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
- `-12.25`
- `-5`
- `-1.5`
- `0`
- `2`
Aussie Maths & Science Teachers: Save your time with SmarterEd
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
`C`
`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`
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
\(D\)
\(\text{Positive}\ x\text{-intercept occurs at}\ \Big(\dfrac{1}{3}, 0\Big) \)
\(\text{Find tangent line at}\ \ x=a\ \text{(by CAS):}\)
\(y_{\text{tang}}=\Big(12a^3+33a^2+10a-3\Big)x-a^2\Big(9a^2+22a+5\Big)\)
\(\text{Solve}\ \ \ y_{\text{tang}}\Bigg(\dfrac{1}{3}\Bigg)=0\ \text{for }a:\)
\(a=\dfrac{-\sqrt{7}-4}{3},\ a=\dfrac{\sqrt{7}-4}{3}\text{ or}\ a=\dfrac{1}{3}\)
\(\therefore\ \text{3 solutions exist.}\)
\(\Rightarrow D\)
\(\text{NOTE: Graphical methods could also be used}\)
For the parabola with equation \(y=ax^2+2bx+c\), where \(a, b, c \in R\), the equation of the axis of symmetry is
\(A\)
\(\text{Axis of symmetry}\) | \(=-\dfrac{b}{2a}\) | \((b = 2b)\) |
\(=-\dfrac{2b}{2a}\) | ||
\(=-\dfrac{b}{a}\) |
\(\Rightarrow A\)
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.
i. `f' (x) > 0` (1 mark)
ii. `g' (x) > 0` (1 mark)
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)`
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
`D`
`f′(x) < 0\ \ text(when the gradient of the curve is negative.)`
`=> D`
Let `f: R -> R, \ \ f (x) = (x - 3)(x - 1)(x^2 + 3) and g: R-> R, \ \ g (x) = x^4 − 8x.`
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)` |
`:.\ text(Horizontal translation of 1 unit to the left.)`
c.i. `text(Consider part of the)\ \ f(x)\ text(graph below:)`
`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)`
`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.`
`g′(x)` | `=4x^3-8` |
`4x^3` | `=8` |
`x` | `=2^(1/3)` |
`n` | `=g(2^(1/3))` |
`=2^(4/3)-8xx2^(1/3)` | |
`=-6 xx 2^(1/3)` |
e.i. `gprime(u) = mqquadgprime(v) = −m`
`gprime(u)` | `= −gprime(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)`
`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′(p)(x-p)` |
`y-(p^4-8p)` | `=(4p^3-8)(x-p)` |
`y` | `=(4p^3-8)x -3p^4` |
`text(Solution 2)`
`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,`
`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`
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.
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.
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`.
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.
The second supporting column, `PQ`, has its lowest point at `P (– u, w)`.
For the opening of the bridge, a banner is erected on the bridge, as shown by the shaded region in the diagram below.
a. | `f(x)` | `=60 – 3/80x^2` |
`f′(x)` | `=- 3/40 x` |
`text(At)\ \ x=-40,\ \ f′(x)=3`
`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` |
`(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)`
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)` |
The rule for a function with the graph above could be
`C`
`text(Polynomial Equation:)`
`y = a(x – b)(x – c)^n(x – d),`
`text(where)\ a < 0`
`n = text(even integer)`
`=> C`