Public Examinations in Mathematics – Page 2

December 22, 2022December 22, 2022

202212221242 Solution to 1983-CE-AMATH-I-6

The complex number $z$ satisfies the condition

$|z-(3+i)|=|z-(5+5i)|$ .

If $z=x+yi$ , where $x$ and $y$ are real, find and simplify the relation between $x$ and $y$ . Find also the values of $x$ and $y$ for which $|z|$ is a minimum.

Roughwork.

$\begin{aligned} |(x+yi)-(3+i)| & = |(x+yi)-(5+5i)| \\ |(x-3)+(y-1)i| & = |(x-5)+(y-5)i| \\ \sqrt{(x-3)^2+(y-1)^2} & = \sqrt{(x-5)^2+(y-5)^2} \\ (x-3)^2+(y-1)^2 & = (x-5)^2+(y-5)^2 \\ \cdots & = \cdots \\ x+2y-10 & = 0 \\ \end{aligned}$

Substituting $10-2y$ for $x$ ,

$\begin{aligned} |z| (=|x+yi|) & = \sqrt{x^2+y^2} \\ & = \sqrt{(10-2y)^2+y^2} \\ & = \sqrt{5y^2-40y+100} \\ \frac{\mathrm{d}}{\mathrm{d}y}\big( |z|\big) & = \frac{\mathrm{d}}{\mathrm{d}y}\sqrt{5y^2-40y+100} \\ 0 & =: \frac{5y-20}{\sqrt{5y^2-40y+100}} \\ y & = 4 \\ x & = 10-2(4) = 2 \\ \end{aligned}$

Ans. $\begin{pmatrix}x = 2 \\ y = 4\end{pmatrix}$

December 21, 2022December 21, 2022

202212211715 Solution to 1968-CE-AMATH-II-X

Find $\displaystyle{\int\frac{\mathrm{d}x}{\sqrt{9-4x^2}}}$

i. by using the substitution $\displaystyle{x=\frac{3}{2}\cos\theta}$ ;
ii. by using the substitution $\displaystyle{x=\frac{3}{2}\sin\theta}$ .

Explain why you appear to get two different answers.

Roughwork.

i.

$\begin{aligned} &\quad\enspace \int \frac{1}{\sqrt{9-4x^2}}\,\mathrm{d}x \\ \dots\, & \textrm{let }x=\frac{3}{2}\cos\theta \textrm{ s.t. }\mathrm{d}x=-\frac{3}{2}\sin\theta\,\mathrm{d}\theta \,\dots \\ & = -\int \frac{3\sin\theta}{2\sqrt{9-4(\frac{3}{2}\cos\theta )^2}}\,\mathrm{d}\theta \\ & = -\int \frac{3\sin\theta}{6\sin\theta}\,\mathrm{d}\theta \\ & = -\frac{\theta}{2} +C_1\textrm{ for some constant}\\ & = -\frac{1}{2}\arccos \bigg(\frac{2x}{3}\bigg) +C_1 \\ \end{aligned}$

ii.

$\begin{aligned} &\quad\enspace \int \frac{1}{\sqrt{9-4x^2}}\,\mathrm{d}x \\ \dots\, & \textrm{let }x=\frac{3}{2}\sin\theta \textrm{ s.t. }\mathrm{d}x=\frac{3}{2}\cos\theta\,\mathrm{d}\theta \,\dots \\ & = \int \frac{3\cos\theta}{2\sqrt{9-4(\frac{3}{2}\sin\theta )^2}}\,\mathrm{d}\theta \\ & = \int \frac{3\cos\theta}{6\cos\theta}\,\mathrm{d}\theta \\ & = \frac{\theta}{2} +C_2\textrm{ for some constant} \\ & = \frac{1}{2}\arcsin \bigg(\frac{2x}{3}\bigg) +C_2 \\ \end{aligned}$

Let’s see if

$\displaystyle{-\frac{1}{2}\arccos\bigg(\frac{2x}{3}\bigg) +C_1 \stackrel{\textrm{?}}{\equiv}\frac{1}{2}\arcsin\bigg(\frac{2x}{3}\bigg)} + C_2$

or, to be rephrased, whether

$\displaystyle{\frac{\mathrm{d}}{\mathrm{d}x}\bigg[\arcsin \bigg(\frac{2x}{3}\bigg) + \arccos\bigg(\frac{2x}{3}\bigg) \bigg]\stackrel{\textrm{?}}{\equiv} 0}$

This is left an exercise for the reader.

December 21, 2022December 21, 2022

202212211654 Solution to 2009-CE-AMATH-II-6

$C$ is a curve with equation $y^3+x^3y=10$ .

(a) Find $\displaystyle{\frac{\mathrm{d}y}{\mathrm{d}x}}$ .
(b) Find the equation of the tangent to the curve $C$ at the point $(1,2)$ .

Roughwork.

$\begin{aligned} y^3+x^3y & = 10 \\ 3y^2\,\mathrm{d}y+x^3\,\mathrm{d}y+3x^2y\,\mathrm{d}x & = 0 \\ -\frac{3x^2y}{x^3+3y^2} = \frac{\mathrm{d}y}{\mathrm{d}x} & = y'(x,y) \\ \end{aligned}$

$\displaystyle{y'(1,2)=-\frac{3(1)^2(2)}{(1)^3+3(2)^2}=-\frac{6}{13}}$

$\begin{aligned} &\quad\enspace \frac{y-2}{x-1} = -\frac{6}{13} \\ & \Rightarrow \big\{ L:\quad 6x+13y-32 = 0\big\} \\ \end{aligned}$

This problem is not to be attempted.

