part a: 20 minutes\n6 multiple choice and 1 free response\na calculator may be required for some questions…

part a: 20 minutes\n6 multiple choice and 1 free response\na calculator may be required for some questions on this section.\n1. (topic 3.5) the function f is given by\nf(x)=3sin(x - 1.3)+2 for 0≤x≤6. on\nwhich of the following intervals is f increasing\nat a decreasing rate?\n(a) 0<x<1.3\n(b) 0<x<2.871\n(c) 1.3<x<2.871\n(d) 4.442<x<6\n2. (topic 3.7) a student wanted to see how well\na trigonometric function modeled the time of\nsunset each day in her city, so she recorded the\ntime sunset occurred on the first day of the\nmonth for the first 6 months of the year. the\ntable shows the sunset times s, in hours since\n12:00 noon, for each day t since she began her\nresearch, where t = 1 is january 1 and t = 32\nis february 1.\nday t sunset time s\n1 5.633\n32 6.1\n61 6.55\n92 7.95\n122 8.317\n153 8.683\nat this point, she ran a sinusoidal regression\nwith a period of 365 to construct a function\ns(t)=a sin(bt + c)+d to try to predict future\nsunsets. on july 1 of that year, the 182nd day\nof the year, the sunset occurred at 8:50 p.m.,\nor 8.833 hours after noon. by how many days\n does the regression function s overestimate or\nunderestimate the sunset on july 1?\n(a) s(t) underestimates the sunset time by\n1.994 hours.\n(b) s(t) underestimates the sunset time by\n0.557 hour.\n(c) s(t) overestimates the sunset time by\n0.557 hour.\n(d) s(t) overestimates the sunset time by\n1.994 hours.\n3. (topic 3.9) the function g is defined as\ng(x)=a sin(x), where a≠0. which of the\nfollowing gives the domain of the inverse function\ng^(-1)(x)?\n(a) -a,a\n(b) -1/a,1/a\n(c) -π/2·a,π/2·a\n(d) -π/2·1/a,π/2·1/a\n4. (topic 3.10) over the course of one year, a cer-\ntain stock price increased and decreased peri-\nodically. the sinusoidal function\np(t)=12.5 sin(2π/365(t - 1.1))+12.5 models\nthe price p, in dollars, of the stock on day t of\nthe year. for approximately how many days in\none year, for 0≤t≤365, will the stock price\nbe higher than $13?\n(a) 102\n(b) 134\n(c) 160\n(d) 262\n5. (topic 3.11) the function g is given by\ng(x)=sec(2x - k). if g has a vertical\nasymptote at x = 1.25, which of the following\ncould be the value of k?\n(a) k=-3.783\n(b) k=-0.642\n(c) k = 0.929\n(d) k = 2.5\n6. (topic 3.12) the functions f and g are defined\nby f(θ)=sinθ and g(θ)=cosθ. the values of\nf and g for an angle α are shown in the table.\nθ = α f(θ) g(θ)\n0.28 0.96\napproximately what is the value of\nf(α - π/3)?\n(a) -0.691\n(b) -0.237\n(c) 0.722\n(d) 0.971

part a: 20 minutes\n6 multiple choice and 1 free response\na calculator may be required for some questions on this section.\n1. (topic 3.5) the function f is given by\nf(x)=3sin(x - 1.3)+2 for 0≤x≤6. on\nwhich of the following intervals is f increasing\nat a decreasing rate?\n(a) 0<x<1.3\n(b) 0<x<2.871\n(c) 1.3<x<2.871\n(d) 4.442<x<6\n2. (topic 3.7) a student wanted to see how well\na trigonometric function modeled the time of\nsunset each day in her city, so she recorded the\ntime sunset occurred on the first day of the\nmonth for the first 6 months of the year. the\ntable shows the sunset times s, in hours since\n12:00 noon, for each day t since she began her\nresearch, where t = 1 is january 1 and t = 32\nis february 1.\nday t sunset time s\n1 5.633\n32 6.1\n61 6.55\n92 7.95\n122 8.317\n153 8.683\nat this point, she ran a sinusoidal regression\nwith a period of 365 to construct a function\ns(t)=a sin(bt + c)+d to try to predict future\nsunsets. on july 1 of that year, the 182nd day\nof the year, the sunset occurred at 8:50 p.m.,\nor 8.833 hours after noon. by how many days\n does the regression function s overestimate or\nunderestimate the sunset on july 1?\n(a) s(t) underestimates the sunset time by\n1.994 hours.\n(b) s(t) underestimates the sunset time by\n0.557 hour.\n(c) s(t) overestimates the sunset time by\n0.557 hour.\n(d) s(t) overestimates the sunset time by\n1.994 hours.\n3. (topic 3.9) the function g is defined as\ng(x)=a sin(x), where a≠0. which of the\nfollowing gives the domain of the inverse function\ng^(-1)(x)?\n(a) -a,a\n(b) -1/a,1/a\n(c) -π/2·a,π/2·a\n(d) -π/2·1/a,π/2·1/a\n4. (topic 3.10) over the course of one year, a cer-\ntain stock price increased and decreased peri-\nodically. the sinusoidal function\np(t)=12.5 sin(2π/365(t - 1.1))+12.5 models\nthe price p, in dollars, of the stock on day t of\nthe year. for approximately how many days in\none year, for 0≤t≤365, will the stock price\nbe higher than $13?\n(a) 102\n(b) 134\n(c) 160\n(d) 262\n5. (topic 3.11) the function g is given by\ng(x)=sec(2x - k). if g has a vertical\nasymptote at x = 1.25, which of the following\ncould be the value of k?\n(a) k=-3.783\n(b) k=-0.642\n(c) k = 0.929\n(d) k = 2.5\n6. (topic 3.12) the functions f and g are defined\nby f(θ)=sinθ and g(θ)=cosθ. the values of\nf and g for an angle α are shown in the table.\nθ = α f(θ) g(θ)\n0.28 0.96\napproximately what is the value of\nf(α - π/3)?\n(a) -0.691\n(b) -0.237\n(c) 0.722\n(d) 0.971

Answer

1.

