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Algebra, MET1 2025 VCAA 9

Consider the functions

\(f: R \backslash\{1\} \rightarrow R, f(x)=\dfrac{w^2}{(x-1)^2}\)

and

\(g: R \rightarrow R, g(x)=(x-w)^2\)

where  \(w \in R\).

  1. If  \(w=-3\), find the four solutions to  \(f(x)=g(x)\).   (3 marks)

  2. Consider the case where  \(w>0\).
    1. Find, in terms of \(w\), the coordinates of the minimum point of the graph of  \(y=(x-1)(x-w)\).   (2 marks)

    2. Hence, or otherwise, find the positive values of \(w\) for which \(f(x)=g(x)\) has exactly three solutions.   (2 marks)

Show Answers Only

a.    \(\text{If} \ \ w=-3, \text{find 4 solutions to} \ \ f(x)=g(x):\)

\(\dfrac{(-3)^2}{(x-1)^2}=(x+3)^2\)

\((x+3)^2(x-1)^2=9\)

\((x+3)(x-1)= \pm 3\)
 

\(\text{Case 1}\)

\(x^2+2 x-3=3\)

\(x^2+2 x-6=0\)

\(x=-1 \pm \sqrt{7}\)
 

\(\text{Case 2}\)

\(x^2+2 x-3\) \(=-3\)
\(x(x+2)\) \(=0\)
\(x=0 \ \ \text{or} \ -2\)

 

b.i.   \(\text{Min TP at} \ \left(\dfrac{w+1}{2}, -\dfrac{1}{4} (w-1)^2\right)\)

b.ii.  \(\text{See worked solutions.}\)

Show Worked Solution

a.    \(\text{If} \ \ w=-3, \text{find 4 solutions to} \ \ f(x)=g(x):\)

\(\dfrac{(-3)^2}{(x-1)^2}=(x+3)^2\)

\((x+3)^2(x-1)^2=9\)

\((x+3)(x-1)= \pm 3\)
 

\(\text{Case 1}\)

\(x^2+2 x-3=3\)

\(x^2+2 x-6=0\)

\(x=-1 \pm \sqrt{7}\)

♦♦ Mean mark (a) 34%.

\(\text{Case 2}\)

\(x^2+2 x-3\) \(=-3\)
\(x(x+2)\) \(=0\)
\(x=0 \ \ \text{or} \ -2\)

 

b.i.  \(\text{Find minimum TP of} \ \ y=(x-1)(x-w)\)

\(\text{Axis of quadratic:}\ \ x=\dfrac{w+1}{2},\ \ (w>0)\)

\(y\) \(=\left(\dfrac{w+1}{2}-1\right)\left(\dfrac{w+1}{2}-w\right)\)
  \(=\left(\dfrac{w-1}{2}\right)\left(\dfrac{1-w}{2}\right)\)
  \(=-\dfrac{1}{4}(w-1)^2\)

 

\(\therefore \ \text{Min TP at} \ \left(\dfrac{w+1}{2}, -\dfrac{1}{4} (w-1)^2\right)\)

♦♦ Mean mark (b.i) 39%.
♦♦♦ Mean mark (b.ii) 11%.
 

b.ii.   \(\text{Solve for \(w\ (w>0)\) where \(f(x)=g(x)\):}\)

\(\dfrac{w^2}{(x-1)^2}\) \(=(x-w)^2\)
\((x-w)^2(x-1)^2\) \(=w^2\)
\([(x-w)(x-1)]^2-w^2\) \(=0\)

 

\(\text{Difference between two squares:}\)

\([(x-w)(x-1)-w][(x-w)(x-1)+w]=0\)
 

\(\text{Case 1}\)

\((x-w)(x-1)-w\) \(=0\)
\(x^2-x-w x+w-w\) \(=0\)
\(x^2-(w+1) x\) \(=0\)
\(x[x-(w+1)]\) \(=0\)

 

\(x=0 \ \ \text{or} \ \ w+1\)
 

\(\text{Case 2}\)

\((x-w)(x-1)+w\) \(=0\)
\(x^2-(w+1) x+2 w\) \(=0\)

 

\(\text{1 solution if} \ \ \Delta=0:\)

