The graph of the even function `y=f(x)` is shown.
The area of the shaded region `A` is `1/2` and the area of the shaded region `B` is `3/2`.
What is the value of `int_(-2)^(2)f(x)\ dx`?
- `4`
- `2`
- `–2`
- `–4`
Calculus, 2ADV C4 2019 MET1 7
The shaded region in the diagram below is bounded by the vertical axis, the graph of the function with rule `f(x) = sin(pix)` and the horizontal line segment that meets the graph at `x = a`, where `1 <= a <= 3/2`.
Let `A(a)` be the area of the shaded region.
Show that `A(a) = 1/pi - 1/pi cos(a pi) - a sin (a pi)`. (3 marks)
Calculus, 2ADV C4 2018 HSC 10 MC
A trigonometric function `f(x)` satisfies the condition
`int_0^pi f(x)\ dx != int_pi^(2pi) f(x)\ dx.`
Which function could be `f(x)`?
- `f(x) = sin (2x)`
- `f(x) = cos (2x)`
- `f(x) = sin (x/2)`
- `f(x) = cos (x/2)`
Show Worked Solution
`text(Consider options A and C)`
`text(Consider options B and D)`
`text(When)\ \ y = cos\ x/2 ,`
`int_0^pi f(x)\ dx != int_pi^(2 pi) f(x)\ dx`
`=> D`
Calculus, 2ADV C4 2016 HSC 13d
The curve `y = sqrt 2 cos (pi/4 x)` meets the line `y = x` at `P(1, 1)`, as shown in the diagram.
Find the exact value of the shaded area. (3 marks)
Calculus, 2ADV C4 2007 HSC 7b
The diagram shows the graphs of `y = sqrt 3 cos x` and `y = sin x`. The first two points of intersection to the right of the `y`-axis are labelled `A` and `B`.
- Solve the equation `sqrt 3 cos x = sin x` to find the `x`-coordinates of `A` and `B`. (2 marks)
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- Find the area of the shaded region in the diagram. (3 marks)
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Calculus, 2ADV C4 2006 HSC 5b
- Show that `d/dx log_e (cos x) = -tan x.` (1 mark)
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The shaded region in the diagram is bounded by the curve `y =tan x` and the lines `y =x` and `x = pi/4.`
Using the result of part (i), or otherwise, find the area of the shaded region. (3 marks)
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Trigonometry, 2ADV T3 2006 HSC 7b
A function `f(x)` is defined by `f(x) = 1 + 2 cos x`.
- Show that the graph of `y = f(x)` cuts the `x`-axis at `x = (2 pi)/3`. (1 mark)
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- Sketch the graph of `y = f(x)` for `-pi <= x <= pi` showing where the graph cuts each of the axes. (3 marks)
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- Find the area under the curve `y = f(x)` between `x = -pi/2` and `x = (2 pi)/3`. (3 marks)
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Show Answers Only
- `text(Proof)\ \ text{(See Worked Solutions)}`
-
- `((7 pi)/6 + sqrt 3 + 1)\ text(u²)`
Show Worked Solution
i. `f(x) = 1 + 2 cos x`
`f(x)\ text(cuts the)\ x text(-axis when)\ f(x) = 0`
| `1 + 2 cos x` | `= 0` |
| `2 cos x` | `=-1` |
| `cos x` | `= -1/2` |
`:. x = (2 pi)/3\ …\ text(as required)`
| ii. |  |
| iii. `text(Area)` | `= int_(-pi/2)^((2 pi)/3) 1 + 2 cos x\ \ dx` |
| `= [x + 2 sin x]_(-pi/2)^((2 pi)/3)` | |
| `= [((2 pi)/3 + 2 sin (2 pi)/3) – ((-pi)/2 + 2 sin (-pi)/2)]` | |
| `= ((2 pi)/3 + 2 xx sqrt 3/2) – ((-pi)/2 +2(- 1))` | |
| `= (2 pi)/3 + sqrt(3) + pi/2 + 2` | |
| `= ((7 pi)/6 + sqrt(3) + 2)\ text(u²)` |
Calculus, 2ADV C3 2010 HSC 9b
Let `y=f(x)` be a function defined for `0 <= x <= 6`, with `f(0)=0`.
The diagram shows the graph of the derivative of `f`, `y = f prime (x)`.
The shaded region `A_1` has area `4` square units. The shaded region `A_2` has area `4` square units.
- For which values of `x` is `f(x)` increasing? (1 mark)
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- What is the maximum value of `f(x)`? (1 mark)
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- Find the value of `f(6)`. (1 mark)
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- Draw a graph of `y =f(x)` for `0 <= x <= 6`. (2 marks)
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Show Answers Only
- `f(x)\ text(is increasing when)\ 0 <= x < 2`
- `text(MAX value of)\ f(x) = 4`
- `-6`
-
Show Worked Solution
| i. | `f(x)\ text(is increasing when)\ \ f prime (x) > 0` |
| `text(From the graph)` | |
| `f(x)\ text(is increasing when)\ 0 <= x < 2` |
| ii. | `f prime (x) = 0\ \ text(when)\ \ x=2` |
| `:.\ text(MAX at)\ \ x = 2` | |
| `int_0^2 f prime (x)\ dx = 4\ \ \ (text(given since)\ A_1 = 4 text{)}` | |
| `text(We also know)` |
| `int_0^2\ f prime (x)\ dx` | `= [f(x)]_0^2` |
| `= f(2)\ – f(0)` | |
| `= f(2)\ \ \ \ text{(since}\ f(0) = 0 text{)}` |
`=> f(2) = 4`
♦♦♦ Parts (ii) and (iii) proved particularly difficult for students with mean marks of 12% and 11% respectively.
`:.\ text(MAX value of)\ \ f(x) = 4`
| iii. | `int_0^4 f prime (x)` | `= A_1\ – A_2` |
| `=0` |
`text(We also know)`
| `int_0^4 f prime (x)\ dx` | `= int_2^4 f prime (x)\ dx + int_0^2 f prime (x)\ dx` |
| `=[f(x)]_2^4 + 4` | |
| `= f(4)\ – f(2) + 4\ \ \ (text(note)\ f(2)=4)` | |
| `=f(4)` |
`=> f(4) = 0`
`text(Gradient = – 3 from)\ \ x = 4\ \ text(to)\ \ x = 6`
| `:.\ f(6)` | `= -3 (6\ – 4)` |
| `= -6` |
♦♦ Mean mark 28%
EXAM TIP: Clearly identify THE EXTREMES when given a defined domain. In this case, the origin is obvious graphically, and the other extreme at `x=6`, is CLEARLY LABELLED!
EXAM TIP: Clearly identify THE EXTREMES when given a defined domain. In this case, the origin is obvious graphically, and the other extreme at `x=6`, is CLEARLY LABELLED!
| iv. |  |
Calculus, 2ADV C4 2011 HSC 6c
The diagram shows the graph `y = 2 cos x` .
- State the coordinates of `P`. (1 mark)
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- Evaluate the integral `int_0^(pi/2) 2 cos x\ dx`. (2 marks)
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- Indicate which area in the diagram, `A`, `B`, `C` or `D`, is represented by the integral
`int_((3pi)/2)^(2pi) 2 cos x\ dx`. (1 mark)
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- Using parts (ii) and (iii), or otherwise, find the area of the region bounded by the curve `y = 2 cos x` and the `x`-axis, between `x = 0` and `x = 2pi` . (1 mark)
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- Using the parts above, write down the value of
`int_(pi/2)^(2pi) 2 cos x\ dx`. (1 mark)
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