Public Examinations in Mathematics – Page 6

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.

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

October 19, 2022October 19, 2022

202210191554 Solution to 1969-CE-AMATH-II-XX

Find the $x$ -coordinate of the point on the curve $y=ax^2+bx+c$ where the tangent is horizontal.

Roughwork.

The gradient $y'(x)$ of the curve $y(x)$ is

$y'(x)=2ax+b$ .

When the tangent is horizontal,

$\begin{aligned} y'(x) & =0 \\ 2ax+b & =0 \\ x & = -\frac{b}{2a} \\ y & = a\bigg( \frac{-b}{2a}\bigg)^2+b\bigg( \frac{-b}{2a}\bigg)+c \\ & = \dots \\ \end{aligned}$

October 19, 2022October 19, 2022

202210191506 Solution to 1970-CE-AMATH-II-XX

Calculate the gradient of the curve $x^2+xy+y^2=1$ at the point $(1,-1)$ .

Roughwork.

Let the function $f$ of variables $x$ and $y$ be

$f(x,y):=x^2+xy+y^2-1=0$

Then,

$\begin{aligned} \Delta f=2x\Delta x+x\Delta y+y\Delta x+2y\Delta y & =0 \\ 2x + x\frac{\Delta y}{\Delta x}+y+2y\frac{\Delta y}{\Delta x} & = 0 \\ (x+2y)\frac{\Delta y}{\Delta x} & = -2x-y \\ \frac{\Delta y}{\Delta x} & = - \frac{2x+y}{x+2y} \\ \end{aligned}$

The gradient $\nabla$ at point $(x=1,y=-1)$ is

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

October 19, 2022October 19, 2022

202210191436 Solution to 1969-CE-AMATH-II-XX

Calculate the gradient of the curve $x=3y+y^2$ at the point $(0,-3)$ .

Roughwork.

Differentiating $x(y)$ wrt to $y$ and taking reciprocal,

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

the gradient $\nabla y(x)$ at point $(x,y) = (0,-3)$ is

$\displaystyle{\frac{\mathrm{d}y}{\mathrm{d}x}\bigg|_{(0,-3)} = \frac{1}{3+2(-3)} = -\frac{1}{3}}$ .

October 19, 2022October 19, 2022

202210191139 Solution to 1973-CE-AMATH-II-XX

Find the area bounded by the curve $y=4x-x^2$ and the straight line $x-y+2=0$ .

Roughwork.

Rewriting,

$\begin{cases} C:y=-x^2+4x \\ L:y=x+2 \\ \end{cases}$

Solving,

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

$\begin{aligned} x=1 \Rightarrow y=(1)+2=3 \\ x=2 \Rightarrow y=(2)+2=4 \\ \end{aligned}$

Now that the points of intersection are $(1,3)$ and $(2,4)$ , plot a graph below.

By either definite integral,

$\textrm{Area} & = \begin{cases} \displaystyle{\int_{1}^{2}\big(y_C(x)-y_L(x)\big)\,\mathrm{d}x }\\ \displaystyle{\int_{3}^{4}\big(x_L(y)-x_C(y)\big)\,\mathrm{d}y }\\ \end{cases}$

The upper integral is easier solvable than the lower one. Exercise.

October 19, 2022October 19, 2022

202210191050 Solution to 1977-CE-AMATH-II-XX

Prove that

$\displaystyle{\cos^4\theta = \frac{1}{8}(\cos 4\theta + 4\cos 2\theta + 3)}$ .

Hence solve the equation

$\cos 4\theta + 4\cos 2\theta +1=0$

for $0^\circ\leqslant\theta\leqslant 360^\circ$ .

Roughwork.

$\begin{aligned} \textrm{RHS} & = \frac{1}{8}(\cos 4\theta + 4\cos 2\theta + 3) \\ & = \frac{1}{8}\big(\cos 2(2\theta ) + 4\cos 2\theta + 3\big) \\ & = \frac{1}{8}\big( (2\cos^22\theta -1) + 4\cos 2\theta + 3\big) \\ & = \frac{1}{8}(2\cos^22\theta +4\cos 2\theta +2) \\ \dots\,\because\enspace & \cos2\theta = 2\cos^2\theta -1\,\dots \\ & = \frac{1}{8}\big( 2(2\cos^2\theta -1)^2+4(2\cos^2\theta -1)+2\big) \\ & = \dots \\ & = \cos^4\theta \\ & = \textrm{LHS} \\ \end{aligned}$

The missing steps are left the reader.

October 18, 2022October 18, 2022

202210181552 Solution to 1971-CE-AMATH-I-XX

Show that the straight line $x+2y+4=0$ touches the curve $y^2=4x$ .

Roughwork.

If there exists some point $(a,b)$ of intersection of the straight line and the curve, we have

$\begin{cases} a+2b+4 = 0 \\ b^2 = 4a \\ \end{cases}$

Substituting $-2b-4$ for $a$ in the second equation, we have

$\begin{aligned} & b^2 = 4(-2b-4) \\ \Leftrightarrow\quad & b^2 + 8b +16 = 0 \\ \Leftrightarrow\quad & (b+4)^2 = 0 \\ \Leftrightarrow\quad & b = -4\quad\textrm{(rep.)} \\ \Rightarrow\quad & a = -2(-4)-4= 4 \\ \end{aligned}$

$\begin{cases} a = 4 \\ b = -4 \\ \end{cases}$

Such point as $(4,-4)$ exists.

$\because$ The intersection of $\begin{cases} L: x+2y+4=0 \\ C: y^2 = 4x \\ \end{cases}$ is non-empty.

$\therefore$ The straight line touches the curve.