Year 12 Methods — Differentiation & Integration Mega Quiz: Worked Solutions

Full worked solutions to the mega quiz. Every question is a past VCAA Examination 2 question. Try each one yourself first, then click Show solution to check your working. All numerical answers here have been checked with a computer algebra system.

Multiple-choice answer key

Q1	Q2	Q3	Q4	Q5	Q6	Q7	Q8
C	E	B	A	E	E	B	B

Section A — Multiple choice

Q1 — VCAA E2 Q9 2018

A tangent to $y = \log_e(2x)$ has gradient $2$. Where does it cross the $y$-axis?

$\dfrac{dy}{dx} = \dfrac{1}{x}$. Gradient $2 \Rightarrow \dfrac{1}{x}=2 \Rightarrow x=\tfrac12$.

Point of contact: $\left(\tfrac12,\ \log_e(2\cdot\tfrac12)\right) = \left(\tfrac12,\ \log_e 1\right) = \left(\tfrac12, 0\right)$.

Tangent: $y - 0 = 2\left(x-\tfrac12\right) \Rightarrow y = 2x - 1$. At $x=0$: $y=-1$.

Answer: C ($-1$)

Q2 — VCAA E2 Q14 2021

Find $k$ so that the average value of $y=\cos\!\left(kx-\tfrac{\pi}{2}\right)$ over $[0,\pi]$ equals the average value of $y=\sin x$ over $[0,\pi]$.

Average of $\sin x$: $\ \dfrac{1}{\pi}\displaystyle\int_0^\pi \sin x\,dx = \dfrac{1}{\pi}\,[-\cos x]_0^\pi = \dfrac{2}{\pi}.$

Note $\cos\!\left(kx-\tfrac{\pi}{2}\right)=\sin(kx)$, so its average is $\ \dfrac{1}{\pi}\displaystyle\int_0^\pi \sin(kx)\,dx = \dfrac{1-\cos(k\pi)}{k\pi}.$

Set equal to $\dfrac{2}{\pi}$: $\ 1-\cos(k\pi) = 2k$. Testing the options, $k=\tfrac12$ gives $1-\cos\tfrac{\pi}{2}=1=2\cdot\tfrac12$. ✓

Answer: E $\left(k=\tfrac12\right)$

Q3 — VCAA E2 Q15 2017

Rectangle $ABCD$ with $A(0,0),B(u,0),C(u,v),D(0,v)$ and $C$ on $y=-x^3+8$. Find the maximum area.

$C(u,v)$ on the curve $\Rightarrow v=-u^3+8$. Area $A = uv = u(8-u^3) = 8u-u^4.$

$\dfrac{dA}{du} = 8-4u^3 = 0 \Rightarrow u^3 = 2 \Rightarrow u = \sqrt[3]{2}.$

$A = 8\sqrt[3]{2} - \left(\sqrt[3]{2}\right)^4 = 8\sqrt[3]{2} - 2\sqrt[3]{2} = 6\sqrt[3]{2}.$

Answer: B $\left(6\sqrt[3]{2}\right)$

Q4 — VCAA E2 Q6 2020

Given the graph of $y=f'(x)$, choose the corresponding graph of $y=f(x)$.

The given graph of $y=f′(x)$: positive, then negative between its two $x$-intercepts, then positive.

The matching $y=f(x)$ (option A): local max, then local min, then increasing.

Read the sign of $f'$: $f$ is increasing where $f'>0$, decreasing where $f'<0$, and has a stationary point where $f'=0$ (crosses the axis).

The given $f'$ is positive, then negative (between its two $x$-intercepts), then positive again. So $f$ goes: increasing → local maximum → decreasing → local minimum → increasing.

Where $f'$ has its turning point but stays positive, $f$ has a point of inflection (not a stationary point). The only option with a maximum, then a minimum, then rising is graph A.

Answer: A

Q5 — VCAA E2 Q9 2016

Given $\dfrac{d}{dx}\!\left(xe^{kx}\right) = (kx+1)e^{kx}$, find $\displaystyle\int xe^{kx}\,dx$.

Integrate the given result: $\ xe^{kx} = \displaystyle\int (kx+1)e^{kx}\,dx = k\!\int xe^{kx}\,dx + \int e^{kx}\,dx.$

Rearrange: $\ \displaystyle\int xe^{kx}\,dx = \frac{1}{k}\left(xe^{kx} - \int e^{kx}\,dx\right)$ (this is option D’s form).

