Let \(f(x)=e^{-x} \sin x\). --- 6 WORK AREA LINES (style=lined) --- --- 0 WORK AREA LINES (style=lined) --- a. \( \Big(\dfrac{\pi}{4},\dfrac{1}{\sqrt2 \times e^{\frac{\pi}{4}}}\Big)\ \text{and}\ \Big(\dfrac{5\pi}{4},\dfrac{-1}{\sqrt2 \times e^{\frac{5\pi}{4}}}\Big)\) b. a. \(f(x)=e^{-x} \sin x\) \(f^{′}(x)=e^{-x} \cos x-e^{-x} \sin x = e^{-x}( \cos x-\sin x) \) \(\text{SPs when}\ f^{′}(x)=0: \) \(e^{-x}=0\ \ \rightarrow \ \text{no solution} \) \(x=\dfrac{\pi}{4}, \dfrac{5\pi}{4} \) \(f(\dfrac{\pi}{4})=e^{-\frac{\pi}{4}}\sin \frac{\pi}{4}=\dfrac{1}{\sqrt2 \times e^{\frac{\pi}{4}}} \) \(f(\dfrac{5\pi}{4})=e^{-\frac{5\pi}{4}}\sin \frac{5\pi}{4}=\dfrac{-1}{\sqrt2 \times e^{\frac{5\pi}{4}}} \) \(\therefore\ \text{SPs at}\ \Big(\dfrac{\pi}{4},\dfrac{1}{\sqrt2 \times e^{\frac{\pi}{4}}}\Big)\ \text{and}\ \Big(\dfrac{5\pi}{4},\dfrac{-1}{\sqrt2 \times e^{\frac{5\pi}{4}}}\Big)\)
\(\cos x-\sin x\)
\(=0\)
\(1-\tan x\)
\(=0\)
\(\tan\)
\(=1\)
b. \(\ x\text{-intercepts at}\ \ x=0, \pi,\ 2\pi \)
Calculus, 2ADV C3 2022 HSC 27
Let `f(x)=xe^(-2x)`.
It is given that `f^(′)(x)=e^(-2x)-2xe^(-2x)`.
- Show that `f^(″)(x)=4(x-1)e^(-2x)`. (2 marks)
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- Find any stationary points of `f(x)` and determine their nature. (2 marks)
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- Sketch the curve `y=x e^{-2 x}`, showing any stationary points, points of inflection and intercepts with the axes. (3 marks)
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Show Answers Only
- `text{Proof (See Worked Solutions)}`
- `text{Max S.P. at}\ \ (1/2, 1/(2e))`
-
Show Worked Solution
a. `f^(′)(x)=e^(-2x)-2xe^(-2x)`
| `f^(″)(x)` | `=-2e^(-2x)-[2x(-2)e^(-2x)+2e^(-2x)]` | |
| `=-2e^(-2x)+4xe^(-2x)-2e^(-2x)` | ||
| `=4xe^(-2x)-4e^(-2x)` | ||
| `=4(x-1)e^(-2x)\ \ text{… as required}` |
b. `text{S.P.’s occur when}\ \ f^(′)(x)=0`
| `e^(-2x)-2xe^(-2x)` | `=0` | |
| `e^(-2x)(1-2x)` | `=0` |
`e^(-2x)=0\ \ =>\ \ text{No solution}`
`1-2x=0\ \ =>\ \ x=1/2`
| `f^(″)(1/2)` | `=4(1/2-1)e^(-2xx1/2)` | |
| `=-2e^(-1)<0\ \ =>\ text{MAX}` |
`f(1/2)=1/2e^(-1)=1/(2e)`
c. `text{POI occurs when}\ \ f^(″)(x)=0`
`f^(″)(x)=4(x-1)e^(-2x)=0\ \ =>\ \ x=1`
`text{Test for change in concavity:}`
`f^(″)(0)=4(0-1)e^0=-4`
`f^(″)(2)=4(2-1)e^(-4)>0\ \ =>\ text{Concavity changes}`
`:.\ text{POI exists at}\ \ (1,1/e^2)`
`text{As}\ \ x->oo\ \ =>\ \ y->0^+`
Mean mark part (c) 54%.
