Additional Mathematics – Hong Kong Certificate of Education (HKCE) – Page 5

November 4, 2022November 4, 2022

202211041114 Solution to 2005-CE-AMATH-II-1

Find $\displaystyle{\int (2x-3)^7\,\mathrm{d}x}$ .

Roughwork.

$\begin{aligned} &\quad\, \int (2x-3)^7\,\mathrm{d}x \\ \dots\enspace & \textrm{let }u=2x-3\textrm{ s.t. }\mathrm{d}x=\frac{\mathrm{d}u}{2} \\ & = \int u^7\bigg( \frac{\mathrm{d}u}{2}\bigg) \\ & = \frac{1}{2} \int u^7\,\mathrm{d}u \\ & = \frac{1}{2}\bigg(\frac{u^8}{8}\bigg) +C\textrm{ for some constant} \\ & = \frac{u^8}{16} + C \\ & = \frac{(2x-3)^8}{16} +C \\ \end{aligned}$

November 4, 2022November 4, 2022

202211041009 Solution to 2005-CE-AMATH-II-10

(a) Show that

$\displaystyle{\frac{\mathrm{d}}{\mathrm{d}x}[x(x+1)^n]=(x+1)^{n-1}[(n+1)x+1]}$ ,

where $n$ is a rational number.
(b) The slope at any point $(x,y)$ of a curve $C$ is given by

$\displaystyle{\frac{\mathrm{d}y}{\mathrm{d}x}=(x+1)^{2004}(2006x+1)}$ .

If $C$ passes through the point $(-1,1)$ , find the equation of $C$ .

Roughwork.

(a)

Beginning by integration on the right hand side,

$\begin{aligned} &\quad\, \int (x+1)^{n-1}[(n+1)x+1]\,\mathrm{d}x \\ & = \int (x+1)^{n-1}[(n+1)(x+1)-n]\,\mathrm{d}x \\ & = (n+1)\int (x+1)^n\,\mathrm{d}x - n\int (x+1)^{n-1}\,\mathrm{d}x \\ & = (n+1)\frac{(x+1)^{n+1}}{n+1} - n\frac{(x+1)^{(n-1)+1}}{(n-1)+1} + C \\ & = (x+1)^{n+1}-(x+1)^n + C \\ & = (x+1)^n(x+1) - (x+1)^n + C \\ & = x(x+1)^n + C \textrm{ for some constant}\\ \end{aligned}$

or by differentiation on the left hand side,

$\begin{aligned} &\quad\,\frac{\mathrm{d}}{\mathrm{d}x}[x(x+1)^n] \\ & = x\cdot \frac{\mathrm{d}}{\mathrm{d}x}\big( (x+1)^n\big) + \frac{\mathrm{d}}{\mathrm{d}x} (x)\cdot (x+1)^n \\ & = x\cdot (n)(x+1)^{n-1}(1) + 1\cdot (x+1)^n \\ & = (nx)(x+1)^{n-1}+(x+1)(x+1)^{n-1} \\ & = (x+1)^{n-1}[(n+1)x+1] \\ \end{aligned}$

(b)

$\begin{aligned} y & = \int \frac{\mathrm{d}y}{\mathrm{d}x}\,\mathrm{d}x \\ & = \int (x+1)^{(2005)-1}[( (2005)+1)x+1] \\ y(x) & \stackrel{\textrm{(a)}}{=} x(x+1)^{2005}+C\textrm{ for some constant} \\ \end{aligned}$

Substituting $-1$ for $x$ and $1$ for $y(x)$ ,

$\begin{aligned} 1 & = (-1)((-1)+1)^{2005} + C \\ C & = 1 \\ \end{aligned}$

In conclusion, the equation of $C$ is

$y=x(x+1)^{2005}+1$ .

November 4, 2022November 4, 2022

202211041001 Solution to 2007-CE-AMATH-II-1

Find $\displaystyle{\int\frac{x^4+1}{x^2}\,\mathrm{d}x}$ .

Roughwork.

$\begin{aligned} &\quad\, \int\frac{x^4+1}{x^2}\,\mathrm{d}x\\ & = \int (x^2+x^{-2})\,\mathrm{d}x \\ & = \frac{x^3}{3} + \frac{x^{-1}}{(-1)} + C\textrm{ for some constant} \\ & = \frac{x^3}{3} - \frac{1}{x} + C \\ \end{aligned}$

November 4, 2022November 4, 2022

202211040951 Solution to 2008-CE-AMATH-II-1

Find $\displaystyle{\int (8x+5)^{250}\,\mathrm{d}x}$ .

Roughwork.