\(\Delta\) \(=[-(w+1)]^2-4 \times 1 \times 2 w\)
  \(=w^2+2 w+1-8 w\)
  \(=w^2-6 x+1\)

 

\(\text{If} \ \ \Delta=0:\)

\(w\) \(=\dfrac{6 \pm \sqrt{(-6)^2-4 \times 1 \times 1}}{2}\)
  \(=\dfrac{6 \pm \sqrt{32}}{2}\)
  \(=3+2 \sqrt{2}\ \ (\text{reject}\ w=3-2 \sqrt{2}\ \ \text{as}\ \ w>0)\)

 

\(\therefore f(x)=g(x) \ \ \text{has exactly 3 solutions}\ (w>0).\)

Functions, MET2 2024 VCAA 1

Consider the function  \( f: R \rightarrow R, f(x)=(x+1)(x+a)(x-2)(x-2 a) \text { where } a \in R \text {. } \)

  1. State, in terms of \(a\) where required, the values of \(x\) for which \(f(x)=0\).  (1 mark
  1. Find the values of \(a\) for which the graph of \(y=f(x)\) has
      
     i. exactly three \(x\)-intercepts.   (2 marks)

    ii. exactly four \(x\)-intercepts.   (1 mark)

  1. Let \(g\) be the function \(g: R \rightarrow R, g(x)=(x+1)^2(x-2)^2\), which is the function \(f\) where \(a=1\).
      
      i. Find \(g^{\prime}(x)\)   (1 mark)

     ii. Find the coordinates of the local maximum of \(g\).   (1 mark)

    iii. Find the values of \(x\) for which \(g^{\prime}(x)>0\).   (1 mark)

     iv. Consider the two tangent lines to the graph of \(y=g(x)\) at the points where
    \(x=\dfrac{-\sqrt{3}+1}{2}\) and \(x=\dfrac{\sqrt{3}+1}{2}\). Determine the coordinates of the point of intersection of these two tangent lines.   (2 marks)

  1. Let \(g\) remain as the function \(g: R \rightarrow R, g(x)=(x+1)^2(x-2)^2\), which is the function \(f\) where \(a=1\).

    Let \(h\) be the function \(h: R \rightarrow R, h(x)=(x+1)(x-1)(x+2)(x-2)\), which is the function \(f\) where \(a=-1\).
      
     i. Using translations only, describe a sequence of transformations of \(h\), for which its image would have a local maximum at the same coordinates as that of \(g\).   (1 mark)

    ii. Using a dilation and translations, describe a different sequence of transformations of \(h\), for which its image would have both local minimums at the same coordinates as that of \(g\).   (2 marks)

Show Answers Only

a.    \(x=-1, x=a, x=2, x=2a\)

bi.  \(a=0, -2, -\dfrac{1}{2}\)

bii. \(R\ \backslash\left\{ -2, -\dfrac{1}{2}, 0, 1\right\}\)

ci.  \(g^{\prime}(x)=2(x-2)(x+1)(2x-1)\)

cii. \(\left(\dfrac{1}{2} , \dfrac{81}{16}\right)\)

ciii. \(x\in\left(-1, \dfrac{1}{2}\right)\cup (2, \infty)\)

civ. \(\left(\dfrac{1}{2}, \dfrac{27}{4}\right)\)

di.   \(\text{Translate }\dfrac{1}{2}\ \text{unit to the right and }\dfrac{81}{16}-4=\dfrac{17}{16}\ \text{units upwards.}\)

dii.  \(\text{Combination is a dilation of }h(x)\ \text{by a factor of}\ \dfrac{3}{\sqrt{10}}\ \text{followed by a }\)

\(\text{translation of }\dfrac{1}{2} \ \text{a unit to the right and an upwards translation of}\ \dfrac{9}{4} \ \text{units}\)

Show Worked Solution

a.    \(x=-1, x=a, x=2, x=2a\)

bi.  \(a=0, -2, -\dfrac{1}{2}\)

bii. \(\text{The solution must be all }R\ \text{except those that give 3 or less solutions.}\)

