A function that has a range of \([6,12]\) is
- \(f: R \rightarrow R, \ f(x)=6+3 \cos (9 x)\)
- \(f: R \rightarrow R, \ f(x)=6+6 \cos (3 x)\)
- \(f: R \rightarrow R, \ f(x)=9-3 \cos (6 x)\)
- \(f: R \rightarrow R, \ f(x)=9-6 \cos (3 x)\)
Aussie Maths & Science Teachers: Save your time with SmarterEd
A function that has a range of \([6,12]\) is
\(C\)
\(\text{By trial and error,}\)
\(\text{Consider option C:}\ \ f(x)=9-3 \cos (6 x)\)
\(\text{Since}\ \ -1 \leqslant \cos (6 x) \leqslant 1\)
\(\text{Range is} \ \ [-3+9,3+9]=[6,12]\)
\(\Rightarrow C\)
Let \(f:[0,2 \pi] \rightarrow R, f(x)=2 \cos (2 x)+1\).
--- 2 WORK AREA LINES (style=lined) ---
--- 8 WORK AREA LINES (style=lined) ---
--- 0 WORK AREA LINES (style=lined) ---
a. \(\text{Amplitude}=2 \ \ \text{about} \ \ y=1.\)
\(\text{Range of } f(x):\ -1 \leqslant y \leqslant 3\)
| b. | \(2 \cos (2 x)+1\) | \(=0\) |
| \(\cos (2 x)\) | \(=-\dfrac{1}{2}\) |
\(\text{Base angle}=\dfrac{\pi}{3}\)
\(2x=\pi-\dfrac{\pi}{3}, \pi+\dfrac{\pi}{3}, \cdots\)
\(2x=\dfrac{2 \pi}{3}, \dfrac{4 \pi}{3}, \dfrac{8 \pi}{3}, \dfrac{10 \pi}{3}\)
\(x=\dfrac{\pi}{3}, \dfrac{2 \pi}{3}, \dfrac{4 \pi}{3}, \dfrac{5 \pi}{3}\)
Consider the function `g: R -> R, \ g(x) = 2sin(2x).`
--- 2 WORK AREA LINES (style=lined) ---
--- 1 WORK AREA LINES (style=lined) ---
--- 6 WORK AREA LINES (style=lined) ---
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 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`
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`
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`
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`
State the range and period of the function
`h: R -> R,\ \ h(x) = 4 + 3 cos ((pi x)/2)`. (2 marks)
--- 5 WORK AREA LINES (style=lined) ---
`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`
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`