$\begin{aligned} &\quad\, \int (8x+5)^{250}\,\mathrm{d}x \\ \dots\enspace &\textrm{let }u=8x+5\textrm{ s.t. }\mathrm{d}x=\frac{\mathrm{d}u}{8}\enspace\dots \\ & = \int u^{250}\,\bigg(\frac{\mathrm{d}u}{8}\bigg) \\ & = \frac{1}{8}\bigg(\frac{u^{251}}{251}\bigg) +C\textrm{ for some constant} \\ & = \frac{(8x+5)^{251}}{2008}+C \\ \end{aligned}$

November 4, 2022November 4, 2022

202211040914 Solution to 2009-CE-AMATH-II-1

Find

(a) $\displaystyle{\int (4x+1)^2\,\mathrm{d}x}$ ,
(b) $\displaystyle{\int \sin 3\theta\cos\theta\,\mathrm{d}\theta}$ .

Roughwork.

(a)

$\begin{aligned} & \quad\, \int (4x+1)^2\,\mathrm{d}x \\ \dots\enspace & \textrm{let }u=4x+1\textrm{ s.t. }\mathrm{d}x=\frac{\mathrm{d}u}{4}\enspace\dots \\ & = \int u^2\,\bigg(\frac{\mathrm{d}u}{4}\bigg) \\ & = \frac{u^3}{12}+C \textrm{ for some constant}\\ & = \frac{(4x+1)^3}{12} + C \\ \end{aligned}$

(b)

$\begin{aligned} & \quad\, \int \sin 3\theta\cos\theta\,\mathrm{d}\theta \\ & = \int \bigg(\frac{e^{3\mathrm{i}\theta}-e^{-3\mathrm{i}\theta}}{2\mathrm{i}}\bigg)\bigg(\frac{e^{\mathrm{i}\theta}+e^{-\mathrm{i}\theta}}{2}\bigg)\,\mathrm{d}\theta \\ & = \int \frac{e^{4\mathrm{i}\theta}+e^{2\mathrm{i}\theta}-e^{-2\mathrm{i}\theta}-e^{-4\mathrm{i}\theta}}{4\mathrm{i}} \,\mathrm{d}\theta \\ & = \frac{1}{2}\int \bigg( \frac{e^{4\mathrm{i}\theta}-e^{-4\mathrm{i}\theta}}{2\mathrm{i}} + \frac{e^{2\mathrm{i}\theta}-e^{-2\mathrm{i}\theta}}{2\mathrm{i}} \bigg) \,\mathrm{d}\theta \\ & = \frac{1}{2}\int (\sin 4\theta +\sin 2\theta )\,\mathrm{d}\theta \\ & = \frac{1}{8}\int \sin 4\theta\,\mathrm{d}(4\theta ) + \frac{1}{4}\int\sin 2\theta\,\mathrm{d}(2\theta ) \\ & = -\frac{1}{8}\cos 4\theta - \frac{1}{4}\cos 2\theta +C\textrm{ for some constant} \\ \end{aligned}$

October 20, 2022October 24, 2022

202210201704 Solution to 1971-CE-AMATH-II-XX

A particle is projected vertically from the ground so that its velocity (in $\mathrm{ft/sec}$ ) in the first $4$ seconds is given by

$V_1=128-32t\qquad (0\leqslant t\leqslant 4)$

and after the fourth second by

$V_2=3t^2-32t+80\qquad (4\leqslant t)$

where $t$ is the time (in seconds) after projection.

i. Show that $V_1=V_2$ when $t=4$ .
ii. Calculate the height of the particle when $t=4$ .
iii. Calculate the height of the particle when $t=5$ .
iv. Find the value of $t$ when the acceleration of the particle is $40\,\mathrm{ft/sec^2}$ .

Roughwork.

i.

When $t=4$ :

$\begin{aligned} V_1 & = 128-32(4) \\ & = 0 \\ V_2 & = 3(4)^2-32(4)+80 \\ & = 0 \\ V_1 & = V_2 \\ \end{aligned}$

ii.

$\begin{aligned} h(t\in [0,4]) & = \int_{0}^{4}V_1\,\mathrm{d}t \\ & = \int_{0}^{4}(128-32t)\,\mathrm{d}t \\ & = \bigg[ 128\bigg(\frac{t^{(0)+1}}{(0)+1}\bigg) - 32\bigg(\frac{t^{(1)+1}}{(1)+1}\bigg)\bigg]\bigg|_{0}^{4} \\ & = [128(4)-16(4)^2]-[128(0)-16(0)^2] \\ & = 256\,\mathrm{ft} \\ \end{aligned}$

iii.

$\begin{aligned} h(t\in [0,5]) & = h(t\in[0,4]) + h(t\in[4,5]) \\ & = 256\,\mathrm{ft} + \int_{4}^{5}V_2\,\mathrm{d}t \\ & = 256\,\mathrm{ft} + \int_{4}^{5}(3t^2-32t+80)\,\mathrm{d}t \\ & = 256\,\mathrm{ft} + \big[ t^3-16t^2+80t\big]\big|_{4}^{5} \\ & = 256\,\mathrm{ft} -3\,\mathrm{ft} \\ & = 253\,\mathrm{ft} \\ \end{aligned}$

iv.