Calculus, 2ADV C3 2019 MET1 4
Given the function `f(x) = log_e (x-3) + 2`,
- State the domain and range of `f(x)`. (1 mark)
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- i. Find the equation of the tangent to the graph of `f(x)` at `(4, 2)`. (2 marks)
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ii. On the axes below, sketch the graph of the function `f(x)`, labelling any asymptote with its equation.
Also draw the tangent to the graph of `f(x)` at `(4, 2)`. (4 marks)
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Show Worked Solution
| a. | `text(Domain)` | `: \ x > 3` |
| `text(Range)` | `: \ y in R` |
| b.i. | `g(x)` | `= log_e (x-3) + 2` |
| `g^{\prime}(x)` | `= 1/(x-3)` | |
| `g^{\prime}(4)` | `= 1` |
`text(Equation of tangent),\ m = 1\ \ text(through)\ (4, 2):`
| `y-2` | `= 1(x-4)` |
| `y` | `= x-2` |
| b.ii. |  |
Calculus, 2ADV C3 2004 HSC 9c
Consider the function `f(x) = (log_e x)/x`, for `x > 0`.
- Show that the graph of `y = f(x)` has a stationary point at `x = e`. (2 marks)
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- By considering the gradient on either side of `x = e`, or otherwise, show that the stationary point is a maximum. (1 mark)
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- Use the fact that the maximum value of `f(x)` occurs at `x = e` to deduce that `e^x ≥ x^e` for all `x > 0`. (2 marks)
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Show Worked Solution
i. `f(x) = (log_e x)/x, text(for)\ x > 0`
`text(Using the quotient rule,)`
| `f′(u/v)` | `=(u′v-uv′)/v^2` |
| `f′(x)` | `= (x · d/(dx) (log_e x) − log_e x · d/(dx) (x))/(x^2)` |
| `= (x(1/x) − log_e x · 1)/(x^2)` | |
| `= (1 − log_e x)/(x^2)` |
`text(SP’s occur when)\ \ f′(x)=0`
| `(1 − log_e x)/(x^2)` | `= 0` |
| `1 − log_e x` | `= 0` |
| `log_e x` | `= 1` |
| `:. x` | `= e` |
`:. text(There is a stationary point at)\ \ x = e.`
| ii. |
|
`text(When)\ \ x < e, log_e x<1\ \ \ text{(from graph)}`
| `1 − log_e x` | `> 0` |
| `(1 − log_e x)/(x^2)` | `> 0` |
| `:. f′(x)` | `> 0` |
`text(When)\ x > e, log_e x>1\ \ \ text{(from graph)}`
| `1 − log_e x` | `< 0` |
| `(1 − log_e x)/(x^2)` | `< 0` |
| `:. f′(x)` | `< 0` |
`:.\ text(The stationary point at)\ \ x = e\ \ text(is a maximum.)`
| iii. | `f(e)` | `= (log_e x)/e` |
| `= 1/e` |
`:. f(x)\ \ text(has a maximum value at)\ \ (e,1/e)`
MARKER’S COMMENT: Very challenging for most students. Successful students recognised the link to part (ii).
| `:. (log_e x)/x` | `≤ 1/e` |
| `log_e x` | `≤ x/e` |
| `e log_e x` | `≤ x` |
| `log_e x^e` | `≤ x` |
| `e^(log_e x^e)` | `≤ e^x` |
| `x^e` | `≤ e^x\ \ \ text{(using}\ e^(log x) = x)` |
| `:. e^x` | `≥ x^e \ \ \ text{(for}\ \ x > 0text{)}` |
Calculus, 2ADV C3 2009 HSC 10
`text(Let)\ \ f(x) = x - (x^2)/2 + (x^3)/3`
- Show that the graph of `y = f(x)` has no turning points. (2 marks)
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- Find the point of inflection of `y = f(x)`. (1 mark)
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- i. Show that `1 - x + x^2 - 1/(1 + x) = (x^3)/(1 + x)` for `x != -1`. (1 mark)
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ii. Let `g(x) = ln (1 + x)`.
