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Calculus, 2ADV C4 2023 HSC 32

The curves  \(y=e^{-2 x}\)  and  \(y=e^{-x}-\dfrac{1}{4}\)  intersect at exactly one point as shown in the diagram. The point of intersection has coordinates \(\left(\ln 2, \dfrac{1}{4}\right)\). (Do NOT prove this.)
 

  1. Show that the area bounded by the two curves and the \(y\)-axis, as shaded in the diagram, is  \(\dfrac{1}{4} \ln 2-\dfrac{1}{8}\).  (3 marks)

  2. Find the values of \(k\) such that the curves  \(y=e^{-2 x}\)  and  \(y=e^{-x}+k\)  intersect at two points.  (3 marks)

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a.    \(\text{See Worked Solutions}\)

b.    \( -\dfrac{1}{4} < k < 0 \)

Show Worked Solution

a.     \(A\) \(= \int_0^{\ln2} e^{-2x}-(e^{-x}-\dfrac{1}{4})\ dx\)
    \(=\Big{[}-\dfrac{1}{2} e^{-2x}+e^{-x}+\dfrac{1}{4}x \Big{]}_0^{\ln2} \)
    \(=\Big{[}(-\dfrac{1}{2} e^{-2\ln2}+e^{-\ln2}+\dfrac{1}{4}\ln2)-(-\dfrac{1}{2}e^0+e^0-0)\Big{]} \)
    \(=\Big{[}(-\dfrac{1}{2} e^{\ln{(2^{-2})}}+e^{\ln{(2^{-1})}}+\dfrac{1}{4}\ln2+\dfrac{1}{2}-1)\Big{]} \)
    \(=\Big{[}(-\dfrac{1}{2} e^{\ln \frac{1}{4}}+e^{\ln \frac{1}{2}}+\dfrac{1}{4}\ln2-\dfrac{1}{2}\Big{]} \)
    \(=-\dfrac{1}{2} \times \dfrac{1}{4} +\dfrac{1}{2}+\dfrac{1}{4}\ln2-\dfrac{1}{2} \)
    \(=\dfrac{1}{4}\ln2-\dfrac{1}{8} \)

 
b.    \(\text{Intersection occurs when}\)

\(e^{-2x}\) \(=e^{-x}+k\)  
\(e^{-2x}-e^{-x}-k\) \(=0\)  

 
\(\text{Let}\ X=e^{-x} \)

\(X^2-X-k=0 \)

\(X\) \(=\dfrac{1\pm \sqrt{1^2-4(1)(-k)}}{2} \)  
  \(=\dfrac{1\pm \sqrt{1+4k}}{2} \)  

 

\(\text{2 solutions}\ \Rightarrow\ \Delta >0 \)

\(1+4k>0\ \ \Rightarrow \ k>-\dfrac{1}{4} \)

\(\text{Since}\ X=e^{-x} >0 \)

\(\dfrac{1-\sqrt{1+4k}}{2}\) \(>0\)  
\(1-\sqrt{1+4k}\) \(>0\)  
\(\sqrt{1+4k}\) \(<1\)  
\(k\) \(<0\)  

 
\(\therefore\ -\dfrac{1}{4} < k < 0 \)

♦♦♦ Mean mark (b) 14%.

Calculus, 2ADV C4 2021 HSC 28

The region bounded by the graph of the function  `f(x) = 8 - 2^x`  and the coordinate axes is shown
 

  1. Show that the exact area of the shaded region is given by  `24 - 7/ln2`.  (3 marks)

  2. A new function  `g(x)`  is found by taking the graph of   `y = -f(-x)`  and translating it by 5 units to the right.
  3. Sketch the graph of  `y = g(x)`  showing the `x`-intercept and the asymptote.  (2 marks)

  4. Hence, find the exact value of  `int_2^5 g(x)\ dx`.  (1 mark)

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  1. `text(See Worked Solution)`
  2.  
  3. `7/(ln2) – 24`
Show Worked Solution

a.   `xtext(-intercept occurs when)`

`8 – 2^x = 0 \ => \ x = 3`

`text(Area)` `= int_0^3 8 – 2^x\ dx`
  `= [8x – (2^x)/(ln2)]_0^3`
  `= 24 – 8/(ln 2) – (0 – 1/(ln2))`
  `= 24 – 8/(ln2) + 1/(ln2)`
  `= 24 – 7/(ln2)\ \ text(u²)`

♦ Mean mark part (b) 48%.

b.