\(\therefore\ R\ \backslash\left\{ -2, -\dfrac{1}{2}, 0, 1\right\}\)

ci.    \(g^{\prime}(x)\) \(=2(2-x)(x+1)^2+(x-2)^22(x+1)\)
    \(=2(x-2)(x+1)(x+1+x+2)\)
    \(=2(x-2)(x+1)(2x-1)\)

  
cii.  \(\text{When  }g^{\prime}(x)=0, x=2, -1, \dfrac{1}{2}\)

\(\text{From graph local maximum occurs when }x=\dfrac{1}{2}\)

\(g\left(\dfrac{1}{2}\right)\) \(=\left(\dfrac{1}{2}+1\right)^2\left(\dfrac{1}{2}-2\right)^2\)
  \(=\dfrac{9}{4}\times \dfrac{9}{4}=\dfrac{81}{16}\)

  
\(\therefore\ \text{Local maximum at}\ \left(\dfrac{1}{2} , \dfrac{81}{16}\right)\)

ciii. \(\text{From graph}\ g^{\prime}(x)>0\ \text{when }x\in\left(-1, \dfrac{1}{2}\right)\cup (2, \infty)\)

civ.  \(\text{Use CAS to find tangent lines and solve to find intersection.}\)

\(\text{Point of intersection of tangent lines}\ \left(\dfrac{1}{2}, \dfrac{27}{4}\right)\)

di.  \(\text{Local maximum of }g(x)\ \rightarrow\left(\dfrac{1}{2}, \dfrac{81}{16}\right)\)

\(\text{From CAS local maximum of }h(x)\ \rightarrow \left(0, 4\right)\)

\(\therefore\ \text{Translate }\dfrac{1}{2}\ \text{unit to the right and }\dfrac{81}{16}-4=\dfrac{17}{16}\ \text{units upwards.}\)

dii. \(\text{Using CAS to solve }g^{\prime}(x)=0\ \text{and }h^{\prime}(x)=0\)

\(\text{Local Minimums for }g(x)\ \text{at }(-1, 0)\ \text{and }(2, 0)\ \text{which are 3 apart.}\)

\(\text{Local minimums for at }h(x)\ \text{at }\left(-\sqrt{\dfrac{5}{2}}, -\dfrac{9}{4}\right)\ \text{and }\left(\sqrt{\dfrac{5}{2}}, -\dfrac{9}{4}\right)\)

\(\therefore\ \text{Combination is a dilation of }h(x)\ \text{by a factor of}\ \dfrac{3}{\sqrt{10}}\ \text{followed by a }\)

\(\text{translation of }\dfrac{1}{2} \ \text{a unit to the right and an upwards translation of}\ \dfrac{9}{4} \ \text{units.}\)

 

Functions, MET2 2022 VCAA 5 MC

The largest value of `a` such that the function `f:(-\infty, a] \rightarrow R, f(x)=x^2+3 x-10`, where `f` is one-to-one, is

  1. `-12.25`
  2. `-5`
  3. `-1.5`
  4. `0`
  5. `2`
Show Answers Only

`C`

Show Worked Solution
`f(x)` `= x^2+3 x-10`  
`f^{\prime}(x)` `= 2x + 3`  

  
Turning point when `f^{\prime}(x) = 0`

`\ 2x+3 = 0  \rightarrow  x = -3/2`

`:.\ a = – 1.5`

`=>C`

Calculus, MET2 2023 VCAA 14 MC

A polynomial has the equation  \(y=x(3x-1)(x+3)(x+1)\).

The number of tangents to this curve that pass through the positive \(x\)-intercept is

  1. 0
  2. 1
  3. 2
  4. 3
  5. 4
Show Answers Only

\(D\)

Show Worked Solution

\(\text{Positive}\ x\text{-intercept occurs at}\ \Big(\dfrac{1}{3}, 0\Big) \)

\(\text{Find tangent line at}\ \ x=a\ \text{(by CAS):}\)

\(y_{\text{tang}}=\Big(12a^3+33a^2+10a-3\Big)x-a^2\Big(9a^2+22a+5\Big)\)

\(\text{Solve}\ \ \ y_{\text{tang}}\Bigg(\dfrac{1}{3}\Bigg)=0\ \text{for }a:\)

\(a=\dfrac{-\sqrt{7}-4}{3},\ a=\dfrac{\sqrt{7}-4}{3}\text{ or}\ a=\dfrac{1}{3}\)

\(\therefore\ \text{3 solutions exist.}\)

\(\Rightarrow D\)

\(\text{NOTE: Graphical methods could also be used}\)


♦♦♦ Mean mark 29%.