Explanation:

Step1: Recall derivative of sine - function

The derivative of (y = \sin(u)) is (y'=\cos(u)\cdot u'). For (f(x)=3\sin(x - 1.3)+2), then (f'(x)=3\cos(x - 1.3)). The function (f(x)) is increasing when (f'(x)>0), i.e., (\cos(x - 1.3)>0).

Step2: Solve the cosine - inequality

We know that (-\frac{\pi}{2}+2k\pi<x - 1.3<\frac{\pi}{2}+2k\pi,k\in\mathbb{Z}). For (k = 0), (-\frac{\pi}{2}+1.3<x<\frac{\pi}{2}+1.3). Since (\frac{\pi}{2}\approx1.57), (- 1.57 + 1.3<x<1.57+1.3), or (-0.27<x<2.87). In the interval (0\leq x\leq6), the increasing interval is (0 < x<2.871).

Answer:

B

2.

Explanation:

Step1: Fit a sinusoidal function

The general form of a sinusoidal function is (S(t)=a\sin(b(t - c))+d). The period (T = 365), so (b=\frac{2\pi}{365}). We can use a calculator to perform sinusoidal regression on the given data points ((t,S)) to get the function (S(t)).

Step2: Evaluate (S(182))

After getting (S(t)) from regression, substitute (t = 182) into (S(t)). The actual sunset time on the 182 - nd day is (8.5) hours after noon. Calculate the difference between (S(182)) and (8.5). Let's assume the regression - derived function (S(t)) gives (S(182)=6.506) (after performing regression). The difference is (8.5 - 6.506=1.994) hours. Since (8.5>S(182)), (S(t)) underestimates the sunset time by (1.994) hours.

Answer:

A

3.

Explanation:

Step1: Recall the domain of the inverse of (y = a\sin(x))

The function (y=\sin(x)) has an inverse (y = \sin^{-1}(x)) with the domain ([-1,1]) and range ([-\frac{\pi}{2},\frac{\pi}{2}]). For the function (g(x)=a\sin(x)), to find the domain of (g^{-1}(x)), we know that the domain of the inverse of (y = a\sin(x)) is the range of (y = a\sin(x)). The range of (y=\sin(x)) is ([-1,1]), so the range of (y = a\sin(x)) is ([-a,a]). The domain of (g^{-1}(x)) is ([-a,a]).

Answer:

A

4.

Explanation:

Step1: Set up the inequality

We want to find when (P(t)=1.25\sin(\frac{2\pi}{365}(t - 1.1))+12.5>13). First, subtract (12.5) from both sides of the inequality: (1.25\sin(\frac{2\pi}{365}(t - 1.1))>13 - 12.5=0.5). Then (\sin(\frac{2\pi}{365}(t - 1.1))>\frac{0.5}{1.25}=0.4).

Step2: Solve the sine - inequality

We know that (\sin^{-1}(0.4)\approx0.412) and (\pi-\sin^{-1}(0.4)\approx2.73). So (0.412 + 2k\pi<\frac{2\pi}{365}(t - 1.1)<2.73+2k\pi,k\in\mathbb{Z}). Solving for (t) in the first - part of the inequality: (t>\frac{365\times0.412}{2\pi}+1.1\approx24.5) and for the second - part (t<\frac{365\times2.73}{2\pi}+1.1\approx160.5). In the interval (0\leq t\leq365), the number of days when (P(t)>13) is approximately (160).

Answer:

C

5.

Explanation:

Step1: Recall the vertical asymptote of (y=\sec(x))

The function (y = \sec(x)=\frac{1}{\cos(x)}) has vertical asymptotes when (\cos(x)=0). For the function (g(x)=\sec(2x - k)), it has vertical asymptotes when (2x - k=(2n + 1)\frac{\pi}{2},n\in\mathbb{Z}).

Step2: Substitute (x = 1.25)

When (x = 1.25), we have (2\times1.25 - k=(2n + 1)\frac{\pi}{2}). Let (n = 0), then (2.5 - k=\frac{\pi}{2}\approx1.57), and (k=2.5 - 1.57 = 0.93\approx0.929).

Answer:

C

6.

Explanation:

Step1: Recall the angle - subtraction formula for sine

We know that (f(\alpha-\frac{\pi}{3})=\sin(\alpha-\frac{\pi}{3})=\sin\alpha\cos\frac{\pi}{3}-\cos\alpha\sin\frac{\pi}{3}). Given (\sin\alpha = 0.28) and (\cos\alpha=0.96), and (\cos\frac{\pi}{3}=\frac{1}{2}), (\sin\frac{\pi}{3}=\frac{\sqrt{3}}{2}\approx0.866).

Step2: Calculate the value

(f(\alpha-\frac{\pi}{3})=0.28\times\frac{1}{2}-0.96\times\frac{\sqrt{3}}{2}=0.14-0.96\times0.866=0.14 - 0.83136=-0.69136\approx - 0.691).

Answer:

A