The period of the function `f(x)=3 \ cos (2 x+\pi)` is
- `2 \pi`
- `\pi`
- `\frac{2\pi}{3}`
- `2`
- `3`
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The period of the function `f(x)=3 \ cos (2 x+\pi)` is
`B`
Period | `= (2pi)/n` | `(n = 2)` |
`= (2pi)/2` | ||
`= pi` |
`=>B`
Consider the function \(f:[-a\pi, a\pi] \rightarrow\ R, f(x)=\sin(ax)\), where \(a\) is a positive integer.
The number of local minima in the graph of \(y=f(x)\) is always equal to
\(E\)
\(\text{Check graphs for different values of }a\)
\(a=1\to \text{1 local minimum}\)
\(a=2\to \text{4 local minimums}\)
\(a=3\to \text{9 local minimums}\)
\(\therefore\ \text{Number of local minimums is always equal to }a^2\)
\(\Rightarrow E\)
The amplitude, \(A\), and the period, \(P\), of the function \(f(x)=-\dfrac{1}{2}\sin(3x+2\pi)\) are
\(E\)
\(\text{Period:}\ \ P= \dfrac{2\pi}{n}= \dfrac{2\pi}{3}\)
\(\text{Amplitude}:\ \ A=\bigg|-\dfrac{1}{2}\bigg|=\dfrac{1}{2}\)
\(\Rightarrow E\)
Part of the graph of `f: R to R , \ f(x) = sin (x/2) + cos(2x)` is shown below.
Consider the set of functions of the form `g_a : R to R, \ g_a (x) = sin(x/a) + cos(ax)`, where `a` is a positive integer.
a. `text{By inspection, graph begins to repeat after 4pi.}`
`text{Period}\ = 4 pi`
b. `text{By CAS: Sketch}\ \ f(x) = sin (x/2) + cos(2x)`
`f_min = -1.722`
c. `text{If} \ \ f(x)\ \ text{is reflected in the} \ y text{-axis and translated} \ 2 pi \ text{to the right} => text{same graph}`
`f(x) = f(-x + h) = f(2 pi – x)`
`:. h = 2 pi`
d. `f(x) = sin(x/2) + cos(2x)`
`g_a(x) = sin(x/a) + cos(2a)`
`g_a(x) = f(x) \ \ text{when} \ \ a = 2`
e.i. `int g_a (x)\ dx`
`= -a cos (x/a) + {sin (ax)}/{a} , \ (c = 0)`
e.ii. `int_0^{2a pi} g_a(x)\ dx`
`= {sin (2a^2 pi)}/{a}`
`= 0 \ \ (a ∈ ZZ^+)`
`text{When integral = 0, areas above and below the} \ x text{-axis are equal.}`
f. `g_a (x) = sin(x/a) + cos (ax)`
`-1 <= sin(x/a) <= 1 \ \ text{and}\ \ -1 <= cos(ax) <= 1`
`:. -2 <= g_a (x) <= 2`
g. `text{Sketch}\ \ g_a (x) \ \ text{by CAS}`
`text{Minimum access at}\ \ a = 1`
`g_a(x)_min = – sqrt2`
The period of the function with rule `y = tan((pix)/2)` is
`B`
`n= pi/2`
`text{Period} = pi/n = 2`
`=> B`
Consider the function `g: R -> R, \ g(x) = 2sin(2x).`
a. `text(S)text(ince) -1<sin(2x)<1,`
`text(Range)\ g(x) = [–2,2]`
b. `text(Period) = (2pi)/n = (2pi)/2 = pi`
c. | `2sin(2x)` | `=sqrt3` |
`sin(2x)` | `=sqrt3/2` | |
`2x` | `=pi/3, (2pi)/3, pi/3 + 2pi, (2pi)/3 + 2pi, …` | |
`x` | `=pi/6, pi/3, pi/6+pi, pi/3+pi, …` |
`:.\ text(General solution)`
`= pi/6 + npi, pi/3 + npi\ \ \ (n in ZZ)`
The wind speed at a weather monitoring station varies according to the function
`v(t) = 20 + 16sin ((pi t)/(14))`
where `v` is the speed of the wind, in kilometres per hour (km/h), and `t` is the time, in minutes, after 9 am.
A sudden wind change occurs at 10 am. From that point in time, the wind speed varies according to the new function
`v_1(t) = 28 + 18 sin((pi(t - k))/(7))`
where `v_1` is the speed of the wind, in kilometres per hour, `t` is the time, in minutes, after 9 am and `k ∈ R^+`. The wind speed after 9 am is shown in the diagram below.