Algebra, MET2 2023 VCAA 2 MC

For the parabola with equation  \(y=ax^2+2bx+c\), where \(a, b, c \in R\), the equation of the axis of symmetry is

  1. \(x=-\dfrac{b}{a}\)
  2. \(x=-\dfrac{b}{2a}\)
  3. \(y=c\)
  4. \(x=\dfrac{b}{a}\)
  5. \(x=\dfrac{b}{2a}\)
Show Answers Only

\(A\)

Show Worked Solution
\(\text{Axis of symmetry}\) \(=-\dfrac{b}{2a}\)      \((b = 2b)\)
  \(=-\dfrac{2b}{2a}\)  
  \(=-\dfrac{b}{a}\)  

 
\(\Rightarrow A\)

♦ Mean mark 53%.

Graphs, MET2-NHT 2019 VCAA 1

Parts of the graphs of  `f(x) = (x-1)^3(x + 2)^3`  and  `g(x) = (x-1)^2(x + 2)^3`  are shown on the axes below.
 


 

The two graphs intersect at three points,  (–2, 0),  (1, 0)  and  (`c`, `d`). The point  (`c`, `d`)  is not shown in the diagram above.

  1. Find the values of `c` and `d`.   (2 marks)

  2. Find the values of `x` such that  `f(x) > g(x)`.   (1 mark)

  3. State the values of `x` for which
    1. `f^{'}(x) > 0`   (1 mark)

    2. `g^{'}(x) > 0`   (1 mark)

  4. Show that  `f(1 + m) = f(–2-m)`  for all  `m`.   (1 mark)

  5. Find the values of `h` such that  `g(x + h) = 0`  has exactly one negative solution.   (2 marks)

  6. Find the values of `k` such that  `f(x) + k = 0`  has no solutions.   (1 mark)

Show Answers Only
  1. `c = 2 \ , \ d = 64`
  2. `(–∞, –2) \ ∪ \ (2, ∞)`
  3. i.  `(–(1)/(2), 1) \ ∪ \ (1, ∞)`
    ii. `(–∞, –2) \ ∪ \ (–2, –(1)/(5)) \ ∪ \ (1, ∞)`
  4. `text{Proof (Show Worked Solution)}`
  5.  `-2< h <=1`
  6. `(729)/(64)`
Show Worked Solution

a.    `text(Solve:) \ \ f(x) = g(x)`

`x = 1 , \ –2 \ text(and) \ 2`
 
`f(2) = 1^3 xx 4^3 = 64`
 
`text(Intersection at) \ (2, 64)`

`:. \ c = 2 \ , \ d = 64`

 

b.    `text(Using the graph and intersection at) \ (2, 64):`

`f(x) > g(x) \ \ text(for) \ \ (–∞, –2) \ ∪ \ (2, ∞)`

 

c.i.  `f'(x) > 0 \ \ text(for) \ \ (–(1)/(2), 1) \ ∪ \ (1, ∞)`

c.ii.  `g'(x) > 0 \ \ text(for) \ \ (–∞, –2) \ ∪ \ (–2, –(1)/(5)) \ ∪ \ (1, ∞)`

 

d.     `f(1 + m)` `= (1 + m-1)^3 (1 + m + 2)^3`
  `= m^3 (m + 3)^3`

 

`f(–2-m)` `= (–2-m -1)^3 (-2-m + 2)^3`
  `= (-m-3)^3 (-m)^3`
  `= (–1)^3 (m + 3)^3 (–1)^3 m^3`
  `= m^3 (m + 3)^3`

 

e.     `g(x + h)` `= (x + h-1)^2(x + h + 2)^3`
  `= underbrace{(x-(1 -h))^2}_{text(+ve solution)} * underbrace{(x-(h-2))^3}_{text(–ve solution)}`

 
`1-h ≥ 0 \ \ => \ \ h ≤ 1`

`-h-2 < 0 \ \ =>\ \ h > -2`

`:. -2< h <=1`

 

f.    `f(x) \ \ text(minimum S.P. when) \ \ f ′(x) = 0 \ =>  \ x =-(1)/(2)`

`text(S.P. at) \ \ (-(1)/(2) \ , \ -(729)/(64))`

`:. \ text(No solution if) \ \ k > (729)/(64)`

Graphs, MET2 2017 VCAA 2 MC

Part of the graph of a cubic polynomial function  `f` and the coordinates of its stationary points are shown below.
 