Use the result in part c.i. to show that `f prime (x) >= g prime (x)` for all `x >= 0`. (2 marks)
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- Sketch the graphs of `y = f(x)` and `y = g(x)` for `x >= 0`. (2 marks)
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- Show that `d/(dx) [(1 + x) ln (1 + x) - (1 + x)] = ln (1 + x)`. (2 marks)
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- Find the area enclosed by the graphs of `y = f(x)` and `y = g(x)`, and the straight line `x = 1`. (2 marks)
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Show Answers Only
- `text{Proof (See Worked Solutions)}`
- `(1/2, 5/12)`
- i. `text{Proof (See Worked Solutions)}`
ii. `text{Proof (See Worked Solutions)}`
-
- `text{Proof (See Worked Solutions)}`
- `1 5/12\ – 2ln2\ \ text(u²)`
Show Worked Solution
| a. | `f(x) = x\ – (x^2)/2 + (x^3)/3` |
♦♦ Mean mark 28% for all of Q10 (note that data for each question part is not available).
`text(Turning points when)\ f prime (x) = 0`
`f prime (x) = 1\ – x + x^2`
`x^2\ – x + 1 = 0`
| `text(S)text(ince)\ \ Delta` | `= b^2\ – 4ac` |
| `= (–1)^2\ – 4 xx 1 xx 1` | |
| `= -3 < 0 => text(No solution)` |
`:.\ f(x)\ text(has no turning points)`
| b. | `text(P.I. when)\ f″(x) = 0` |
| `f″(x)` | `= -1 + 2x = 0` |
| `2x` | `= 1` |
| `x` | `= 1/2` |
`text(Check for change in concavity)`
| `f″(1/4)` | `= -1/2 < 0` |
| `f″(3/4)` | `= 1/2 > 0` |
`=>\ text(Change in concavity)`
`:.\ text(P.I. at)\ \ x = 1/2`
| `f(1/2)` | `= 1/2\ – ((1/2)^2)/2 + ((1/2)^3)/3` |
| `= 1/2\ – 1/8 + 1/24` | |
| `= 5/12` |
`:.\ text(Point of Inflection at)\ (1/2, 5/12)`
| c.i. | `text(Show)\ 1\ – x + x^2\ – 1/(1 + x) = (x^3)/(1 + x),\ \ \ x != -1` | |
| `text(LHS)` | `= (1+x)/(1+x)\ – (x(1+x))/(1+x) + (x^2(1+x))/((1+x))\ – 1/(1+x)` |
| `= (1 + x\ – x\ – x^2 + x^2 + x^3\ – 1)/(1+x)` | |
| `= (x^3)/(1+x)\ \ \ text(… as required)` |
| c.ii. | `text(Let)\ g(x) = ln(1+x)` |
| `g prime (x) = 1/(1 + x)` |
| `f prime (x)\ – g prime (x)` | `= 1\ – x + x^2\ – 1/(1+x)` |
| `= (x^3)/(1 + x)\ \ text{(using part (i))}` |
`text(S)text(ince)\ (x^3)/(1 + x) >= 0\ text(for)\ x >= 0`
| `f prime (x)\ – g prime (x) >= 0` |
| `f prime (x) >= g prime (x)\ text(for)\ x >= 0` |
MARKER’S COMMENT: When 2 graphs are drawn on the same set of axes, you must label them.
| d. |
|
| e. | `text(Show)\ d/(dx) [(1 + x) ln (1 + x)\ – (1 + x)] = ln (1 + x)` |
| `text(Using)\ d/(dx) uv=uv′+vu′` |
| `text(LHS)` | `= (1+x) xx 1/(1 + x) + ln(1+x)xx1 + – 1` |
| `= 1+ ln(1+x)\ – 1` | |
| `= ln(1+x)` | |
| `=\ text(RHS … as required)` |
| f. | `text(Area)` | `= int_0^1 f(x)\ – g(x)\ dx` |
| `= int_0^1 (x\ – (x^2)/2 + (x^3)/3\ – ln(x+1))\ dx` | ||
| `= [x^2/2\ – x^3/6 + (x^4)/12\ – (1 + x) ln (1+x) + (1+x)]_0^1` | ||
| `text{(using part (e) above)}` | ||
| `= [(1/2 – 1/6 + 1/12 – (2)ln2 + 2) – (ln1 + 1)]` | ||
| `= 5/12\ – 2ln2 + 2\ – 1` | ||
| `= 1 5/12\ – 2 ln 2\ \ text(u²)` |