Additional Mathematics – Hong Kong Certificate of Education (HKCE) – Page 3

December 13, 2022December 13, 2022

202212131316 Solution to 1983-CE-AMATH-II-3

Use the substitution

$u=1+3x^2$

to evaluate

$\displaystyle{\int_{0}^{1}x^3\sqrt{1+3x^2}\,\mathrm{d}x}$ .

Roughwork.

$\begin{aligned} & \quad\enspace \int_{0}^{1}x^3\sqrt{1+3x^2}\,\mathrm{d}x \\ \dots &\textrm{ let }x=\frac{\tan\theta}{\sqrt{3}}\textrm{ so that }\dots \\ \dots &\textrm{ }\mathrm{d}x=\frac{1}{\sqrt{3}}\sec^2\theta\,\mathrm{d}\theta\textrm{ }\dots \\ \dots &\textrm{ }x=0\Leftrightarrow \theta = 0\textrm{ and }x=1\Leftrightarrow \theta =\frac{\pi}{3}\textrm{ }\dots \\ & = \int_{0}^{\frac{\pi}{3}}\bigg(\frac{\tan\theta}{\sqrt{3}}\bigg)^3\sqrt{1+\tan^2\theta}\,\bigg(\frac{1}{\sqrt{3}}\sec^2\theta\,\mathrm{d}\theta\bigg) \\ & = \frac{1}{9}\int_{0}^{\frac{\pi}{3}}\sec^3\theta\tan^3\theta\,\mathrm{d}\theta \\ \dots & \textrm{ but }\frac{\mathrm{d}(\sec^3\theta )}{\mathrm{d}\theta}=3\sec^2\theta\cdot\sec\theta\tan\theta\textrm{ so }\dots \\ & = \frac{1}{27}\int_{0}^{\frac{\pi}{3}}\tan^2\theta\,\mathrm{d}(\sec^3\theta )\\ \end{aligned}$

Whoops! Read and follow the instructions.

$\begin{aligned} & \quad\enspace \int_{0}^{1}x^3\sqrt{1+3x^2}\,\mathrm{d}x \\ \dots &\textrm{ let }u=1+3x^2\textrm{ then }\mathrm{d}x=\frac{\mathrm{d}u}{6x}\textrm{ }\dots \\ \dots &\textrm{ then }x=0\Leftrightarrow u=1\textrm{ and }x=1\Leftrightarrow u=4\textrm{ }\dots \\ & = \int_{1}^{4}x^3\sqrt{u}\,\bigg(\frac{\mathrm{d}u}{6x}\bigg) \\ & = \frac{1}{6}\int_{1}^{4}\bigg(\frac{u-1}{3}\bigg)\sqrt{u}\,\mathrm{d}u \\ & = \frac{1}{18}\int_{1}^{4}\big( u^{\frac{3}{2}}-u^{\frac{1}{2}}\big)\,\mathrm{d}u \\ & = \frac{1}{18}\bigg[ \frac{2u^{\frac{5}{2}}}{5}-\frac{2u^{\frac{3}{2}}}{3}\bigg]\bigg|_{1}^{4} \\ & = \cdots \\ \end{aligned}$

This problem is not to be attempted.

December 12, 2022December 13, 2022

202212121714 Solution to 1974-CE-AMATH-II-X

A bridge $ABC$ is in the form of the parabola given by the equation

$8y=16-x^2$

as shown in the figure.

The slope of the bridge at the point $P$ is $0.5$ . Find the coordinate of $P$ .

Roughwork.

$\begin{aligned} \bigg|\frac{\mathrm{d}y}{\mathrm{d}x}\bigg| & = \bigg|-\frac{1}{4}x\bigg| \\ 0.5 & = \frac{1}{4}|x| \\ x & = \pm 2 \\ 8y & = 16-(\pm 2)^2 \\ y & = 1.5 \\ \end{aligned}$

Answer. $P(2,1.5)$ .

December 12, 2022December 21, 2022

202212121040 Solution to 1968-CE-AMATH-II-X

Find the maximum and minimum values of $y$ on the curve

$y^2=x(1-x)^2$ .

Sketch the curve.

Roughwork. (true-negative/false-positive example)

All computations in the enclosed section below involving higher-order derivatives than the first are wrong owing to the mistaken assumption

$\times :\quad\displaystyle{\bigg(\frac{\mathrm{d}y}{\mathrm{d}x}\bigg)^2=\frac{(\mathrm{d}y)^2}{(\mathrm{d}x)^2}}$

which was to be detested. Show

$\begin{aligned} y^2 & = x(1-x)^2 \\ & = x(1-2x+x^2) \\ y^2 & = x^3-2x^2+x \\ \end{aligned}$