`y = f(–x) -> text(reflect)\ \ y = f(x)\ \ text(in the)\ ytext(-axis)`

`y = -f(–x) -> text(reflect)\ \ y = f(–x)\ \ text(in the)\ xtext(-axis)`
 

♦♦♦ Mean mark part (c) 13%.

c.   `int_2^5 g(x)\ dx\ \ text{is the same area as found in part (a)}`

`text(except it is below the)\ xtext(-axis.)`

`:. int_2^5 g(x)\ dx = 7/(ln2) – 24`

Calculus, 2ADV C4 2019 HSC 16c

The diagram shows the region  `R`, bounded by the curve  `y = x^r`, where  `r >= 1`, the `x`-axis and the tangent to the curve at the point  `(1, 1)`.
 

  1. Show that the tangent to the curve at  `(1, 1)`  meets the `x`-axis at
     
         `qquad ((r - 1)/r, 0)`.  (2 marks)

  2. Using the result of part (i), or otherwise, show that the area of the region  `R`  is
     
         `qquad (r - 1)/(2r (r + 1))`.  (2 marks)

  3. Find the exact value of  `r`  for which the area of  `R`  is a maximum.  (3 marks)

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  1. `text(Proof)\ text{(See Worked Solutions)}`
  2. `text(Proof)\ text{(See Worked Solutions)}`
  3. `r = 1 + sqrt 2`
Show Worked Solution
i.    `y` `= x^r`
  `(dy)/(dx)` `= r x^(r – 1)`

 
`text(When)\ \ x = 1, \ (dy)/(dx) = r`

♦♦ Mean mark part (i) 31%.

`text(Equation of tangent:)`

`y – 1` `= r(x – 1)`
`y` `= rx – r + 1`

 
`text(When)\ \ y = 0:`

`rx – r + 1` `= 0`
`rx` `= r – 1`
`x` `= (r – 1)/r`

 
`:.\ text(T)text(angent meets)\ x text(-axis at)\ \ ((r – 1)/r, 0)`

 

ii.   `text(Area under curve)`

♦♦♦ Mean mark part (ii) 21%.

`= int_0^1 x^r`

`= [1/(r + 1) ⋅ x^(r + 1)]_0^1`

`= 1/(r + 1) xx 1^(r + 1) – 0`

`= 1/(r + 1)`

 
`text(Area under tangent)`

`= 1/2 xx b xx h`

`= 1/2 (1 – (r – 1)/r) xx 1`

`= 1/2 (1 – (r – 1)/r)`
 

`:. R` `= 1/(r + 1) – 1/2(1 – (r – 1)/r)“
  `= 1/(r + 1) – 1/(2r) [r – (r – 1)]`
  `= 1/(r + 1) – 1/(2r)`
  `= (2r – (r + 1))/(2r(r + 1))`
  `= (r – 1)/(2r(r + 1))`

 

iii.    `R` `= (r – 1)/(2r(r + 1)) = (r – 1)/(2r^2 + 2r)`
  `(dR)/(dr)` `= ((2r^2 + 2r) xx 1 – (r – 1)(4r + 2))/(2r^2 + 2r)^2`
    `= (2r^2 + 2r – 4r^2 – 2r + 4r + 2)/(2r^2 + 2r)^2`
    `= (-2r^2 + 4r + 2)/(2r^2 + 2r)^2`
    `= (-2(r^2 – 2r – 1))/(2r^2 + 2r)^2`

 

`text(Find)\ \ r\ \ text(when)\ \ (dR)/(dr) = 0:`

♦♦ Mean mark part (iii) 23%.