 

`f′(x) < 0`  for the interval

  1. `(0,3)`
  2. `(−oo,−5) ∪ (0,3)`
  3. `(−oo,−3) ∪ (5/3,oo)`
  4. `(−3,5/3)`
  5. `((−400)/27,36)`
Show Answers Only

`D`

Show Worked Solution

`f′(x) < 0\ \ text(when the gradient of the curve is negative.)`

`=> D`

Calculus, MET2 2014 VCAA 5

Let  `f: R -> R, \ \ f (x) = (x-3)(x-1)(x^2 + 3)  and  g: R-> R, \ \ g (x) = x^4-8x.`

  1. Express  `x^4-8x`  in the form  `x(x-a) ((x + b)^2 + c)`.   (2 marks)

  2. Describe the translation that maps the graph of  `y = f (x)`  onto the graph of  `y = g (x)`.   (1 mark)

  3. Find the values of `d` such that the graph of  `y = f (x + d)` has
    1. one positive `x`-axis intercept.   (1 mark)

    2. two positive `x`-axis intercepts.   (1 mark)

  4. Find the value of `n` for which the equation  `g (x) = n`  has one solution.   (1 mark)

  5. At the point  `(u, g(u))`, the gradient of  `y = g(x)`  is `m` and at the point `(v, g(v))`, the gradient is  `-m`, where `m` is a positive real number.
    1. Find the value of  `u^3 + v^3`.   (2 marks)

    2. Find `u` and `v` if  `u + v = 1`.   (1 mark)

    1. Find the equation of the tangent to the graph of  `y = g(x)`  at the point  `(p, g(p))`.   (1 mark)

    2. Find the equations of the tangents to the graph of  `y = g(x)`  that pass through the point with coordinates  `(3/2, -12)`.   (3 marks)

Show Answers Only
  1. `x(x-2)((x + 1)^2 + 3)`
  2. `text(See Worked Solutions)`
  3.  i. `[1,3)`
  4. ii. `d < 1`
  5. `-6 xx 2^(1/3)`
  6.  i. `4`
  7. ii. `u = (sqrt5 + 1)/2,quad v = (-sqrt5 + 1)/2`
  8. i. `y = 4(p^3-2)x-3p^4`
  9. ii. `y = -8x;quady = 24x-48`
Show Worked Solution

a.  `text(Solution 1)`

`g(x) = x^4-8x`

`= x(x-2)(x^2 + 2x + 4)\ \ \ text([by CAS])`

`= x(x-2)((x + 1)^2 + 3)`
 

`text(Solution 2)`

`x^4-8x` `=x(x^3-2^3)`
  `=x(x-2)(x^2 +2x+4)`
  `=x(x-2)(x^2 +2x+1+3)`
  `=x(x-2)((x+1)^2+3)`

 

b.    `f(x + 1)` `= ((x + 1)-3)((x + 1)-1)((x + 1)^2 + 3)`
    `= x(x-2)((x + 1)^2 + 3)`
    `= g(x)`

 

♦ Mean mark (b) 37%.

`:.\ text(Horizontal translation of 1 unit to the left.)`
 

c.i.  `text(Consider part of the)\ \ f(x)\ text(graph below:)`

♦♦♦ Mean mark 7%.

 
met2-2014-vcaa-sec5-answer
 

`text(For one positive)\ xtext(-axis intercept, translate at least)`

`text(1 unit left, but not more than 3 units left.)`

`:. d ∈ [1,3)`
 

c.ii.   `text(Translate less than 1 unit left, or translate)`

♦♦♦ Mean mark part (c)(ii) 19%.