$\begin{aligned} 2y\,\mathrm{d}y & = 3x^2\,\mathrm{d}x-4x\,\mathrm{d}x+\mathrm{d}x \\ \frac{\mathrm{d}y}{\mathrm{d}x} & = \frac{3x^2-4x+1}{2y} \\ 2(\mathrm{d}y)^2 & = 6x(\mathrm{d}x)^2 - 4(\mathrm{d}x)^2 \\ \bigg(\frac{\mathrm{d}y}{\mathrm{d}x}\bigg)^2 & = 3x-2 \\ \end{aligned}$

$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}x}\bigg(\frac{\mathrm{d}y}{\mathrm{d}x}\bigg)^2 & = 2\bigg(\frac{\mathrm{d}y}{\mathrm{d}x}\bigg)\cdot \frac{\mathrm{d}^2y}{\mathrm{d}x^2} \\ 3 & = 2\bigg(\frac{3x^2-4x+1}{2y}\bigg)\cdot \frac{\mathrm{d}^2y}{\mathrm{d}x^2} \\ \frac{\mathrm{d}^2y}{\mathrm{d}x^2} & = \frac{3y}{3x^2-4x+1} \\ \end{aligned}$

$\begin{aligned} y'(x) & = 0 \\ 3x^2-4x+1 & = 0 \\ (3x-1)(x-1) & = 0 \\ x & = \frac{1}{3},1 \\ \end{aligned}$

$\begin{aligned} y^2 & = \bigg(\frac{1}{3}\bigg)\bigg(1-\bigg(\frac{1}{3}\bigg)\bigg)^2 \\ y & =\pm\frac{2\sqrt{3}}{9}\\ y^2 & = (1)(1-(1))^2 \\ y & = 0 \\ \end{aligned}$

$\begin{pmatrix}x\\y\end{pmatrix} = \Bigg\{ \begin{pmatrix}\frac{1}{3}\\ -\frac{2\sqrt{3}}{9} \end{pmatrix}, \begin{pmatrix}\frac{1}{3}\\ \frac{2\sqrt{3}}{9} \end{pmatrix}, \begin{pmatrix}1\\0\end{pmatrix}\Bigg\}$ are extrema.

$\begin{aligned} y''(x) & = 0 \\ 3y & = 0 \\ \pm\sqrt{x(1-x)^2} & = 0 \\ x(1-x)^2 & = 0 \\ x & = 0,1 \\ \end{aligned}$

$\begin{pmatrix}x\\y\end{pmatrix} = \bigg\{ \begin{pmatrix}0\\ 0 \end{pmatrix}, \begin{pmatrix}1\\ 0\end{pmatrix}\bigg\}$ are inflexions.

To answer whether

$\displaystyle{\frac{\mathrm{d}^2y}{\mathrm{d}x^2}}\bigg|_{\big( \frac{1}{3},\pm\frac{2\sqrt{3}}{9}\big)}$

is positive or negative is critical to yielding the maximum and the minimum values of the function.

The correction is left a $\textrm{\scriptsize{MUST}}$ for the author.

December 12, 2022December 12, 2022

202212121004 Solution to 1968-CE-AMATH-II-X

Find $\displaystyle{\frac{\mathrm{d}y}{\mathrm{d}x}}$ if

$ax^2+2hxy+by^2+2gx+2fy+c=0$ .

Roughwork.

$\begin{aligned} ax^2+2hxy+by^2+2gx+2fy+c & =0 \\ 2ax\,\mathrm{d}x+2h(x\,\mathrm{d}y+y\,\mathrm{d}x)+2by\,\mathrm{d}y+2g\,\mathrm{d}x+2f\,\mathrm{d}y & = 0 \\ 2ax + 2h\bigg( x\frac{\mathrm{d}y}{\mathrm{d}x}+y\bigg) +2by\frac{\mathrm{d}y}{\mathrm{d}x}+2g+2f\frac{\mathrm{d}y}{\mathrm{d}x} & = 0 \\ (ax+hy+g) + (hx+by+f)\frac{\mathrm{d}y}{\mathrm{d}x} & = 0 \\ -\frac{ax+hy+g}{hx+by+f} & = y' \\ \end{aligned}$

This problem is not to be attempted.

December 12, 2022December 12, 2022

202212120958 Solution to 1967-CE-AMATH-II-X

A spherical balloon is of diameter $x$ . By what approximate amount does the volume increases, if the diameter is increased by a very small amount $\Delta x$ ?

Roughwork.

$\begin{aligned} V & = \frac{4}{3}\pi x^3 \\ \Delta V & = 4\pi x^2\,\Delta x \\ \end{aligned}$

December 6, 2022December 6, 2022

202212061712 Solution to 1967-CE-AMATH-I-X

If the population of Hong Kong is $4$ million in $1967$ and it increases by $3\%$ each year, what will be the estimated population in $1987$ ? Give your answer correct to $4$ significant figures.

Roughwork.

