2024 AP AP Precalculus Practice Paper with Answers

(A) (i) First evaluate \( g(0) \):
\[ g(0) = 3\ln(0 + 5) - 0.5(0)^2 - 1 = 3\ln(5) - 1 \approx 3.828314 \]
Since \( g(0) \approx 3.828314 \) lies on the interval \([2, 5]\), we use the linear piece of \( f \) connecting \((2, 2)\) and \((5, -4)\).
The slope of this segment is \( m = \dfrac{-4 - 2}{5 - 2} = -2 \).
The equation of the line is \( y - 2 = -2(x - 2) \implies f(x) = -2x + 6 \).
Evaluating \( f \) at \( g(0) \):
\[ f(g(0)) = -2(3.828314) + 6 = -7.656628 + 6 \approx -1.657 \]

(ii) To find all values of \( x \) for which \( g(x) = 2 \), solve:
\[ 3\ln(x + 5) - 0.5x^2 - 1 = 2 \implies 3\ln(x + 5) - 0.5x^2 - 3 = 0 \]
Using a graphing calculator to find the zeros on the domain \( x > -5 \), the solutions are:
\[ x \approx -1.337 \quad \text{and} \quad x \approx 2.461 \]

(B) (i) As \( x \to -\infty \), the leading terms of the numerator and denominator dominate:
\[ \lim_{x \to -\infty} h(x) = \lim_{x \to -\infty} \dfrac{5x^2 - 3}{2x^2 - 8} = \lim_{x \to -\infty} \dfrac{5 - \frac{3}{x^2}}{2 - \frac{8}{x^2}} = \dfrac{5}{2} = 2.5 \]

(ii) Factor the denominator: \( 2x^2 - 8 = 2(x - 2)(x + 2) \).
As \( x \to 2^+ \), the numerator approaches \( 5(2)^2 - 3 = 17 > 0 \).
The denominator approaches \( 2(0^+)(4) = 0^+ \).
Therefore:
\[ \lim_{x \to 2^+} h(x) = \infty \]

(C) (i) No, \( f \) is not invertible on \([-4, 5]\).
Reason: A function has an inverse function only if each output corresponds to exactly one input. On the graph of \( f \), there are multiple distinct input values that have the same output value; for instance, \( f(1) = 1 \) and \( f(-2) = 1 \). Thus, the inverse relation maps a single input to multiple outputs, which violates the definition of a function.

(ii) On the interval \([-1, 2]\), the line segment passes through \((-1, -1)\) and \((2, 2)\), so its equation is \( f(x) = x \).
To find \( f^{-1}(0) \), solve \( f(x) = 0 \):
\[ x = 0 \implies f^{-1}(0) = 0 \]

Part (A): 2 points
- 1 point for the correct value \( f(g(0)) = -1.657 \) (or exact equivalent \( 8 - 6\ln(5) \)).
- 1 point for both correct solutions \( x = -1.337 \) and \( x = 2.461 \).

Part (B): 2 points
- 1 point for the limit notation and correct value: \( \displaystyle \lim_{x \to -\infty} h(x) = \frac{5}{2} \) (or \( 2.5 \)).
- 1 point for the correct one-sided limit: \( \displaystyle \lim_{x \to 2^+} h(x) = \infty \) (or "does not exist with infinite behavior").

Part (C): 2 points
- 1 point for "No" with an explanation that correctly references the definition of a function (e.g., multiple inputs map to the same output / the inverse maps one input to multiple outputs). Note: Mentioning only the "horizontal line test" without referencing function definition receives 0 points.
- 1 point for the correct value \( f^{-1}(0) = 0 \).

Part (A):

(i) Substituting \(t = 2\) and \(C(2) = 14.50\) into \(C(t) = a + b\ln(t+1)\):
$$a + b\ln(2+1) = 14.50 \implies a + b\ln(3) = 14.50$$

Substituting \(t = 8\) and \(C(8) = 28.20\):
$$a + b\ln(8+1) = 28.20 \implies a + b\ln(9) = 28.20$$

(ii) Since \(\ln(9) = \ln(3^2) = 2\ln(3)\), we have:
$$a + 2b\ln(3) = 28.20$$
Subtracting the first equation from the second:
$$b\ln(3) = 28.20 - 14.50 = 13.70$$
$$b = \frac{13.70}{\ln(3)} \approx 12.47029... \approx 12.470$$

Then:
$$a = 14.50 - b\ln(3) = 14.50 - 13.70 = 0.800$$

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Part (B):

(i) The average rate of change from \(t = 2\) to \(t = 8\) is:
$$\text{AROC} = \frac{C(8) - C(2)}{8 - 2} = \frac{28.20 - 14.50}{6} = \frac{13.70}{6} \approx 2.283\text{ mg/L per minute}$$

(ii) Using the point \((2, 14.50)\) and the average rate of change:
$$L(5) = 14.50 + 2.28333...(5 - 2) = 14.50 + 6.85 = 21.35\text{ mg/L}$$
(Equivalently, using \((8, 28.20)\): \(L(5) = 28.20 - 2.28333...(8 - 5) = 21.35\text{ mg/L}\).)