December 21, 2022December 21, 2022

202212211306 Solution to 1987-CE-AMATH-I-2

Let $x=y+\sin y$ . Find $\displaystyle{\frac{\mathrm{d}y}{\mathrm{d}x}}$ and $\displaystyle{\frac{\mathrm{d}^2y}{\mathrm{d}x^2}}$ in terms of $y$ .

Roughwork.

$\begin{aligned} x & = y+\sin y \\ \mathrm{d}x & = \mathrm{d}y+\cos y\,\mathrm{d}y \\ \frac{\mathrm{d}y}{\mathrm{d}x} & = \frac{1}{1+\cos y} \\ \bigg(\frac{\mathrm{d}y}{\mathrm{d}x}\bigg)^2 & = \frac{1}{(1+\cos y)^2} \\ \frac{\mathrm{d}}{\mathrm{d}x}\bigg(\frac{\mathrm{d}y}{\mathrm{d}x}\bigg)^2 & = 2\bigg(\frac{\mathrm{d}y}{\mathrm{d}x}\bigg)\bigg(\frac{\mathrm{d}^2y}{\mathrm{d}x^2}\bigg) \\ \end{aligned}$

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

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

$\begin{aligned} \quad\enspace \frac{\mathrm{d}^2y}{\mathrm{d}x^2} & = \frac{1+\cos y}{2}\cdot \frac{2\sin y}{(1+\cos y)^4} \\ & = \frac{\sin y}{(1+\cos y)^3} \\ \end{aligned}$

Stretching the focus, compromise the better for it.

December 21, 2022December 21, 2022

202212211204 Solution to 1972-CE-AMATH-I-X

$F(x)\equiv A(x-1)^2+B(x-1)+C$ where $A$ , $B$ , and $C$ are constants. Find the values of $A$ , $B$ , and $C$ , given that $(x+2)$ is a factor of $F(x)$ and that the remainders when $F(x)$ is divided by $(x-1)$ and $(x+1)$ are $9$ and $-11$ respectively.

Roughwork.

$\begin{aligned} & \quad\enspace (x+2)\textrm{ is a factor of }F(x) \\ & \Leftrightarrow F(x)=(x+2)\cdot G(x) \\ & \Leftrightarrow F(-2)=0 \\ & \Leftrightarrow A((-2)-1)^2+B((-2)-1)+C=0 \\ \end{aligned}$

$\begin{aligned} & \quad\enspace \mathrm{9}\textrm{ remains when dividing }F(x)\textrm{ by }(x-1) \\ &\Leftrightarrow F(1)=C=9 \\ \end{aligned}$

$\begin{aligned} & \quad\enspace \mathrm{-11}\textrm{ remains when dividing }F(x)\textrm{ by }(x+1) \\ &\Leftrightarrow F(-1)=A((-1)-1)^2+B((-1)-1)+C=-11 \\ \end{aligned}$

This problem is not to be attempted.

December 20, 2022December 20, 2022

202212201209 Solution to 2001-CE-AMATH-I-10

Two lines $L_1:x+y-5=0$ and $L_2:2x-3y=0$ intersect at a point $A$ . Find the equations of the two lines passing through $A$ whose distances from the origin are equal to $2$ .

Roughwork.

Solving for $(a,b)$ point of intersection:

$\begin{cases} a+b-5 = 0 \\ 2a-3b =0 \\ \end{cases}$

we have it $A(3,2)$ .

Let the equations of the two lines be

$\begin{cases} l_1: & a_1x+b_1y+c_1 = 0 \\ l_2: & a_2x+b_2y+c_2 = 0 \\ \end{cases}$

Despite the formula

$\displaystyle{\textrm{distance}(ax+by+c=0,(x_0,y_0))=\frac{|ax_0+by_0+c|}{\sqrt{a^2+b^2}}}$

_{Wikipedia on Distance from a point to a line}

let’s rely on first principles. So draw a picture.

And the rest is left an exercise for the reader.

December 14, 2022December 14, 2022

202212141640 Solution to 1983-CE-AMATH-II-5

Find the equation of the two lines which are both parallel to the line

$3x-2y=0$

and tangent to the ellipse

$4x^2+y^2=16$ .

Roughwork.

The slope of the line is

$\begin{aligned} 3x-2y & =0 \\ 3\,\mathrm{d}x-2\,\mathrm{d}y & = 0 \\ \textrm{Slope }m =\frac{\mathrm{d}y}{\mathrm{d}x} & = \frac{3}{2} \\ \end{aligned}$

and the gradient of the ellipse

$\begin{aligned} 4x^2+y^2 & =16 \\ 8x\,\mathrm{d}x + 2y\,\mathrm{d}y & = 0 \\ m(x,y) =\frac{\mathrm{d}y}{\mathrm{d}x} & = -\frac{4x}{y} \\ \end{aligned}$

Plugging

$\begin{aligned} \frac{3}{2} & = -\frac{4x}{y} \\ y & = -\frac{8}{3}x \\ \end{aligned}$

into

$\begin{aligned} 4x^2+\bigg(-\frac{8}{3}x\bigg)^2 & = 16 \\ x & = \pm \frac{6}{5} \\ \end{aligned}$

so are the points of contact

$\begin{pmatrix} x\\y \end{pmatrix} = \Bigg\{\begin{pmatrix} 6/5 \\ -16/5 \end{pmatrix} , \begin{pmatrix} -6/5 \\ 16/5 \end{pmatrix} \Bigg\}$

From

$\displaystyle{\frac{y\mp (6/5)}{x\pm (16/5)}=\frac{3}{2}}$

follow the equations

$3x-2y\pm 12 = 0$

as requested.

This problem is not to be attempted.

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.