$\begin{aligned} & \quad\enspace (4\times 10^6)(1+3\%)^{(1987-1967)} \\ & = (4\times 10^6)(1+0.03)^{20} \\ & = (4\times 10^6)\sum_{n=0}^{20}\begin{pmatrix}20\\ n\end{pmatrix}(1)^{20-n}(0.03)^n \\ & = (4\times 10^6)\sum_{n=0}^{20}\begin{pmatrix}20\\ n\end{pmatrix}(0.03)^n \\ & = \cdots \\ \end{aligned}$

This problem is not to be attempted.

November 23, 2022November 23, 2022

202211231608 Solution to 1982-CE-AMATH-I-5

Show that the tangents to the curve

$y=x^3-9x^2+30x+4$

cannot be parallel to the $x$ -axis.

Roughwork.

The slope $y'(x)$ of tangent to the curve is

$\begin{aligned} y'(x) & = \frac{\mathrm{d}y}{\mathrm{d}x} \\ & = \frac{\mathrm{d}}{\mathrm{d}x}(x^3-9x^2+30x+4) \\ & = 3x^2-18x+30 \\ \end{aligned}$

Assume to the contrary that such tangent as $y'(x)=0$ exists. Then

$\begin{aligned} 3x^2-18x+30 & = 0 \\ \frac{-(-18)\pm\sqrt{(-18)^2-4(3)(30)}}{2(3)} & = x \\ \end{aligned}$

where the discriminant

$\Delta = -36<0$ ,

implies there are no real roots. Contradiction, which was to be demonstrated.

November 23, 2022November 23, 2022

202211231536 Solution to 1986-CE-AMATH-I-4

Find the equation of the tangent to the curve

$x^2+xy+y^2=7$

at the point $(2,1)$ .

Roughwork.

$\begin{aligned} x^2+xy+y^2-7 & = 0 \\ 2x\,\mathrm{d}x+x\,\mathrm{d}y+y\,\mathrm{d}x + 2y\,\mathrm{d}y & = 0 \\ 2x + x\frac{\mathrm{d}y}{\mathrm{d}x}+y+2y\frac{\mathrm{d}y}{\mathrm{d}x} & = 0 \\ -\frac{2x+y}{x+2y} & = y'(x) \\ \end{aligned}$

The slope of tangent at $(x=2,y=1)$ is

$y'(x)|_{(2,1)}=\displaystyle{-\frac{2(2)+(1)}{(2)+2(1)}=-\frac{5}{4}}$ .

and its equation

$\begin{aligned} & \quad\enspace \frac{y-1}{x-2} = -\frac{5}{4} \\ & \Rightarrow 5x + 4y = 14 \\ \end{aligned}$

November 22, 2022November 22, 2022

202211221145 Solution to 1972-CE-AMATH-I-XX

The equation of the ellipse shown in the figure below is given by

$9x^2+ky^2=144$ .

Find the value of $k$ and the coordinates of the point $A$ .

Roughwork.

$\begin{aligned} 9x^2+ky^2-144 & = 0 \\ 18x\,\mathrm{d}x+2ky\,\mathrm{d}y & = 0 \\ \frac{\mathrm{d}y}{\mathrm{d}x} & = -\frac{9x}{ky} \\ \end{aligned}$

The equation of a standard ellipse centred at the origin with width $2a$ and height $2b$ is analytically

$\displaystyle{\frac{x^2}{a^2}+\frac{y^2}{b^2}=1}$ .

_{Wikipedia on Ellipse}

$\begin{aligned} 9x^2+ky^2 & = 144 \\ \frac{x^2}{(4)^2} + \frac{y^2}{(12/\sqrt{k})^2} & = 1 \\ a & = 4 \\ b & = \frac{12}{\sqrt{k}} \\ \end{aligned}$

This problem is not to be attempted.

November 22, 2022November 22, 2022

202211220908 Solution to 1999-CE-AMATH-I-1

Find

(a) $\displaystyle{\frac{\mathrm{d}}{\mathrm{d}x}\sin (x^2+1)}$ ,
(b) $\displaystyle{\frac{\mathrm{d}}{\mathrm{d}x}\bigg[\frac{\sin (x^2+1)}{x}\bigg]}$ .

Roughwork.

Let $g(x)=\sin x$ and $h(x)=x^2+1$ s.t.

$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}x}\sin (x^2+1) & = \frac{\mathrm{d}}{\mathrm{d}x}(g\circ h)(x) \\ & = (g\circ h)'(x) \\ & = (g'\circ h)\cdot h'(x) \\ & = \frac{\mathrm{d}g(h(x))}{\mathrm{d}h(x)}\cdot\frac{\mathrm{d}h(x)}{\mathrm{d}x} \\ & = \cos (h(x))\cdot 2x \\ & = 2x\cos (x^2+1) \end{aligned}$

Next, let $f(x)=x^{-1}$ s.t.

$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}x}\bigg[\frac{\sin (x^2+1)}{x}\bigg] & = \frac{\mathrm{d}}{\mathrm{d}x}\Big( g(h(x))\cdot f(x)\Big) \end{aligned}$

The remaining are left the reader as an exercise.