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Part (C):

The graph of \(C(t) = 0.800 + 12.470\ln(t+1)\) is concave down on the interval \([2, 8]\) because the rate of change of \(C(t)\) is decreasing (since \(C'(t) = \frac{12.470}{t+1}\) decreases as \(t\) increases). For a function that is concave down, any secant line segment connecting two points on the graph lies entirely below the graph of the function on the interval between the two points.

Therefore, the linear estimate \(L(5)\) is an underestimate of the actual model value \(C(5)\).

Part (A): 2 points
* 1 point for writing two correct equations in terms of \(a\) and \(b\).
* 1 point for finding both values \(a = 0.800\) (or \(0.8\)) and \(b = 12.470\) (accept \(12.470\) or \(12.471\) due to rounding/truncation).

Part (B): 2 points
* 1 point for the average rate of change with valid supporting computation: \(\frac{28.20 - 14.50}{8 - 2} = 2.283\).
* 1 point for the linear estimation at \(t = 5\) showing work: \(14.50 + 2.283(3) = 21.35\) (accept \(21.349\) to \(21.350\)).

Part (C): 2 points
* 1 point for correctly stating that \(C(t)\) is concave down on \([2, 8]\) (or that the rate of change of \(C(t)\) is decreasing).
* 1 point for concluding underestimate with a justification referencing that the secant line lies below the curve for a concave down function.

(A)
The time between consecutive minimum and maximum values represents half a period:
\(\frac{\text{Period}}{2} = 11 - 3 = 8 \implies \text{Period} = 16 \text{ seconds}\).

The quarter-period interval is \(\frac{16}{4} = 4 \text{ seconds}\).
The midline value is \( d = \frac{4.8 + 1.2}{2} = 3.0 \text{ meters}\).

Starting at \( t = 3 \):
- \( J = (3, 1.2) \) [Minimum]
- \( K = (3 + 4, 3.0) = (7, 3.0) \) [Midline, increasing]
- \( P = (7 + 4, 4.8) = (11, 4.8) \) [Maximum]
- \( Q = (11 + 4, 3.0) = (15, 3.0) \) [Midline, decreasing]
- \( R = (15 + 4, 1.2) = (19, 1.2) \) [Minimum]

(B)
- Amplitude: \( a = \frac{4.8 - 1.2}{2} = 1.8 \)
- Midline: \( d = \frac{4.8 + 1.2}{2} = 3 \)
- Frequency parameter: \( b = \frac{2\pi}{\text{Period}} = \frac{2\pi}{16} = \frac{\pi}{8} \)
- Horizontal shift: A cosine maximum occurs at \( t = 11 \), so \( b(t + c) = 0 \implies 11 + c = 0 \implies c = -11 \).

Therefore, \( a = 1.8 \), \( b = \frac{\pi}{8} \), \( c = -11 \), \( d = 3 \).

(C)
(i) Average rate of change on \([3, 7]\):
\[ \frac{h(7) - h(3)}{7 - 3} = \frac{3.0 - 1.2}{4} = \frac{1.8}{4} = 0.45 = \frac{9}{20} \text{ meters per second} \]

(ii) On the interval \( 7 < t < 11 \), the graph of \( h \) is concave down (or transitioning from its maximum positive slope at \( t = 7 \) toward a slope of 0 at the peak \( t = 11 \)). Therefore, the rate of change of \( h \) is decreasing on \( (7, 11) \).

