Part of the graph of \(f:[-\pi, \pi] \rightarrow R, f(x)=x \sin (x)\) is shown below. --- 8 WORK AREA LINES (style=lined) --- --- 3 WORK AREA LINES (style=lined) --- --- 3 WORK AREA LINES (style=lined) --- --- 3 WORK AREA LINES (style=lined) --- --- 0 WORK AREA LINES (style=lined) --- a. \(\dfrac{\sqrt{3}\pi^2}{6}\) bi. \(x\cos(x)+\sin(x)\) bii. \(\left[-\dfrac{\pi}{3}+\dfrac{\sqrt{3}}{3},\ 1\right]\) biii. \(\text{In the interval }\left[\dfrac{\pi}{2}, \dfrac{2\pi}{3}\right],\ f^{\prime}(x)\ \text{changes from positive to negative.}\) \(f^{\prime}(x)=0\ \text{at some point in the interval.}\) \(\therefore\ \text{A stationary point must exist in the given range.}\) b.ii. \(\text{Gradient in given range gradually decreases.}\) \(\text{Range of}\ f^{\prime}(x)\ \text{will be defined by the endpoints.}\) \(f^{\prime}(x)=0\ \text{at some point in the interval.}\) \(\therefore\ \text{A stationary point must exist in the given range.}\) \(f^{\prime}(-\pi)=-\pi\cos(-\pi)+\sin(-\pi)=\pi\) \(\therefore\ \text{Endpoints are }\ (-\pi,\ \pi)\ \text{and}\ (\pi,\ -\pi).\)
a.
\(A\)
\(=\dfrac{\pi}{3}\times\dfrac{1}{2}\left(f(0)+2f\left(\dfrac{\pi}{3}\right)+2f\left(\dfrac{2\pi}{3}\right)+f(\pi)\right)\)
\(=\dfrac{\pi}{6}\left(0+2\times \dfrac{\pi}{3}\sin\left(\dfrac{\pi}{3}\right)+2\times \dfrac{2\pi}{3}\sin\left(\dfrac{2\pi}{3}\right)+\pi\sin({\pi})\right)\)
\(=\dfrac{\pi}{6}\left(2\times \dfrac{\pi}{3}\times\dfrac{\sqrt{3}}{2}+2\times\dfrac{2\pi}{3}\times \dfrac{\sqrt{3}}{2}+0\right)\)
\(=\dfrac{\pi}{6}\left(\dfrac{2\pi\sqrt{3}}{6}+\dfrac{4\pi\sqrt{3}}{6}\right)\)
\(=\dfrac{\pi}{6}\times \dfrac{6\pi\sqrt{3}}{6}\)
\(=\dfrac{\sqrt{3}\pi^2}{6}\)
bi.
\(f(x)\)
\(=x\sin(x)\)
\(f^{\prime}(x)\)
\(=x\cos(x)+\sin(x)\)
\(f^{\prime}\left(\dfrac{\pi}{2}\right)\)
\(=1\)
\(f^{\prime}\left(\dfrac{2\pi}{3}\right)\)
\(=\dfrac{2\pi}{3}\left(\dfrac{-1}{2}+\dfrac{\sqrt{3}}{2}\right)=-\dfrac{\pi}{3}+\dfrac{\sqrt{3}}{3}\)
\(\therefore\ \text{Range of }\ f^{\prime} (x)\ \text{is}\quad\left[-\dfrac{\pi}{3}+\dfrac{\sqrt{3}}{3},\ 1\right]\)
♦♦♦ Mean mark (b.iii.) 12%.
