AP Precalculus -3.2 Sine, Cosine, and Tangent- FRQ Exam Style Questions - Effective Fall 2023
AP Precalculus -3.2 Sine, Cosine, and Tangent- FRQ Exam Style Questions – Effective Fall 2023
AP Precalculus -3.2 Sine, Cosine, and Tangent- FRQ Exam Style Questions – AP Precalculus- per latest AP Precalculus Syllabus.
▶️ Answer/Explanation
(A) Exponential model \(P(t) = ae^{bt}\)
- Given: \(P(1) = 145\), \(P(2) = 115\).
- Divide the equations: \(\frac{P(2)}{P(1)} = \frac{115}{145} = \frac{23}{29} \approx 0.7931 = e^{b(2-1)} = e^b\).
- So \(b = \ln\left(\frac{23}{29}\right) \approx \ln(0.7931) \approx -0.2318\).
- Use \(P(1) = ae^{b\cdot1} = 145\): \(a = 145 \cdot e^{-b} = 145 \cdot \frac{29}{23} \approx 145 \cdot 1.2609 \approx 182.83\).
- Alternatively, compute directly: \(a = P(1) e^{-b} = 145 \cdot \frac{29}{23} = \frac{145 \cdot 29}{23} = \frac{4205}{23} \approx 182.826\).
- Thus, \(P(t) \approx 182.826 e^{-0.2318 t}\).
(B) i. Average rate of change from \(t = 0\) to \(t = 4\)
- \(P(0) = a \approx 182.826\).
- \(P(4) = 182.826 e^{-0.2318 \cdot 4} \approx 182.826 e^{-0.9272} \approx 182.826 \cdot 0.3956 \approx 72.34\).
- Average rate = \(\frac{P(4) – P(0)}{4 – 0} \approx \frac{72.34 – 182.826}{4} = \frac{-110.486}{4} \approx -27.622\) pellets per minute.
- Meaning: On average over the first 4 minutes, the dog ate about 27.6 pellets per minute.
(B) ii. Comparison of average rates
- For an exponential decay \(P(t) = ae^{bt}\) with \(b < 0\), the function is decreasing and concave up (since \(P”(t) = ab^2 e^{bt} > 0\)).
- In a concave up decreasing function, the average rate of change over an interval decreases in magnitude (becomes less negative) as the interval moves to the right.
- Thus, the average rate from \(t = 4\) to \(t = t_a\) will be greater (less negative) than that from \(t = 0\) to \(t = 4\).
- Reason: The instantaneous rate (derivative) is negative but increasing (becoming less negative), so later intervals have a smaller rate of loss.
(C) Limit as \(t \to \infty\)
- \(\lim_{t \to \infty} P(t) = \lim_{t \to \infty} ae^{bt} = 0\) because \(b < 0\).
- Context: The model predicts the pellets approach zero but never actually reach it. In reality, the dog will finish all pellets in finite time, so the model is not perfectly accurate for very large \(t\). However, it is a good approximation for the decay process while pellets remain.
- Thus, the model does not fully make sense as \(t \to \infty\) because it predicts the dog never finishes, while in reality the bowl becomes empty at some finite time.
▶️ Answer/Explanation
(A)(i) Rewrite \(g(x)\)
The function is given by \(g(x) = 3 \ln x – \frac{1}{2} \ln x\).
Combine the like terms:
\(g(x) = \left(3 – \frac{1}{2}\right) \ln x\)
\(g(x) = \frac{5}{2} \ln x\)
Apply the power property of logarithms, \(a \ln b = \ln(b^a)\):
\(g(x) = \ln\left(x^{5/2}\right)\)
(A)(ii) Rewrite \(h(x)\)
The function is given by \(h(x) = \frac{\sin^2 x – 1}{\cos x}\).
Recall the Pythagorean identity: \(\sin^2 x + \cos^2 x = 1\).
Rearrange the identity to isolate the numerator expression: \(\sin^2 x – 1 = -\cos^2 x\).
Substitute this into the function:
\(h(x) = \frac{-\cos^2 x}{\cos x}\)
Simplify the expression by canceling one \(\cos x\) term:
\(h(x) = -\cos x\)
(B)(i) Solve \(j(x) = 0\)
Set the function equal to zero: \(2(\sin x)(\cos x) = 0\).
Divide both sides by 2:
\(\sin x \cos x = 0\)
By the zero product property, either \(\sin x = 0\) or \(\cos x = 0\).
Case 1: \(\sin x = 0\). In the interval \(\left[0, \frac{\pi}{2}\right]\), \(x = 0\).
Case 2: \(\cos x = 0\). In the interval \(\left[0, \frac{\pi}{2}\right]\), \(x = \frac{\pi}{2}\).
The solutions are \(x = 0\) and \(x = \frac{\pi}{2}\).
(B)(ii) Solve \(k(x) = 3e\)
Set the function equal to \(3e\):
\(8e^{(3x)} – e = 3e\)
Add \(e\) to both sides:
\(8e^{(3x)} = 4e\)
Divide both sides by 8:
\(e^{(3x)} = \frac{4e}{8}\)
\(e^{(3x)} = \frac{e}{2}\)
Take the natural logarithm (\(\ln\)) of both sides:
\(\ln\left(e^{3x}\right) = \ln\left(\frac{e}{2}\right)\)
Use logarithm properties to simplify (\(\ln(e^a) = a\) and \(\ln(a/b) = \ln a – \ln b\)):
\(3x = \ln e – \ln 2\)
Since \(\ln e = 1\):
\(3x = 1 – \ln 2\)
Divide by 3:
\(x = \frac{1 – \ln 2}{3}\)
(C) Find input values for \(m(x) = \frac{9}{2}\)
Set the function equal to \(\frac{9}{2}\):
\(\cos(2x) + 4 = \frac{9}{2}\)
Subtract 4 from both sides (note that \(4 = \frac{8}{2}\)):
\(\cos(2x) = \frac{9}{2} – \frac{8}{2}\)
\(\cos(2x) = \frac{1}{2}\)
The reference angle for cosine equal to \(\frac{1}{2}\) is \(\frac{\pi}{3}\).
The general solution for \(2x\) is:
\(2x = \frac{\pi}{3} + 2\pi n\) or \(2x = -\frac{\pi}{3} + 2\pi n\) (where \(n\) is an integer).
Solve for \(x\) by dividing by 2:
\(x = \frac{\pi}{6} + \pi n\) or \(x = -\frac{\pi}{6} + \pi n\)
Combining these, the input values are \(x = \pm \frac{\pi}{6} + \pi n\) for any integer \(n\).