Part (A): 2 points
- 1 point for the coordinates of the extrema: \( J = (3, 1.2) \), \( P = (11, 4.8) \), \( R = (19, 1.2) \)
- 1 point for the coordinates of the midline points: \( K = (7, 3) \), \( Q = (15, 3) \)

Part (B): 2 points
- 1 point for the values of \( a = 1.8 \) (or \(\frac{9}{5}\)) and \( d = 3 \)
- 1 point for the values of \( b = \frac{\pi}{8} \) and \( c = -11 \) (or equivalent valid phase shift consistent with the given constraints)

Part (C): 2 points
- 1 point for the average rate of change: \( 0.45 \) or \( \frac{9}{20} \) (units not required)
- 1 point for the answer 'decreasing' with a valid reason referencing the concavity of the graph (concave down) or slopes decreasing from positive to 0

(A)(i)
\[\log_3(2x + 5) - \log_3(x - 1) = 2\]
Using the quotient property of logarithms:
\[\log_3\left(\frac{2x + 5}{x - 1}\right) = 2\]
Converting to exponential form:
\[\frac{2x + 5}{x - 1} = 3^2 = 9\]
\[2x + 5 = 9(x - 1)\]
\[2x + 5 = 9x - 9\]
\[7x = 14 \implies x = 2\]
Checking domain restrictions: \(2(2) + 5 = 9 > 0\) and \(2 - 1 = 1 > 0\), so \(x = 2\) is valid.

(A)(ii)
\[e^{2x} - 5e^x - 14 = 0\]
Let \(u = e^x\). The equation becomes quadratic:
\[u^2 - 5u - 14 = 0\]
\[(u - 7)(u + 2) = 0\]
So \(u = 7\) or \(u = -2\).
Since \(e^x > 0\) for all real \(x\), \(e^x = -2\) has no real solutions.
\[e^x = 7 \implies x = \ln(7)\]

(B)(i)
Using the double-angle identity \(\sin(2x) = 2\sin(x)\cos(x)\):
\[f(x) = 2\left(\sin(x)\cos(x)\right)^2 = 2\left(\frac{1}{2}\sin(2x)\right)^2 = 2\cdot \frac{1}{4}\sin^2(2x) = \frac{1}{2}\sin^2(2x)\]
Using the power-reducing identity \(\sin^2(\theta) = \frac{1 - \cos(2\theta)}{2}\) with \(\theta = 2x\):
\[f(x) = \frac{1}{2}\left(\frac{1 - \cos(4x)}{2}\right) = \frac{1}{4} - \frac{1}{4}\cos(4x)\]

(B)(ii)
Rewriting in terms of sine and cosine:
\[g(x) = \frac{\sin(x)}{\cos(x)} + \frac{\cos(x)}{\sin(x)} = \frac{\sin^2(x) + \cos^2(x)}{\sin(x)\cos(x)}\]
Using the Pythagorean identity \(\sin^2(x) + \cos^2(x) = 1\):
\[g(x) = \frac{1}{\sin(x)\cos(x)} = \frac{2}{2\sin(x)\cos(x)} = \frac{2}{\sin(2x)} = 2\csc(2x)\]

(C)
Isolating the cosine term:
\[\cos\left(3x - \frac{\pi}{4}\right) = \frac{\sqrt{3}}{2}\]
The general solutions for the argument are:
\[3x - \frac{\pi}{4} = \frac{\pi}{6} + 2\pi k \quad \text{or} \quad 3x - \frac{\pi}{4} = -\frac{\pi}{6} + 2\pi k, \quad k \in \mathbb{Z}\]
For the first branch:
\[3x = \frac{\pi}{4} + \frac{\pi}{6} + 2\pi k = \frac{5\pi}{12} + 2\pi k \implies x = \frac{5\pi}{36} + \frac{2\pi}{3}k\]
For the second branch:
\[3x = \frac{\pi}{4} - \frac{\pi}{6} + 2\pi k = \frac{\pi}{12} + 2\pi k \implies x = \frac{\pi}{36} + \frac{2\pi}{3}k\]
where \(k\) is any integer.

Part (A)(i): 1 point
- 1 point for the correct solution \(x = 2\) with supporting algebraic work.

Part (A)(ii): 1 point
- 1 point for the correct solution \(x = \ln(7)\) and identifying/discarding the extraneous value \(e^x = -2\).

Part (B)(i): 1 point
- 1 point for the correct expression \(\frac{1}{4} - \frac{1}{4}\cos(4x)\) (or equivalent constants \(a = \frac{1}{4}, b = -\frac{1}{4}, c = 4\)).

Part (B)(ii): 1 point
- 1 point for the correct expression \(2\csc(2x)\).

Part (C): 2 points
- 1 point for setting up the correct equations for the argument: \(3x - \frac{\pi}{4} = \pm\frac{\pi}{6} + 2\pi k\) (or \(\frac{\pi}{6} + 2\pi k\) and \(\frac{11\pi}{6} + 2\pi k\)).
- 1 point for all correct general solutions: \(x = \frac{5\pi}{36} + \frac{2\pi}{3}k\) and \(x = \frac{\pi}{36} + \frac{2\pi}{3}k\) for any integer \(k\).