biii. \(\text{In the interval }\left[\dfrac{\pi}{2}, \dfrac{2\pi}{3}\right],\ f^{\prime}(x)\ \text{changes from positive to negative.}\)
c. \(f^{\prime}(\pi)=\pi\cos(\pi)+\sin(\pi)=-\pi\)
Calculus, MET2 2024 VCAA 15 MC
The points of inflection of the graph of \(y=2-\tan \left(\pi\left(x-\dfrac{1}{4}\right)\right)\) are
- \(\left(k+\dfrac{1}{4}, 2\right), k \in Z\)
- \(\left(k-\dfrac{1}{4}, 2\right), k \in Z\)
- \(\left(k+\dfrac{1}{4},-2\right), k \in Z\)
- \(\left(k-\dfrac{3}{4},-2\right), k \in Z\)
Show Worked Solution
\(\text{The graph of}\ \ y=-\tan(\pi x)\ \text{is translated 2 units upwards and then}\)
\(\text{translated }\dfrac{1}{4}\ \text{units to the right to become the graph shown below.}\)
\(\text{Hence all points of inflection lie of the line}\ y=2\)
\(\rightarrow\ \text{Eliminate Options C and D}\)
\(\text{Consider Option A:}\)
\(\text{When}\ \ k=0, x=0+\dfrac{1}{4}=\dfrac{1}{4}\)
\(\text{then}\ \ y=2-\tan \left(\pi\left(\dfrac{1}{4}-\dfrac{1}{4}\right)\right)=2\)
\(\text{Similarly for any}\ k \in Z\ \text{a point of inflection will be found at }\ \left(k+\dfrac{1}{4}, 2\right)\)
\(\Rightarrow A\)
Calculus, MET2 2024 VCAA 13 MC
The function \(f:(0, \infty) \rightarrow R, f(x)=\dfrac{x}{2}+\dfrac{2}{x}\) is mapped to the function \(g\) with the following sequence of transformations:
- dilation by a factor of 3 from the \(y\)-axis
- translation by 1 unit in the negative direction of the \(y\)-axis.
The function \(g\) has a local minimum at the point with the coordinates
- \((6,1)\)
- \(\left(\dfrac{2}{3}, 1\right)\)
- \((2,5)\)
- \(\left(2,-\dfrac{1}{3}\right)\)
Show Worked Solution
\(\text{Dilate by a factor of 3 from the}\ y\text{-axis:}\)
\(f(x) \rightarrow f_1(x)=\dfrac{\frac{x}{3}}{2}+\dfrac{2}{\frac{x}{3}}=\dfrac{x}{6}+\dfrac{6}{x}\)
\(\text{Translate 1 unit down:}\)
\(f_1(x) \rightarrow g(x)=\dfrac{x}{6}+\dfrac{6}{x}-1\)
\(g'(x)=\dfrac{1}{6}-6x^{-2}\)
♦ Mean mark 45%.
\(\Rightarrow A\)
Calculus, MET2 EQ-Bank 2
Jac and Jill have built a ramp for their toy car. They will release the car at the top of the ramp and the car will jump off the end of the ramp. The cross-section of the ramp is modelled by the function \(f\), where \(f(x)= \begin{cases}\displaystyle \ 40 & 0 \leq x<5 \\ \dfrac{1}{800}\left(x^3-75 x^2+675 x+30\ 375\right) & 5 \leq x \leq 55\end{cases}\) \(f(x)\) is both smooth and continuous at \(x=5\). The graph of \(y=f(x)\) is shown below, where \(x\) is the horizontal distance from the start of the ramp and \(y\) is the height of the ramp. All lengths are in centimetres. --- 2 WORK AREA LINES (style=lined) --- --- 1 WORK AREA LINES (style=lined) --- --- 1 WORK AREA LINES (style=lined) --- --- 1 WORK AREA LINES (style=lined) --- Jac and Jill decide to use two trapezoidal supports, each of width \(10 cm\). The first support has its left edge placed at \(x=5\) and the second support has its left edge placed at \(x=15\). Their cross-sections are shown in the graph below. --- 5 WORK AREA LINES (style=lined) --- --- 4 WORK AREA LINES (style=lined) --- --- 6 WORK AREA LINES (style=lined) --- --- 3 WORK AREA LINES (style=lined) ---