`text(right.)`

`:. d < 1`
 

d.   `text(If)\ \ g(x)=n\ \ text(has one solution, then it)`

`text(will occur when)\ \ g^{′}(x)=0  and  x>0.`

♦♦♦ Mean mark 17%.
`g^{′}(x)` `=4x^3-8`
`4x^3` `=8`
`x` `=2^(1/3)`

 

met2-2014-vcaa-sec5-answer1
 

`n` `=g(2^(1/3))`
  `=2^(4/3)-8xx2^(1/3)`
  `=-6 xx 2^(1/3)`

 

e.i.   `gprime(u) = mqquadgprime(v) = −m`

♦♦ Mean mark (e.i.) 28%.
`g^{prime}(u)` `= -g^{prime}(v)`
`4u^3-8` `= -(4v^3-8)`
`4u^3 + 4v^3` `= 16`
`:. u^3 + v^3` `= 4`

 

e.ii.   `u^3 + v^3 = 4\ …\ (1)`

♦♦♦ Mean mark (e.ii.) 10%.

`u + v = 1\ …\ (2)`

`text(Solve simultaneous equation for)\ \ u > 0:`

`:. u = (sqrt5 + 1)/2,quad v = (-sqrt5 + 1)/2`
 

f.i.   `text(Solution 1)`

`text(Using the point-gradient formula,)`

`y-g(p)` `=g^{prime}(p)(x-p)`
`y-(p^4-8p)` `=(4p^3-8)(x-p)`
`y` `=(4p^3-8)x -3p^4`

 
`text(Solution 2)`

♦♦ Mean mark (f.i.) 22%.

`y = 4(p^3-2)x-3p^4`

`text([CAS: tangentLine)\ (g(u),x,p)]`

 

f.ii.   `text(Sub)\ (3/2,-12)\ text(into tangent equation,)`

`text(Solve:)\ \ -12 = 4(p^3-2)(3/2)-3p^4\ \ text(for)\ p,`

♦♦♦ Mean mark (f.ii.) 14%.

`p = 0\ \ text(or)\ \ p = 2`

`text(When)\ \ p = 0:`    `y` `= 4(-2)x`
    `= -8x`
`text(When)\ \ p = 2:`    `y` `= 4(2^3-2)x-3(2)^4`
    `= 24x-48`

 
`:. text(Equations are:)\ \ y =-8x, \ y = 24x-48`

Calculus, MET2 2015 VCAA 2

A city is located on a river that runs through a gorge.

The gorge is 80 m across, 40 m high on one side and 30 m high on the other side.

A bridge is to be built that crosses the river and the gorge.

A diagram for the design of the bridge is shown below.
 

 VCAA 2015 2a

The main frame of the bridge has the shape of a parabola. The parabolic frame is modelled by  `y = 60-3/80x^2`  and is connected to concrete pads at  `B (40, 0)`  and  `A (– 40, 0).`

The road across the gorge is modelled by a cubic polynomial function.

  1. Find the angle, `theta`, between the tangent to the parabolic frame and the horizontal at the point  `(– 40, 0)` to the nearest degree.   (2 marks)

The road from `X` to `Y` across the gorge has gradient zero at  `X (– 40, 0)`  and at  `Y (40, 30)`, and has equation  `y = x^3/(25\ 600)-(3x)/16 + 35`.

  1. Find the maximum downwards slope of the road. Give your answer in the form  `-m/n`  where `m` and `n` are positive integers.   (2 marks)

Two vertical supporting columns, `MN` and `PQ`, connect the road with the parabolic frame.

The supporting column, `MN`, is at the point where the vertical distance between the road and the parabolic frame is a maximum.

  1. Find the coordinates `(u, v)` of the point `M`, stating your answers correct to two decimal places.   (3 marks)

The second supporting column, `PQ`, has its lowest point at  `P (– u, w)`.

  1. Find, correct to two decimal places, the value of `w` and the lengths of the supporting columns `MN` and `PQ`.   (3 marks)

For the opening of the bridge, a banner is erected on the bridge, as shown by the shaded region in the diagram below.