Now $\displaystyle\int e^{kx}\,dx = \frac{e^{kx}}{k}$, so $\ \displaystyle\int xe^{kx}\,dx = \frac{1}{k}\left(xe^{kx} - \frac{e^{kx}}{k}\right) = \frac{1}{k^2}\left(kxe^{kx} - e^{kx}\right)+c.$

Answer: E $\left(\dfrac{1}{k^2}\left(kxe^{kx}-e^{kx}\right)+c\right)$

Q6 — VCAA E2 Q9 2020

If $\displaystyle\int_4^8 f(x)\,dx = 5$, find $\displaystyle\int_0^2 f\big(2(x+2)\big)\,dx$.

Let $u = 2(x+2) = 2x+4$, so $du = 2\,dx$, i.e. $dx = \tfrac12\,du$.

Limits: $x=0 \Rightarrow u=4$; $x=2 \Rightarrow u=8$.

$\displaystyle\int_0^2 f(2x+4)\,dx = \frac12\int_4^8 f(u)\,du = \frac12 (5) = \frac52.$

Answer: E $\left(\tfrac52\right)$

Q7 — VCAA E2 Q15 2020

Find the average value over $[-2a,a]$ of the piecewise-linear function through $(-2a,2a)$, $(0,-a)$ and $(a,a)$.

The function over $[-2a,\,a]$ (drawn with $a=1$).

Left piece $(-2a,2a)\to(0,-a)$: gradient $\dfrac{-a-2a}{0-(-2a)} = -\tfrac32$, so $y = -\tfrac32 x - a$.

Right piece $(0,-a)\to(a,a)$: gradient $\dfrac{a-(-a)}{a-0} = 2$, so $y = 2x - a$.

$\displaystyle\int_{-2a}^{0}\!\left(-\tfrac32 x - a\right)dx = a^2$, $\displaystyle\int_{0}^{a}(2x-a)\,dx = 0$. Total $= a^2$.

Average $= \dfrac{1}{a-(-2a)}\times a^2 = \dfrac{a^2}{3a} = \dfrac{a}{3}.$

Answer: B $\left(\tfrac{a}{3}\right)$

Q8 — VCAA E2 Q17 2017

For the even function $f$ with $x$-intercepts $a,b,c,d$, write the area bound by the curve and the $x$-axis on $[a,d]$.

Even function with intercepts $a,b,c,d$: area $=$ (green) $+$ (red, taken positive) $+$ (green).

From the graph: $f>0$ on $(a,b)$, $f<0$ on $(b,c)$, and $f>0$ on $(c,d)$.

Area must be positive everywhere, so on the middle piece (where $f$ is below the axis) we take the negative of the integral: $$\text{Area} = \int_a^b f(x)\,dx \;-\; \int_b^c f(x)\,dx \;+\; \int_c^d f(x)\,dx.$$

Answer: B

Section B — Extended response

Q9 (13 marks) — VCAA E2 Q4 2020

$f(x) = 2xe^{\,1-x^2},\quad 0 \le x \le 3.$

First, the derivative (product rule): $f'(x) = 2e^{1-x^2} + 2x\cdot(-2x)e^{1-x^2} = 2e^{1-x^2}\left(1-2x^2\right) = \left(2-4x^2\right)e^{1-x^2}.$

a. $f'(1) = (2-4)e^{0} = \mathbf{-2}.$

b. $\tan\theta = -2$ and $\theta$ obtuse $\Rightarrow \theta = 180^\circ - \tan^{-1}(2) = 180^\circ - 63.43^\circ \approx \mathbf{117^\circ}.$

c. $f'(p) = \left(2-4p^2\right)e^{\,1-p^2}.$

d.i. Perpendicular to gradient $-2$ means gradient $\tfrac12$. Solve $\left(2-4p^2\right)e^{1-p^2}=\tfrac12$ (CAS) $\Rightarrow p \approx \mathbf{0.655}.$

d.ii. Tangent at $x=1$: through $(1,2)$ with gradient $-2$, i.e. $y=-2x+4$. Tangent at $x=p$: through $(0.655,\,2.318)$ with gradient $\tfrac12$. They intersect at $\approx \mathbf{(0.80,\ 2.39)}.$

e.i. $f(0)=0$, so the first segment joins $(0,0)$ to $Q\!\left(n,\,2ne^{1-n^2}\right)$. Gradient $= \dfrac{2ne^{1-n^2}}{n}=2e^{1-n^2}$, so $\ y = 2e^{\,1-n^2}\,x.$

e.ii. Second segment joins $Q\!\left(n, 2ne^{1-n^2}\right)$ to $\left(3,\,6e^{-8}\right)$: $$y - 6e^{-8} = \frac{6e^{-8} - 2ne^{1-n^2}}{3-n}\,(x-3).$$

e.iii. Equal areas between $f$ and each segment $\Leftrightarrow$ area under the curve equals area under the two-segment path: $\displaystyle\int_0^3 f\,dx = \tfrac12\,n\,f(n) + \tfrac12\big(f(n)+f(3)\big)(3-n).$ Solving (CAS) with $1