a. \(\text{Using CAS: Define f(x) then}\) \(\text{OR}\quad f^{\prime}(x)= \begin{cases}\displaystyle \ 0 & 0 < x<5 \\ \dfrac{1}{800}\left(3x^2-150 x+675 \right) & 5 \leq x < 55\end{cases}\) b.i \(\text{Using CAS: Find }f^”(x)\ \text{then solve }=0\) \(\therefore\ \text{Point of inflection at }(25, 20)\) b.ii \(\text{Gradient function strictly increasing for }x\in [25, 55]\) b.iii \(\text{Gradient function strictly decreasing for }x\in [5, 25]\) c. \(\text{Using CAS:}\)
\(\therefore\ \text{The sloped edges of the trapeziums would be above }f(x)\) \(\text{making the trapezium rule approximation greater than the}\) \(\text{actual area.}\) e.i \(f(55)=\dfrac{35}{4}\ \text{(Using CAS)}\) \(\text{and }f(55)=g(55)\rightarrow g(55)=\dfrac{35}{4}\) \(\text{and }g'(55)=f'(55)\) \(\rightarrow\ f'(55)=\dfrac{1}{800}(3\times 55^2-150\times 55+675)=\dfrac{15}{8}\) \(\rightarrow\ -\dfrac{55}{8}+b=\dfrac{15}{8}\) \(\therefore b=\dfrac{35}{4}\quad (2)\) \(\text{Sub (2) into (1)}\) \(c=\dfrac{3165}{16}-55\times\dfrac{35}{4}\) \(c=-\dfrac{4535}{16}\) \(\therefore\ b=\dfrac{35}{4}\ \text{or}\ 8.75 , c=-\dfrac{4535}{16}\ \text{or}\ -283.4375\) e.ii \(\text{Using CAS: Solve }g(x)=0|x>55\) \(\text{Horizontal distance}=\sqrt{365}+70-55=34.104\dots\approx 34.10\ \text{cm (2 d.p.)}\)
Show Worked Solution
\(\text{Area}\)
\(=\dfrac{h}{2}\Big(f(5)+2f(15)+f(25)\Big)\)
\(=\dfrac{10}{2}\Big(40+2\times 33.75+20\Big)=637.5\)
\(\therefore\ \text{Estimate : Exact}\)
\(=637.5:650\)
\(=\dfrac{637.5}{650}\times 100\)
\(=98.0769\%\approx 98.1\%\)
d. \(\text{In the interval }x\in [25, q]\text{ the curve is concave up.}\)
\(\therefore -\dfrac{1}{16}\times 55^2+55b+c\)
\(=\dfrac{35}{4}\)
\(c\)
\(=\dfrac{35}{4}+\dfrac{3025}{16}-55b\)
\(c\)
\(=\dfrac{3165}{16}-55b\quad (1)\)
\(g'(x)=-\dfrac{x}{8}+b\)
Calculus, MET2 2022 VCAA 16 MC
The function `f(x)=\frac{1}{3} x^3+m x^2+n x+p`, for `m, n, p \in R`, has turning points at `x=-3` and `x=1` and passes through the point (3, 4).
The values of `m, n` and `p` respectively are
- `m=0, \quad n=-\frac{7}{3}, p=2`
- `m=1, n=-3, \quad p=-5`
- `m=-1, n=-3, \quad p=13`
- `m=\frac{5}{4}, \quad n=\frac{3}{2}, \quad p=-\frac{83}{4}`
- `m=\frac{5}{2}, \quad n=6, \quad p=-\frac{91}{2}`
Calculus, MET2 2023 VCAA 5 MC
Which one of the following functions has a horizontal tangent at \((0, 0)\)?
- \(y=x^{-\frac{1}{3}}\)
- \(y=x^{\frac{1}{3}}\)
- \(y=x^{\frac{2}{3}}\)
- \(y=x^{\frac{4}{3}}\)
- \(y=x^{\frac{3}{4}}\)
Calculus, MET2-NHT 2019 VCAA 13 MC
The graph of `f(x) = x^3 - 6(b - 2)x^2 + 18x + 6` has exactly two stationary points for
- `1 < b < 2`
- `b = 1`
- `b = (4 ± sqrt6)/(2)`
- `(4 - sqrt6)/(2) ≤ b ≤ (4 + sqrt6)/(2)`
- `b < (4 - sqrt6)/(2) \ or \ b > (4 + sqrt6)/(2)`
Calculus, MET2 2018 VCAA 17 MC
The turning point of the parabola `y = x^2 - 2bx + 1` is closest to the origin when
- `b = 0`
- `b = -1 or b = 1`
- `b = -1/sqrt 2 or b = 1/sqrt 2`
- `b = 1/2 or b = -1/2`
- `b = 1/4 or b = -1/4`
Calculus, MET2 2018 VCAA 5 MC
Consider `f(x) = x^2 + p/x,\ x != 0,\ p in R`.
There is a stationary point on the graph of `f` when `x = -2`.
The value of `p` is
- −16
- –8
- 2
- 8
- 16
Calculus, MET2 2017 VCAA 11 MC
The function `f : R → R, \ f (x) = x^3 + ax^2 + bx` has a local maximum at `x = –1` and a local minimum at `x = 3`.
