21​)

Explanation: For a point on the unit circle, the coordinates are given by (cos⁡θ,sin⁡θ)(\cos\theta, \sin\theta)(cosθ,sinθ). The angle θ=2π3\theta = \frac{2\pi}{3}θ=32π​ is in Quadrant II, where the x-coordinate (cosine) is negative and the y-coordinate (sine) is positive. The reference angle is π3\frac{\pi}{3}3π​. Thus, cos⁡(2π3)=−cos⁡(π3)=−12\cos(\frac{2\pi}{3}) = -\cos(\frac{\pi}{3}) = -\frac{1}{2}cos(32π​)=−cos(3π​)=−21​ and sin⁡(2π3)=sin⁡(π3)=32\sin(\frac{2\pi}{3}) = \sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2}sin(32π​)=sin(3π​)=23​​. The coordinates of PPP are (−12,32)(-\frac{1}{2}, \frac{\sqrt{3}}{2})(−21​,23​​).

Question 23

The total processing power of computers in a research lab doubles every 3 years. The current total processing power is designated as P0P_0P0​.

Which of the following functions P(t)P(t)P(t) models the lab's total processing power ttt years from now?

  1. P(t)=P0(2)t/3P(t) = P_0(2)^{t/3}P(t)=P0​(2)t/3 (correct answer)
  2. P(t)=P0(2)3tP(t) = P_0(2)^{3t}P(t)=P0​(2)3t
  3. P(t)=P0(1+23t)P(t) = P_0(1 + \frac{2}{3}t)P(t)=P0​(1+32​t)
  4. P(t)=P0(3)t/2P(t) = P_0(3)^{t/2}P(t)=P0​(3)t/2

Explanation: The initial value is P0P_0P0​. The growth factor is 2, but this growth occurs over a period of 3 years. The number of 3-year periods that have passed in ttt years is t/3t/3t/3. Therefore, the growth factor of 2 should be applied t/3t/3t/3 times. The model is P(t)=P0(2)t/3P(t) = P_0(2)^{t/3}P(t)=P0​(2)t/3.

Question 24

Which of the following is a parametrization of the hyperbola given by (y+3)24−(x−1)236=1\frac{(y+3)^2}{4} - \frac{(x-1)^2}{36} = 14(y+3)2​−36(x−1)2​=1?

  1. x(t)=1+2tan⁡(t),y(t)=−3+6sec⁡(t)x(t) = 1 + 2\tan(t), y(t) = -3 + 6\sec(t)x(t)=1+2tan(t),y(t)=−3+6sec(t)
  2. x(t)=1+6sec⁡(t),y(t)=−3+2tan⁡(t)x(t) = 1 + 6\sec(t), y(t) = -3 + 2\tan(t)x(t)=1+6sec(t),y(t)=−3+2tan(t)
  3. x(t)=−1+36tan⁡(t),y(t)=3+4sec⁡(t)x(t) = -1 + 36\tan(t), y(t) = 3 + 4\sec(t)x(t)=−1+36tan(t),y(t)=3+4sec(t)
  4. x(t)=1+6tan⁡(t),y(t)=−3+2sec⁡(t)x(t) = 1 + 6\tan(t), y(t) = -3 + 2\sec(t)x(t)=1+6tan(t),y(t)=−3+2sec(t) (correct answer)

Explanation: The standard parametrization for a vertical hyperbola (y−k)2b2−(x−h)2a2=1\frac{(y-k)^2}{b^2} - \frac{(x-h)^2}{a^2} = 1b2(y−k)2​−a2(x−h)2​=1 is x(t)=h+atan⁡(t)x(t) = h + a\tan(t)x(t)=h+atan(t) and y(t)=k+bsec⁡(t)y(t) = k + b\sec(t)y(t)=k+bsec(t). Here, the center is (h,k)=(1,−3)(h,k) = (1, -3)(h,k)=(1,−3), a=36=6a = \sqrt{36} = 6a=36​=6, and b=4=2b = \sqrt{4} = 2b=4​=2. So, x(t)=1+6tan⁡(t)x(t) = 1 + 6\tan(t)x(t)=1+6tan(t) and y(t)=−3+2sec⁡(t)y(t) = -3 + 2\sec(t)y(t)=−3+2sec(t).

Question 25

The function fff is given by f(x)=2(3)xf(x) = 2(3)^xf(x)=2(3)x, and the function ggg is given by g(x)=−2(3)xg(x) = -2(3)^xg(x)=−2(3)x. Which statement best describes the relationship between the graphs of fff and ggg?

  1. The graph of ggg is a reflection of the graph of fff across the y-axis.
  2. The graph of ggg is a reflection of the graph of fff across the x-axis. (correct answer)
  3. The graph of ggg is a vertical shift of the graph of fff by -4 units.
  4. The graph of ggg is a horizontal shift of the graph of fff.

Explanation: The function g(x)g(x)g(x) can be written as g(x)=−f(x)g(x) = -f(x)g(x)=−f(x). A transformation of the form y=−f(x)y=-f(x)y=−f(x) reflects the graph of y=f(x)y=f(x)y=f(x) across the x-axis. Each y-coordinate on the graph of fff is replaced by its opposite on the graph of ggg.