December 2022 – Page 3

December 14, 2022December 14, 2022

202212141214 Solution to 1999-CE-AMATH-II-5

A family of straight lines is given by the equation

$y-3+k(x-y+1)=0$ ,

where $k$ is real.

(a) Find the equation of a line $L_1$ in the family whose $x$ -intercept is $5$ .
(b) Find the equation of a line $L_2$ in the family which is parallel to the $x$ -axis.
(c) Find the acute angle between $L_1$ and $L_2$ .

Roughwork.

(a)

Substituting $(5,0)$ for the $x$ -intercept:

$\begin{aligned} (0)-3+k((5)-(0)+1) & = 0 \\ k & = \frac{1}{2} \\ \Longrightarrow \enspace L_1: \big\{ x+y-2 &=0 \big\} \\ \end{aligned}$

(b)

Differentiating on both hand sides,

$\begin{aligned} \mathrm{d}y+k(\mathrm{d}x-\mathrm{d}y) & =0 \\ \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{k}{k-1} & = 0 \\ k & = 0 \\ \Longrightarrow \enspace L_2: \big\{ y & = 3 \big\} \\ \end{aligned}$

(c) $45^\circ$ .

This problem is not to be attempted.

December 13, 2022December 13, 2022

202212131615 Solution to 1989-CE-AMATH-I-6

$p$ and $q$ are real numbers such that

$(p+qi)^2=21-20i$ .

Find the values of $p$ and $q$ . Hence write down the two square roots of $21-20i$ .

Roughwork.

$\begin{aligned} \textrm{LHS} & = (p+qi)^2 \\ & = (p^2-q^2)+(2pq)i \\ \textrm{RHS} & = (21)+(-20)i \\ \end{aligned}$

$\begin{cases} p^2-q^2=21 \\ pq = -10 \end{cases}$

$\therefore$ $(p,q)=(\pm 5,\mp 2)$ .

December 13, 2022December 13, 2022

202212131544 Solution to 1975-CE-AMATH-I-X

A rod $PQ$ of length $4$ units slides with $P$ on the $x$ -axis and $Q$ on the $y$ -axis. $R$ is the point on $PQ$ such that $PR:RQ=1:3$ . Find the equation of the locus of $R$ .

Roughwork.

Let $P=(a,0)$ and $Q=(0,b)$ where $a,b\in [0,4]$ and $a^2+b^2=(4)^2$ . The locus of $R(x,y)$ is

$\begin{aligned} &\quad\enspace \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} \frac{3}{4}a \\ \frac{1}{4}b \end{pmatrix} \\ &\Longrightarrow \begin{pmatrix} a \\ b \end{pmatrix} = \begin{pmatrix} \frac{4}{3}x \\ 4y \end{pmatrix} \\ &\Longrightarrow \bigg(\frac{4}{3}x\bigg)^2+ (4y)^2 = 16 \\ &\Longrightarrow \enspace R:\big\{ x^2+9y^2-9=0\big\} \\ \end{aligned}$

This problem is not to be attempted.

December 13, 2022December 13, 2022

202212131509 Solution to 1985-CE-AMATH-II-6

Find the equations of the two tangents drawn from the point $(-1,0)$ to the parabola $y^2=4x$ .

Roughwork.

$\begin{aligned} y^2 & = 4x \\ 2y\,\mathrm{d}y & = 4\,\mathrm{d}x \\ \frac{\mathrm{d}y}{\mathrm{d}x} & = \frac{2}{y} \\ \end{aligned}$

Let $(a_1,b_1)$ and $(a_2,b_2)$ be two points at which the two tangents touches the parabola. Then,

$\displaystyle{\frac{2}{b_i} =\frac{\mathrm{d}y}{\mathrm{d}x}\bigg|_{(a_i,b_i)} =\frac{b_i-(0)}{a_i-(-1)}}$

By solving two equations in two unknowns:

$\begin{cases} b_i^2 = 2a_i+2 \\ b_i^2 = 4a_i \\ \end{cases} \Longrightarrow\quad \begin{pmatrix} a_i\\b_i \end{pmatrix}_{i=1,2} = \Bigg\{ \begin{pmatrix} 1\\ 2 \end{pmatrix} , \begin{pmatrix} 1 \\ -2 \end{pmatrix} \Bigg\}$

This problem is not to be attempted.

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 9, 2022December 21, 2022

202212091158 Solution to 1965-HL-PHY-3

(a) State the three different types of heat transfer, and give a simple example for each type of heat transfer.
(b) Calculate the amount of heat required to change $10\,\mathrm{g}$ of ice in a sealed container of volume $1\,\mathrm{L}$ and pressure at $1\,\mathrm{atm}$ at $-20\,^\circ\mathrm{C}$ to steam at $120\,^\circ\mathrm{C}$ . (Neglect the heat absorbed by the air and the container. The specific heat of ice and steam is $0.5\,\mathrm{cal\,g^{-1}\,^\circ C^{-1}}$ ). What is the pressure inside the container expressed in atmospheric pressure, when the $10\,\mathrm{g}$ of ice is completely changed to steam at $120\,^\circ\mathrm{C}$ ?

Roughwork.

(a) Skip it.

(b) Stuck. Any equation of state $f(P,V,T)=0$ as may well be applicable to solids and liquids as the ideal gas law $PV=nRT$ to gases, lacking the enthalpies of fusion and vaporisation unknown?

(to be continued)