VCAA 2015 2ai

  1. Find the `x`-coordinates, correct to two decimal places, of `E` and `F`, the points at which the road meets the parabolic frame of the bridge.   (3 marks)

  2. Find the area of the banner (shaded region), giving your answer to the nearest square metre.   (1 mark)

Show Answers Only
  1. `72^@`
  2. `−3/16`
  3. `M (2.49,34.53)`
  4. `w = 35.47\ text(m);quadMN = 25.23\ text(m);quadPQ = 24.30\ text(m)`
  5. `x_E = -23.71;quadx_F = 28.00`
  6. `870\ text(m²)`
Show Worked Solution
a.    `f(x)` `=60-3/80x^2`
  `f′(x)` `=- 3/40 x`

 

`text(At)\ \ x=-40,\ \ f′(x)=3`

♦ Mean mark part (a) 40%.
`tan theta` `= 3`
`:. theta` `= tan^(−1)(3)`
  `=71.56…`
  `= 72^@`

 

b.    `g(x)` `= (x^3)/(25\ 600)-3/16 x + 35`
  `g′(x)` `=(3x^2)/(25\ 600)-3/16`
     

`text(S)text(ince)\ \ 3x^2>=0\ \ text(for all)\ \ x,`

`text(Max downwards slope occurs at)\ \ x= 0.`

`gprime(0) = −3/16` 

 

c.   `text(Let)\ \ V=\ text(distance)\ MN`

`V` `= (60-3/80x^2)-((x^3)/(25\ 600)-3/16x + 35)`
`(dV)/(dx)` `=-3/40 x-(3x^2)/(25\ 600) + 3/16`
♦ Mean mark part (c) 39%.
`(dV)/(dx)` `= 0quadtext(for)\ x ∈ [−40,40]`
 `x` `= 2.490…`

 
`text(When)\ \ x=2.490…,\ \ g(2.490…) = 34.533…`

`:. M (2.49,34.53)`

♦♦ Mean mark part (d) 29%.
MARKER’S COMMENT: Many students didn’t work to a sufficient number of decimal places.

 
d.   `P(-2.49, w)`

`text(When)\ \ x=-2.490…,\ \ g(-2.490…) = 35.47…`

`:. w = 35.47\ \ text{(2 d.p.)}`

`V_(MN)` `=(60-3/80(2.49…)^2)-(((2.49…)^3)/(25\ 600)-3/16(2.49…) + 35)`
  `=25.23\ text{m  (2 d.p.)}`
`V_(PQ)` `=(60-3/80(-2.49…)^2)-(((-2.49…)^3)/(25\ 600)-3/16(-2.49…) + 35)`
  `=24.30\ text{m  (2 d.p.)}`

 

e.   `text(Intersection occurs when:)`

`f(x) = g(x)quadtext(for)\ x ∈ (−40,40)`

`60-3/80x^2=(x^3)/(25\ 600)-3/16x + 35\ \ text([by CAS])`
 

`x_E = -23.7068… = -23.71\ text{(2 d.p.)}`

`x_F = 27.9963… = 28.00\ text{(2 d.p.)}`
 

f.     `text(Area)` `= int_-23.71^28.00 (f(x)-g(x))dx`
    `=int_-23.71^28.00 (60-3/80x^2-((x^3)/(25\ 600)-3/16x + 35))\ dx`
    `=869.619…\ \ text([by CAS])`
    `= 870\ text(m²)`

Graphs, MET2 2015 VCAA 3 MC

VCAA 2015 3mc
 

The rule for a function with the graph above could be

  1. `y = -2(x + b)(x - c)^2(x - d)`
  2. `y = 2(x + b)(x - c)^2(x - d)`
  3. `y = -2(x - b)(x - c)^2(x - d)`
  4. `y = 2(x - b)(x - c)(x - d)`
  5. `y = -2(x - b)(x + c)^2(x + d)`
Show Answers Only

`C`

Show Worked Solution

`text(Polynomial Equation:)`

♦♦♦ Mean mark 20% (lowest in 2015 exam).
MARKER’S COMMENT: `b` is negative. If `b= – 2` for example, the factor is `(x-(– 2))“=(x+2)`.

`y = a(x – b)(x – c)^n(x – d),`

`text(where)\ a < 0`

`n = text(even integer)` 

`=>   C`