2025 AP Precalculus free-response questions, solved

A (i) $f(1) = 1.75$, so $h(1) = g(1.75) = -0.167(1.75)^3 + (1.75)^2 - 1.834 = 0.333$.

A (ii) $f(0) = 3.5$, so $f^{-1}(3.5) = 0$.

B (i) $x = -1.233$, $x = 1.578$ and $x = 5.643$

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

C (i) Exponential.

C (ii) The inputs in the table go up by 1 each step, and each output is 0.5 times the one before: $\frac{7}{14} = \frac{3.5}{7} = \frac{1.75}{3.5} = \frac{0.875}{1.75} = 0.5$. Over equal-length input intervals the outputs change by the same factor, so $f$ is best modeled by an exponential function.

A (i) $D(0) = 25$, $D(2) = 30$ and $D(4) = 34$ give $c = 25$, $4a + 2b + c = 30$ and $16a + 4b + c = 34$.

A (ii) With $c = 25$: $4a + 2b = 5$ and $16a + 4b = 9$. Doubling the first and subtracting it from the second gives $8a = -1$, so $a = -0.125$, then $b = 2.75$ and $c = 25$.

B (i) $\frac{D(4) - D(0)}{4 - 0} = \frac{34 - 25}{4} = 2.25$ thousand plays per month

B (ii) $25 + 2.25(1.5) = 28.375$ thousand plays

B (iii) $A_t$ is the $y$-coordinate of a point on the secant line through $(0, D(0))$ and $(4, D(4))$. Since $a < 0$, the graph of $D$ is concave down on $0 < t < 4$, so the secant line lies below the graph of $D$ there. So $A_t < D(t)$ for $0 < t < 4$.

C The total number of plays since release can never decrease. $D$ has its absolute maximum at $t = -\frac{b}{2a} = 11$ and decreases after that, so $D$ can only model the total plays up to $t = 11$: the right end of its domain is $t = 11$.

A 200 cycles a second means one cycle takes $\frac{1}{200}$ second, so the labeled points are $\frac{1}{800}$ second apart. X starts at its highest point, so F is at $t = 0$. F $(0, 2)$, G $\left(\frac{1}{800}, 0\right)$, J $\left(\frac{2}{800}, -2\right)$, K $\left(\frac{3}{800}, 0\right)$, P $\left(\frac{4}{800}, 2\right)$.

B $a = 2$, $d = 0$, $b = \frac{2\pi}{1/200} = 400\pi$. The first peak is at $t = 0$, so $c = \frac{1}{4} \cdot \frac{1}{200} - 0 = \frac{1}{800}$. That gives $h(t) = 2\sin\left(400\pi\left(t + \frac{1}{800}\right)\right)$. Check: $h(0) = 2\sin\frac{\pi}{2} = 2$.

C (i) d: $h$ is negative and decreasing.

C (ii) The graph of $h$ is concave up on $(t_1, t_2)$, and the rate of change of $h$ is increasing.

A (i) $2\log_3 x = 4$, so $\log_3 x = 2$ and $x = 3^2 = 9$.

A (ii) $4\cos^2 x = 3$, so $\cos^2 x = \frac{3}{4}$ and $\cos x = \pm\frac{\sqrt{3}}{2}$. In $\left[0, \frac{\pi}{2}\right)$, cosine is positive, so $x = \frac{\pi}{6}$.

B (i) $3\log_2 2 = \log_2 2^3 = \log_2 8$, so $j(x) = \log_2 x + \log_2 8 = \log_2(8x)$.

B (ii) $\csc^2 x - 1 = \cot^2 x$, so $k(x) = \frac{6}{\tan x\cot^2 x} = \frac{6}{\cot x} = 6\tan x$.

C $e^{2x} - e^x - 12 = (e^x)^2 - e^x - 12 = (e^x - 4)(e^x + 3) = 0$. $e^x = -3$ is impossible because $e^x > 0$, so $e^x = 4$ and $x = \ln 4$.