📺 Walkthrough (e.iii): the two shaded regions between the curve and the segments, and why equal areas gives $n\approx1.088$.

a) $-2$ • b) $117^\circ$ • c) $(2-4p^2)e^{1-p^2}$ • d) $p\approx0.655$, intersect $(0.80,2.39)$ • e) $n\approx1.088$

Q10 (8 marks) — VCAA E2 Q3 2018

A bridge $5$ m high, $110$ m long, over three identical arches $h(x)=5\sin\!\left(\tfrac{(x-\text{start})\pi}{30}\right)$. Arch 1: $[5,35]$, Arch 2: $[40,70]$, Arch 3: $[a,105]$.

a. Each arch is $30$ m wide with a $5$ m gap. Arch 3 spans $[a,105]$ with width $30$, so $a = 105-30 = \mathbf{75}.$

b. $h_3(x)=5\sin\!\left(\tfrac{(x-75)\pi}{30}\right)$ is $h_2$ shifted; $75-40=35$, so a translation of $35$ units in the positive $x$-direction.

c. Area of one arch $= \displaystyle\int_5^{35} 5\sin\!\left(\tfrac{(x-5)\pi}{30}\right)dx = \frac{300}{\pi}$. Stone $=$ rectangle under bridge $-$ three arches: $$5\times110 - 3\times\frac{300}{\pi} = 550 - \frac{900}{\pi} \approx 263.5 \approx \mathbf{264\ \text{m}^2}.$$

d. Gradient of the second bridge $= \tan\!\left(\tfrac{\pi}{90}\right) \approx \mathbf{0.035}.$

e. On Arch 5, $h(x)=5\sin\!\left(\tfrac{(x-40)\pi}{30}\right)$, so $h'(x)=\tfrac{\pi}{6}\cos\!\left(\tfrac{(x-40)\pi}{30}\right)$. Set $h'(x)=\tan\!\left(\tfrac{\pi}{90}\right)$ (CAS) $\Rightarrow x\approx54.36$, $y\approx4.99$. So $P \approx \mathbf{(54.36,\ 4.99)}.$

a) $a=75$ • b) translate $35$ right • c) $264\ \text{m}^2$ • d) $0.035$ • e) $P\approx(54.36,4.99)$

Q11 (11 marks) — VCAA E2 Q2 2020

River banks $f_1(x)=20\cos\!\left(\tfrac{\pi x}{100}\right)+40$ (north) and $f_2(x)=20\cos\!\left(\tfrac{\pi x}{100}\right)+30$ (south), $0\le x\le200$. Swimmer starts at $P(50,30)$.

a. Swimming north, reach the north bank at $x=50$: $f_1(50)=20\cos\tfrac{\pi}{2}+40=40$. Distance $=40-30=\mathbf{10\ \text{m}}.$

b. Swimming east ($y=30$): solve $f_1(x)=30 \Rightarrow \cos\tfrac{\pi x}{100}=-\tfrac12 \Rightarrow x=\tfrac{200}{3}$. Distance $=\tfrac{200}{3}-50=\tfrac{50}{3}\approx\mathbf{16.67\ \text{m}}.$

c. Minimise the distance from $P$ to $\left(x,f_1(x)\right)$ (CAS): minimum $\approx \mathbf{8.5\ \text{m}}$ (near $x\approx54.5$).

d. Surface area $= \displaystyle\int_0^{200}\big(f_1-f_2\big)\,dx = \int_0^{200} 10\,dx = \mathbf{2000\ \text{m}^2}.$ (The banks are always $10$ m apart.)

e. The ‘no swimming’ zone is the river below $y=30$. Splitting where $f_1$ and $f_2$ cross $y=30$ and integrating gives $\approx \mathbf{837\ \text{m}^2}.$

f. New north bank $y=kf_1(x)$. Width north across the river $=kf_1-f_2=(20k-20)\cos\tfrac{\pi x}{100}+(40k-30)$. For $k\ge1$ this is largest when $\cos=1$: $60k-50$. Require $60k-50<20 \Rightarrow k<\tfrac{7}{6}$. So $\mathbf{1\le k<\tfrac{7}{6}}.$

📺 Walkthrough (e): shading the part of the river south of $y=30$ and finding its area, $\approx837\ \text{m}^2$.

a) $10$ m • b) $\tfrac{50}{3}\approx16.67$ m • c) $\approx8.5$ m • d) $2000\ \text{m}^2$ • e) $\approx837\ \text{m}^2$ • f) $1\le k<\tfrac{7}{6}$