2024 AP Precalculus free-response questions, solved

A (i) $f(3) = 1$, so $h(3) = g(1) = 2.916(0.7) = 2.041$.

A (ii) $x = -3$, $x = 0$ and $x = 3$

B (i) $2.916(0.7)^x = 2$ when $x = 1.057$.

B (ii) $\lim_{x \to \infty} g(x) = 0$

C (i) $f$ does not have an inverse function.

C (ii) $f(-3) = f(0) = f(3) = 1$, so the output 1 comes from three different inputs. An inverse would have to send 1 to three different values, and a function gives each input exactly one output, so the inverse of $f$ can’t be a function.

A (i) $G(0) = 40$ and $G(91) = 76$, so $a + b\ln(0 + 1) = 40$ and $a + b\ln(91 + 1) = 76$.

A (ii) $\ln 1 = 0$, so $a = 40$. Then $b = \frac{76 - 40}{\ln 92} = 7.961$.

B (i) $\frac{G(91) - G(0)}{91 - 0} = \frac{76 - 40}{91} = 0.396$ thousand units per day

B (ii) $40 + 0.395604(50 - 0) = 59.780$ thousand units

B (iii) $A_t$ is the $y$-coordinate of a point on the secant line through $(0, G(0))$ and $(91, G(91))$. Since $b > 0$, the graph of $G$ is concave down on $0 < t < 91$, so the secant line lies below the graph of $G$ there. So $A_t < G(t)$ for $0 < t < 91$.

C On day 91 the model and the real sales agree: both are 76 thousand. After day 91 real daily sales decrease, but $G$ keeps increasing, so the difference between them gets larger every day.

A W touches the ground at $t = \frac{1}{2}$ and $t = \frac{5}{2}$, so the period is 2 seconds and the labeled points are $\frac{1}{2}$ second apart. Heights run from 0 to 18, so the midline is 9. W is at the top halfway between the two ground contacts, at $t = \frac{3}{2}$, so that’s F. F $\left(\frac{3}{2}, 18\right)$, G $(2, 9)$, J $\left(\frac{5}{2}, 0\right)$, K $(3, 9)$, P $\left(\frac{7}{2}, 18\right)$.

B $a = 9$, $d = 9$, $b = \frac{2\pi}{2} = \pi$. The first peak is at $t = \frac{3}{2}$, so $c = \frac{2}{4} - \frac{3}{2} = -1$. That gives $h(t) = 9\sin(\pi(t - 1)) + 9$. Check: $h\left(\frac{3}{2}\right) = 9\sin\frac{\pi}{2} + 9 = 18$, and $h\left(\frac{1}{2}\right) = 9\sin\left(-\frac{\pi}{2}\right) + 9 = 0$.

C (i) a: $h$ is positive and increasing.

C (ii) The graph of $h$ is concave down on $(t_1, t_2)$, so the rate of change of $h$ is decreasing.

A (i) $e^{(x + 3)} = 10$, so $x + 3 = \ln 10$ and $x = \ln 10 - 3$.

A (ii) $\arcsin\left(\frac{x}{2}\right) = \frac{\pi}{4}$, so $\frac{x}{2} = \sin\frac{\pi}{4} = \frac{\sqrt{2}}{2}$ and $x = \sqrt{2}$.

B (i) $j(x) = \log_{10}(8x^5 \cdot 2x^2) - \log_{10}(x^9) = \log_{10}\left(\frac{16x^7}{x^9}\right) = \log_{10}\left(\frac{16}{x^2}\right)$

B (ii) $1 - \sin^2 x = \cos^2 x$, so $k(x) = \frac{\cos^2 x}{\sin x} \cdot \frac{1}{\cos x} = \frac{\cos x}{\sin x} = \frac{1}{\tan x}$

C $\cos^{-1}(\tan(2x)) = 0$ means $\tan(2x) = \cos 0 = 1$. So $2x = \frac{\pi}{4} + n\pi$ and $x = \frac{\pi}{8} + \frac{n\pi}{2}$, where $n$ is any integer.