Using this value of `k`, the weather monitoring station sends a signal when the wind speed is greater than 38 km/h.
i. Find the value of `t` at which a signal is first sent, correct to two decimal places. (1 mark)
ii. Find the proportion of one cycle, to the nearest whole percent, for which `v_1 > 38`. (2 marks)
State the values of `a`, `b`, `c` and `d`, in terms of `k` where appropriate. (3 marks)
`v_text(min) \ = 4 \ text(km/h)`
ii. `31text(%)`
a. `text(Amplitude) = 16`
`text{Find Period (n):}`
`(2 pi)/(n)` | `= (pi)/(14)` |
`n` | `= 28` |
b. `v_text(max) = 20 + 16 = 36 \ text(km/h)`
`v_text(min) = 20 – 16 = 4 \ text(km/h)`
c. `v(60)` | `= 20 + 16 sin ((60 pi)/(14))` |
`= 32.5093 \ \ text(km/h)` |
d. `v(t)\ \ text(is always positive.)`
`s(t) = int_0^60 v(t) \ dt`
`v(t)_(avg)` | `= (1)/(60) int_0^60 20 + 16 sin ((pi t)/(14))\ dt` |
`= 20.447` | |
`= 20.45 \ text(km/h) \ \ text((to 2 d.p.))` |
e. `text(S) text{olve (for}\ k text{):} \ \ v(60) = v_1(60)`
`k = 3.4358 \ \ text((to 4 d.p.))`
f.i. `text(S) text(olve for) \ t , \ text(given) \ \ v_1(t) = 38 \ \ text(and) \ \ k = 31.4358`
`=> t = 60.75 \ text(minutes)`
f.ii. `text(S) text(olving for) \ \ v_1(t) = 38 \ , \ k = 31.4358`
`t_1 = 60.75 \ text{(part i)}, \ t_2 = 65.123`
`text(Period of) \ \ v_1 = (2 pi)/(n) = (pi)/(7)\ \ => \ n = 14`
`:. \ text(Proportion of cycle)` | `= (65.123 – 60.75)/(14)` |
`= 0.312` | |
`= 31 text{% (nearest %)}` |
g. `f(x) → g(x)`
`y’ = 28 + 18 sin ({pi(x′ – k)}/{7})`
`x’ = ax + c` | `\ \ \ \ \ \ y′ = by + d` |
`text(Using) \ \ y’ = by + d`
`28 + 18 sin ({pi(x’ – k)}/{7}) = b (20 + 16 sin ({pi x}/{14})) + d`
`text(Equating coefficients of) \ \ sin theta :`
`16b = 18 \ \ \ => \ b = (9)/(8)`
`text(Equating constants:)`
`20 xx (9)/(8) + d = 28 \ \ \ => \ \ d = (11)/(2)`
`(x’ – k)/(7)` | `= (x)/(14)` |
`x’` | `= (x)/(2) + k \ \ => \ \ a = (1)/(2) \ , \ c = k` |
`a = (1)/(2) \ , \ b = (9)/(8) \ , \ c = k \ , \ d = (11)/(2)`
The diagram below shows one cycle of a circular function.
The amplitude, period and range of this function are respectively
`E`
`text(Graph centres around)\ \ y = 1`
`text(Amplitude) \ = 3`
`:. \ text(Range) \ = [1 – 3, 1 + 3] = [-2, 4]`
`text(Period:) = 4`
`=> E`
During a telephone call, a phone uses a dual-tone frequency electrical signal to communicate with the telephone exchange.
The strength, `f`, of a simple dual-tone frequency signal is given by the function `f(t) = sin((pi t)/3) + sin ((pi t)/6)`, where `t` is a measure of time and `t >= 0`.
Part of the graph of `y = f(t)` is shown below
Let `g` be the function obtained by applying the transformation `T` to the function `f`, where
`T([(x), (y)]) = [(a, 0), (0, b)] [(x), (y)] + [(c), (d)]`
and `a, b, c` and `d` are real numbers.
Find the value of `k`. (2 marks)
a. `text(Period) = 12`
b. `t = 0, 4, 6`
c. `f(t) = sin ((pi t)/3) + sin ((pi t)/6)`
`f(t)_max ~~ 1.76\ \ text{(by CAS)}`
d. | `text(Area)` | `= int_0^4 sin((pi t)/3) + sin ((pi t)/6) dt – int_4^6 sin ((pi t)/3) + sin ((pi t)/6) dt` |
`= 15/pi\ text(u²)` |
e. `text(Same area) => f(t)\ text(is reflected in the)\ x text(-axis and)`
`text(translated 6 units to the left.)`
`x′=ax+c`
`y′=by+d`
`text(Reflection in)\ xtext(-axis) \ => \ b=-1, \ d=0`
`text(Translate 6 units to the left) \ => \ a=1, \ c=-6`
`:. a = 1,\ b = -1,\ c = -6,\ d = 0`
f. | `text(Area of rectangle)` | `= 2 xx text(Area between)\ f(t) and x text(-axis)\ \ t in [0, 6]` |
`12k` | `= 2 xx 15/pi` | |
`:. k` | `=5/(2pi)` |
Let `f: R -> R,\ \ f(x) = 3 sin ((2x)/5) - 2`.