The values of `a` and `b` are respectively
- `–2\ text(and)\ \ –3`
- `text(2 and 1)`
- `text(3 and)\ \ –9`
- `–3\ text(and)\ \ –9`
- `– 6\ text(and)\ \ –15`
Calculus, MET2 2007 VCAA 12 MC
Let `f: R -> R` be a differentiable function such that
- `f prime(3) = 0`
- `f prime(x) < 0` when `x < 3` and when `x > 3`
When `x = 3`, the graph of `f` has a
- local minimum
- local maximum
- stationary point of inflection
- point of discontinuity
- gradient of 3
Show Worked Solution
`text(Use a gradient table:)`
`:.\ text(Point of inflection at)\ \ x = 3.`
`=> C`
Calculus, MET2 2009 VCAA 21 MC
A cubic function has the rule `y = f (x)`. The graph of the derivative function `f prime` crosses the `x`-axis at `(2, 0)` and `(– 3, 0)`. The maximum value of the derivative function is 10.
The value of `x` for which the graph of `y = f(x)` has a local maximum is
- `– 2`
- `2`
- `– 3`
- `3`
- `– 1/2`
Show Worked Solution
`text(If)\ \ f(x)\ \ text(is cubic) -> f prime (x)\ \ text(is quadratic)`
`f′(x) >0\ \ text(for)\ \ x in (-3,2), and f′(x) =0\ \ text(at)\ \ x=2.`
`f′(x) <0\ \ text(for)\ \ x>2.`
`:.\ text(Local max occurs when)\ \ x = 2`
`=> B`
Calculus, MET2 2010 VCAA 17 MC
The function `f` is differentiable for all `x in R` and satisfies the following conditions.
- `f prime (x) < 0\ \ text(where)\ \ x < 2`
- `f prime (x) = 0\ \ text(where)\ \ x = 2`
- `f prime (x) = 0\ \ text(where)\ \ x = 4`
- `f prime (x) > 0\ \ text(where)\ \ 2 < x < 4`
- `f prime (x) > 0\ \ text(where)\ \ x > 4`
Which one of the following is true?
- The graph of `f` has a local maximum point where `x = 4.`
- The graph of `f` has a stationary point of inflection where `x = 4.`
- The graph of `f` has a local maximum point where `x = 2.`
- The graph of `f` has a local minimum point where `x = 4.`
- The graph of `f` has a stationary point of inflection where `x = 2.`
Show Worked Solution
`text(A table summary of the gradients is:)`
`:.\ text(Point of Inflection at)\ \ x = 4`
`=> B`
Calculus, MET2 2014 VCAA 4 MC
Let `f` be a function with domain `R` such that `f (5) = 0` and `f prime (x) < 0` when `x != 5.`
At `x = 5`, the graph of `f` has a
- local minimum.
- local maximum
- gradient of 5
- gradient of – 5
- stationary point of inflection.
Show Worked Solution
`:.\ text(S)text(ince concavity doesn’t change,)`
`text(point of inflection at)\ \ x = 5.`
`=> E`
Calculus, MET2 2013 VCAA 19 MC
Part of the graph of a function `f: [0, oo) -> R, \ f(x) = e^(x sqrt 3) sin (x)` is shown below.
The first three turning points are labelled `T_1, T_2` and `T_3.`
The `x`-coordinate of `T_3` is
- `(8 pi)/3`
- `(16 pi)/3`
- `(13 pi)/6`
- `(17 pi)/6`
- `(29 pi)/6`
Calculus, MET2 2013 VCAA 21 MC
The cubic function `f: R -> R, f(x) = ax^3-bx^2 + cx`, where `a, b` and `c` are positive constants, has no stationary points when
- `c > b^2/(4a)`
- `c < b^2/(4a)`
- `c < 4b^2a`
- `c > b^2/(3a)`
- `c < b^2/(3a)`
Calculus, MET2 2012 VCAA 16 MC
The graph of a cubic function `f` has a local maximum at `(a, text{−3)}` and a local minimum at `(b, text{−8)}.`
The values of `c`, such that the equation `f(x) + c = 0` has exactly one solution, are
- `3 < c < 8`
- `c > -3 or c < -8`
- `-8 < c < -3`
- `c < 3 or c > 8`
- `c < -8`
Show Worked Solution
`text(Sketch a possible graph:)`
♦♦ Mean mark 34%.
`text(For one solution:)`
`text(Shift graph up less than 3)`
`c < 3,\ \ text(or)`
`text(Shift graph up more than 8)`
`c > 8`
`=> D`