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Calculus, 2ADV C3 2025 HSC 16

Consider the function  \(f(x)=\dfrac{x^2}{e^x}\).

  1. Find the stationary points of the function and determine their nature.   (4 marks)

  2. A partially completed graph of  \(f(x)=\dfrac{x^2}{e^x}\)  is shown.
  3. Use your answer from part (a) to complete the graph.   (1 mark)


     

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a.   \(\text{MIN at}\ \ (0,0)\)

\(\text{MAX at}\ \ \left(2,\dfrac{4}{e^2}\right)\)

b.   
     

Show Worked Solution
a.     \(f(x)\) \(=\dfrac{x^2}{e^x}\)
  \(f^{\prime}(x)\) \(=2 x \cdot e^x-e^x \cdot x^2\)
    \(=\dfrac{x e^x(2-x)}{e^{2 x}}\)
    \(=\dfrac{x(2-x)}{e^x}\)

 
\(\text{Find} \ x\ \text{when} \ \ f^{\prime}(x)=0:\)

\(x(2-x)=0\)

\(x=0 \ \text {or} \ 2\)

\(\text{When} \ \ x=0 \ \Rightarrow \ f(0)=0\)

\(\text {When}\ \  x=2 \ \Rightarrow \ f(2)=\dfrac{4}{e^2}\)

\(\text {Checking nature of SP’s:}\)

\begin{array}{|c|c|c|c|c|c|}
\hline x & -1 & 0 & 1 & 2 & 3 \\
\hline f^{\prime}(x) & -3 e & 0 & \dfrac{1}{e} & 0 & -\dfrac{3}{e^3} \\
\hline
\end{array}

\(\therefore \ \text{MIN at}\ \ (0,0)\)

\(\quad \ \ \text{MAX at}\ \ \left(2,\dfrac{4}{e^2}\right)\)
 

b.
     

Calculus, 2ADV C3 2023 HSC 30

Let  \(f(x)=e^{-x} \sin x\).

  1. Find the coordinates of the stationary points of \(f(x)\) for  \(0\leq x\leq 2\pi\). You do NOT need to check the nature of the stationary points.  (3 marks)

  2. Without using any further calculus, sketch the graph of  \(y=f(x)\), for  \(0\leq x\leq 2\pi\), showing stationary points and intercepts.  (2 marks)
     


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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.    
         

Show Worked Solution

a.    \(f(x)=e^{-x} \sin x\)

\(f^{′}(x)=e^{-x} \cos x-e^{-x} \sin x = e^{-x}( \cos x-\sin x) \)

\(\text{SPs when}\ f^{′}(x)=0: \)

\(e^{-x}=0\ \ \rightarrow \ \text{no solution} \)

\(\cos x-\sin x\) \(=0\)  
\(1-\tan x\) \(=0\)  
\(\tan\) \(=1\)  
Mean mark (a) 54%.

\(x=\dfrac{\pi}{4}, \dfrac{5\pi}{4} \)
 

\(f(\dfrac{\pi}{4})=e^{-\frac{\pi}{4}}\sin \frac{\pi}{4}=\dfrac{1}{\sqrt2 \times e^{\frac{\pi}{4}}} \)

\(f(\dfrac{5\pi}{4})=e^{-\frac{5\pi}{4}}\sin \frac{5\pi}{4}=\dfrac{-1}{\sqrt2 \times e^{\frac{5\pi}{4}}} \)
 

\(\therefore\ \text{SPs at}\ \Big(\dfrac{\pi}{4},\dfrac{1}{\sqrt2 \times e^{\frac{\pi}{4}}}\Big)\ \text{and}\  \Big(\dfrac{5\pi}{4},\dfrac{-1}{\sqrt2 \times e^{\frac{5\pi}{4}}}\Big)\)

 
b.    \(\ x\text{-intercepts at}\ \ x=0, \pi,\ 2\pi \)
 

 

♦♦ Mean mark (b) 34%.

Calculus, 2ADV C3 2022 HSC 27

Let  `f(x)=xe^(-2x)`.

It is given that  `f^(′)(x)=e^(-2x)-2xe^(-2x)`.

  1. Show that  `f^(″)(x)=4(x-1)e^(-2x)`.  (2 marks)

  2. Find any stationary points of  `f(x)`  and determine their nature.  (2 marks)

  3. Sketch the curve  `y=x e^{-2 x}`, showing any stationary points, points of inflection and intercepts with the axes.  (3 marks)

Show Answers Only
  1. `text{Proof (See Worked Solutions)}`
  2. `text{Max S.P. at}\ \ (1/2, 1/(2e))`
  3.  
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`,

  1. State the domain and range of `f(x)`.  (1 mark)

  2. i.  Find the equation of the tangent to the graph of  `f(x)` at  `(4, 2)`.  (2 marks)

     

    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)
     

Show Answers Only
  1. `x >3`

     

    `y in R`

  2. i.  `y = x-2`

     

    ii. `text(See Worked Solutions)`

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`.

  1. Show that the graph of  `y = f(x)`  has a stationary point at  `x = e`.  (2 marks)

  2. By considering the gradient on either side of  `x = e`, or otherwise, show that the stationary point is a maximum.  (1 mark)

  3. 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)

Show Answers Only
  1. `text(Proof)\ \ text{(See Worked Solutions)}`
  2. `text(Proof)\ \ text{(See Worked Solutions)}`
  3. `text(Proof)\ \ text{(See Worked Solutions)}`
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.   

Geometry and Calculus, 2UA 2004 HSC 9c Answer

`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`

  1. Show that the graph of  `y = f(x)`  has no turning points.   (2 marks)

  2. Find the point of inflection of  `y = f(x)`.     (1 mark)

  3. i. Show that `1 - x + x^2 - 1/(1 + x) = (x^3)/(1 + x)`  for  `x != -1`.   (1 mark)

     

    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)

  1. Sketch the graphs of  `y = f(x)`  and  `y = g(x)`  for  `x >= 0`.    (2 marks)

  2. Show that  `d/(dx) [(1 + x) ln (1 + x) - (1 + x)] = ln (1 + x)`.   (2 marks)

  3. Find the area enclosed by the graphs of  `y = f(x)`  and  `y = g(x)`, and the straight line  `x = 1`.   (2 marks)

Show Answers Only
  1. `text{Proof  (See Worked Solutions)}`
  2. `(1/2, 5/12)`
  3. i. `text{Proof  (See Worked Solutions)}`

     

    ii. `text{Proof  (See Worked Solutions)}` 

  4.  
        Geometry and Calculus, 2UA 2009 HSC 10 Answer
  5. `text{Proof  (See Worked Solutions)}`
  6. `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. 

Geometry and Calculus, 2UA 2009 HSC 10 Answer
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)`