$\begin{aligned} a(t\in [0,4]) & = \frac{\mathrm{d}}{\mathrm{d}t}V_1=-32 \\ a(t\in [4,T]) & = \frac{\mathrm{d}}{\mathrm{d}t}V_2=6t-32 \\ \end{aligned}$

$\begin{aligned} 6t-32 & = 40 \\ t & = 12\,\mathrm{s}\\ \end{aligned}$

October 20, 2022October 28, 2022

202210201648 Solution to 1972-CE-AMATH-II-XX

It is given that $\displaystyle{\frac{\mathrm{d}y}{\mathrm{d}x}=4x-\frac{3}{2}x^2}$ and that $y=10$ when $x=2$ . Find $y$ in terms of $x$ .

Roughwork.

$\begin{aligned} y & = \int\frac{\mathrm{d}y}{\mathrm{d}x}\,\mathrm{d}x \\ & = \int\bigg( 4x-\frac{3}{2}x^2\bigg)\,\mathrm{d}x \\ & = \frac{4x^{(1)+1}}{(1)+1} -\frac{3}{2}\bigg(\frac{x^{(2)+1}}{(2)+1}\bigg) + C\textrm{ for some constant} \\ & = 2x^2+\frac{1}{2}x^3 + C \\ \end{aligned}$

$\begin{aligned} 10 & = 2(2)^2+\frac{1}{2}(2)^3+C \\ C & = -2 \\ \end{aligned}$

In conclusion,

$y(x)=2x^2+\frac{1}{2}x^3-2$ .

October 20, 2022October 20, 2022

202210201553 Solution to 1977-CE-AMATH-I-XX

In the figure below, the complex numbers $z_0$ , $z_1$ , $z_2$ , $z_3$ , and $z_4$ are represented in the Argand diagram by the vertices of a regular pentagon with centre at the origin $O$ . If $z_0=2$ , write $z_2$ in polar form and calculate the value of $(z_2)^5$ .

Recall.

A complex number $z$ in Cartesian form $z=a+\mathrm{i}b$ can also be expressed in the polar form

$z=re^{\mathrm{i}\varphi}=r(\cos\varphi +\mathrm{i}\sin\varphi )$

where $r=\sqrt{a^2+b^2}$ is called the modulus $|z|$ , and $\varphi$ the argument $\textrm{arg}(z)$ , of $z$ .

Roughwork.

Observe that

$|z_0|=|z_1|=|z_2|=|z_3|=|z_4|=2$ .

and

$\displaystyle{\textrm{arg}(z_2)=2\pi\cdot\frac{2}{5}=\frac{4\pi}{5}}$ .

Hence

$z_2=2e^{\mathrm{i}\frac{4\pi}{5}}$ .

The remaining are left the reader as an exercise.

October 20, 2022October 20, 2022

202210201121 Solution to 1977-CE-AMATH-I-XX

Prove, by mathematical induction, that $23^n-1$ is divisible by $11$ .

Roughwork.

Let proposition $P(n)$ be that

$P(n):\enspace 11|23^n-1$ .

First, note

$\begin{aligned} & \quad\enspace P(1):\enspace 11|23^{1}-1 \\ & \Leftrightarrow P(1):\enspace 11|23-1 \\ & \Leftrightarrow P(1):\enspace 11|22 \\ & \Leftrightarrow P(1)\textrm{ is true} \\ \end{aligned}$

that $P(n)$ is true for $n=1$ .

Next, assumed be to $P(n)$ true.

$\begin{aligned} &\quad\enspace P(n+1):\enspace 11|23^{(n+1)}-1 \\ & \Leftrightarrow P(n+1):\enspace 11|23^{n}\cdot 23-1 \\ & \Leftrightarrow P(n+1):\enspace 11|23^n-1+23^n\cdot 22\\ & \dots\textrm{ by assumption }11|23^n-1\textrm{ and for }11|22 \\ & \Leftrightarrow P(n+1)\textrm{ is true} \\ \end{aligned}$

Thus, $P(n)$ is true implies $P(n+1)$ true.

From $P(1)\textrm{ true}\,\vee\, P(n)\Longrightarrow P(n+1)\textrm{ true}$ follows $P(n)$ the proposition is true for all positive integers $n\geqslant 1$ . By the principle of mathematical induction proven is that $11$ divides $23^n-1$ .

October 19, 2022October 19, 2022

202210191606 Solution to 1969-CE-AMATH-II-XX

Differentiate $2(3-4x)^{\frac{5}{6}}$ with respect to $x$ .

Roughwork.

$\begin{aligned} & \quad\, \frac{\mathrm{d}}{\mathrm{d}x}\Big( 2(3-4x)^{\frac{5}{6}}\Big) \\ & = 2\bigg( \frac{5}{6}\bigg) (3-4x)^{\frac{5}{6}-1}(-4) \\ & = -\frac{20}{3}(3-4x)^{-\frac{1}{6}} \\ & = -\frac{20}{3\sqrt[6]{3-4x}} \\ \end{aligned}$