`r^2 – 2r – 1 = 0`

`r` `= (2 +- sqrt((-2)^2 – 4 xx 1 xx (-1)))/2 `
  `= (2 +- sqrt 8)/2`
  `= 1 + sqrt 2\ \ (r >= 1)`

 

  `qquadr qquad` `qquad 1 qquad` `\ \ 1 + sqrt 2\ \ ` `qquad 3 qquad`
  `(dR)/(dr)` `1/4` `0` `-1/144`

 

`:. R_text(max)\ text(occurs when)\ \ r = 1 + sqrt 2`

Calculus, 2ADV C4 SM-Bank 12

Let  `f(x) = 2e^(-x/5)\ \ \ text(for)\ \ x>=0`

A right-angled triangle `OQP` has vertex `O` at the origin, vertex `Q` on the `x`-axis and vertex `P` on the graph of  `f`, as shown. The coordinates of `P` are  `(x, f(x)).`
 

 vcaa-2013-meth-10
 

  1. Find the area, `A`, of the triangle `OPQ` in terms of `x`.  (1 mark)

  2. Find the maximum area of triangle `OQP` and the value of `x` for which the maximum occurs.  (3 marks)

  3. Let `S` be the point on the graph of  `f` on the `y`-axis and let `T` be the point on the graph of  `f` with the `y`-coordinate `1/2`.Find the area of the region bounded by the graph of  `f` and the line segment `ST`.  (2 marks)

     

 

      vcaa-2013-meth-10i

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  1. `x e^(-x/5)`
  2. `5/e\ text(u²)`
     
  3. `25/4 log_e (4) – 15/2\ text(u²)`
Show Worked Solution
i.   `text(Area)` `= 1/2 xx b xx h`
    `= 1/2x(2e^(-x/5))`
  `:. A` `= xe^(-x/5)`

 

ii.   `text(Stationary point when)\ \ (dA)/(dx) = 0,`

♦ Mean mark (Vic) 35%.
`x(-1/5 e^(-x/5)) + e^(-x/5)` `= 0`
`e^(-x/5)(1 – x/5)` `= 0`
`:. x` `= 5\ \ \ \ (e^(-x/5) >0,\ \ text(for all)\ x)`

 

`text(When)\ \ x = 5,\ \ A` `= xe^(-x/5)= 5e^-1`
   

`:. A_max = 5/e\ text(u²,   when)\ \ x = 5`

 

iii.   `text(Find)\ \ S:\ F(0) = 2`

`=>S(0, 2)`

♦♦ Mean mark (Vic) 32%.

 

`text(Find)\ \ T:\ \ \ ` `2e^(-x/5)` `= 1/2`
  `e^(-x/5)` `= 1/4`
  `-x/5` `= log_e (1/4)`
  `x` `= 5 log_e (4)`

`=> T(5log_e(4), 1/2)`

 

vcaa-2013-meth-10ii

`:.\ text(Area)` `= text(Area)\ SOAT – int_0^(5 log_e(4)) (2e^(-x/5)) dx`
  `=1/2h(a+b) + 10 [e^(-x/5)]_0^(5 log_e (4))`
  `= 5/2 log_e (4) (2 + 1/2) + 10 [e^(-log_e (4)) – e^0]`
  `= 25/4 log_e (4) +10 (1/4 – 1)`
  `= 25/4 log_e (4) – 15/2\ text(u²)`

Calculus, 2ADV C4 2010 HSC 4b

The curves  `y=e^(2x)`  and  `y=e^-x`  intersect at the point `(0,1)`  as shown in the diagram.
 

2010 4b
  

Find the exact area enclosed by the curves and the line  `x=2`.          (3 marks)

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`1/2e^4+e^-2-3/2\ \ text(u²)`

Show Worked Solutions
MARKER’S COMMENT: The best responses used only a single integral before any substitution as shown in Worked Solutions.
`text(Area)` `=int_0^2e^(2x)\ \ dx-int_0^2 e^-x\ \ dx`
  `=int_0^2(e^(2x)-e^-x)dx`
  `=[1/2e^(2x)+e^-x]_0^2`
  `=[(1/2e^4+e^-2)-(1/2e^0+e^0)]`
  `=1/2e^4+e^-2-3/2\ \ \ text(u²)`