The period and range of `f` are respectively
`B`
`text(Period)` | `= (2pi)/n` |
`= (2 pi)/(2/5)` | |
`= 5 pi` | |
`text(Range)` | `= [-2 -3, -2 + 3]` |
`= [-5, 1]` |
`=> B`
The graph of `y = tan(ax)`, where `a ∈ R^+`, has a vertical asymptote `x = 3 pi` and has exactly one `x`-intercept in the region `(0, 3 pi)`.
The value of `a` is
`C`
`y = tan(ax)`
`tan x ->\ text(period of)\ pi,\ text(asymptotes at)\ \ x = pi/2, (3 pi)/2`
`tan(x/2) ->\ text(period of)\ 2 pi,\ text(asymptotes at)\ \ x=pi, 3 pi`
`tan(x/2) -> text(has one)\ x text(-intercept of)\ 2 pi\ \ {x: (0, 3 pi)}`
`:. a = 1/2`
`=> C`
Let `f: R -> R,\ f(x) = 4 cos ((2 pi x)/3) + 1`.
The period of this function is
`C`
`n` | `=\ text(period)` |
`(2 pi)/n` | `= (2 pi)/3` |
`n` | `= 3` |
`=> C`
Let `f : R → R, \ f (x) = 5sin(2x) - 1`.
The period and range of this function are respectively
`C`
`text(Period) = (2pi)/2 = pi`
`text(Range)` | `= [−1 – 5, −1 + 5]` |
`= [−6 ,4]` |
`=> C`
The graph shown is `y = A sin bx`.
On the same set of axes, draw the graph `y = 3 sin x + 1` for `0 <= x <= pi`. (2 marks)
a. `A = 4`
b. `text(S)text(ince the graph passes through)\ \ (pi/4, 4)`
`text(Substituting into)\ \ y = 4 sin bx`
`4 sin (b xx pi/4)` | `=4` |
`sin (b xx pi/4)` | `= 1` |
`b xx pi/4` | `= pi/2` |
`:. b` | `= 2` |
c. |
The function with rule `f(x) = 4 tan (x/3)` has period
`D`
`text(Period) ` | `=pi/n` |
`= pi/(1/3)` | |
`= 3 pi` |
`=> D`
Let `f: R -> R,\ f(x) = 1 - 2 cos ({pi x}/2).`
The period and range of this function are respectively
`B`
`text(Period)` | `= (2 pi)/n = (2pi)/(pi/2)=4` |
`text(Amplitude = 2 and median is)\ \ y=1.`
`text(Range)` | `= [1 – 2, quad 1 + 2]` |
`= [−1, 3]` |
`=> B`
The function with rule `f(x) = −3sin((pix)/5)` has period
`=> C`
`text(Period)` | `= (2pi)/n` |
`= (2pi)/(pi/5)` | |
`= 10` |
`=> C`
For the function `f: [– pi, pi] -> R, f(x) = 5 cos (2 (x + pi/3))`
Label endpoints of the graph with their coordinates. (3 marks)
Write down the amplitude and period of the function
`qquad f: R -> R,\ \ f(x) = 4 sin ((x + pi)/3)`. (2 marks)
`text(Amplitude) = 4;\ \ text(Period) = 6 pi`
`text(Amplitude) = 4`
`text(Period):\ \ (2 pi)/n = (2 pi)/(1/3) = 6 pi`
State the range and period of the function
`h: R -> R,\ \ h(x) = 4 + 3 cos ((pi x)/2).` (2 marks)
`text(Range) = [1, 7];\ \ \ text(Period) = 4`
`-1` | `<cos ((pi x)/2)<1` | |
`-3` | `<3cos ((pi x)/2)<3` | |
`1` | `< 4+ 3cos ((pi x)/2)<7` |
`:.\ text(Range:)\ [1, 7]`
`text(Period) = (2pi)/n = (2 pi)/(pi/2) = 4`
The population of wombats in a particular location varies according to the rule `n(t) = 1200 + 400 cos ((pi t)/3)`, where `n` is the number of wombats and `t` is the number of months after 1 March 2013.
a. `text(Period) = (2pi)/n = (2pi)/(pi/3) = 6\ text(months)`
`text(A)text(mplitude) = 400`
b. `text(Max:)\ 1200 + 400 = 1600\ text(wombats)`
`text(Min:)\ 1200 – 400 = 800\ text(wombats)`
c. `n(10) = 1000\ text(wombats)`
d. | `text(Solve)\ n(t)` | `= 1000\ text(for)\ t ∈ [0,12]` |
`t= 2,4,8,10`
`text(S)text(ince the graph starts at)\ \ (0,1600),`
`=> n(t) < 1000\ \ text(for)`
`t ∈ (2,4)\ text(or)\ t ∈ (8,10)`
`:.\ text(Fraction)` | `= ((4 – 2) + (10 – 8))/12` |
`= 1/3\ \ text(year)` |
Trigg the gardener is working in a temperature-controlled greenhouse. During a particular 24-hour time interval, the temperature `(Ttext{°C})` is given by `T(t) = 25 + 2 cos ((pi t)/8), \ 0 <= t <= 24`, where `t` is the time in hours from the beginning of the 24-hour time interval.
Trigg is designing a garden that is to be built on flat ground. In his initial plans, he draws the graph of `y = sin(x)` for `0 <= x <= 2 pi` and decides that the garden beds will have the shape of the shaded regions shown in the diagram below. He includes a garden path, which is shown as line segment `PC.`
The line through points `P((2 pi)/3, sqrt 3/2)` and `C (c, 0)` is a tangent to the graph of `y = sin (x)` at point `P.`
In further planning for the garden, Trigg uses a transformation of the plane defined as a dilation of factor `k` from the `x`-axis and a dilation of factor `m` from the `y`-axis, where `k` and `m` are positive real numbers.
a. `T_text(max)\ text(occurs when)\ \ cos((pit)/8) = 1,`
`T_text(max)= 25 + 2 = 27^@C`
`text(Max occurs when)\ \ t = 0, or 16\ text(h)`
b. | `text(Period)` | `= (2pi)/(pi/8)` |
`= 16\ text(hours)` |
c. `text(Solve:)\ \ 25 + 2 cos ((pi t)/8)=26\ \ text(for)\ t,`
`t` | `= 8/3,40/3,56/3\ \ text(for)\ t ∈ [0,24]` |
`t_text(min)` | `= 8/3` |
d. `text(Consider the graph:)`
`text(Time above)\ 26 text(°C)` | `= 8/3 + (56/3 – 40/3)` |
`= 8\ text(hours)` |
e.i. `(dy)/(dx) = cos(x)`
`text(At)\ x = (2pi)/3,`
`(dy)/(dx)` | `= cos((2pi)/3)= −1/2` |
e.ii. `text(Solution 1)`
`text(Equation of)\ \ PC,`
`y-sqrt3/2` | `=-1/2(x-(2pi)/3)` |
`y` | `=-1/2 x +pi/3 +sqrt3/2` |
`PC\ \ text(passes through)\ \ (c,0),`
`0` | `=-1/2 c +pi/3 + sqrt3/2` |
`c` | `=sqrt3 + (2 pi)/3\ …\ text(as required)` |
`text(Solution 2)`
`text(Equating gradients:)`
`- 1/2` | `= (sqrt3/2 – 0)/((2pi)/3 – c)` |
`-1` | `= sqrt3/((2pi – 3c)/3)` |
`3c – 2pi` | `= 3sqrt3` |
`3c` | `= 3 sqrt3 + 2pi` |
`:. c` | `= sqrt3 + (2pi)/3\ …\ text(as required)` |
f.i. `Xprime ((2pi)/3 m,0)qquadPprime((2pi)/3 m, sqrt3/2 k)qquadCprime ((sqrt3 + (2pi)/3)m, 0)`
`XprimePprime` | `= 10` |
`sqrt3/2 k` | `= 10` |
`:. k` | `= 20/sqrt3` |
`=(20sqrt3)/3` |
`XprimeCprime=30`
`((sqrt3 + (2pi)/3)m) – (2pi)/3 m` | `= 30` |
`:. m` | `= 30/sqrt3` |
`=10sqrt3` |
f.ii. | `Pprime((2pi)/3 m, sqrt3/2 k)` | `= Pprime((2pi)/3 xx 10sqrt3, sqrt3/2 xx 20/sqrt3)` |
`= Pprime((20pisqrt3)/3,10)` |
Let `f: R -> R,\ f(x) = 2sin(3x) - 3.`
The period and range of this function are respectively
`A`
`text(Range:)\ [−3 – 2, −3 + 2]`
`= [−5,−1]`
`text(Period) = (2pi)/n = (2pi)/3`
`=> A`
The function with rule `f(x) = -3 tan(2 pi x)` has period
`C`
`text(Period)` | `= pi/n` |
`=pi/(2 pi)` | |
`= 1/2` |
`=> C`