Public Examinations in Mathematics – Page 7

October 14, 2022October 20, 2022

202210141629 Solution to 1976-CE-AMATH-II-XX

By using

$\displaystyle{\int_{a}^{b}f(x)\,\mathrm{d}x = \int_{a}^{c}f(x)\,\mathrm{d}x + \int_{c}^{b}f(x)\,\mathrm{d}x}$

or otherwise, evaluate

$\displaystyle{\int_{0}^{\pi}\cos^{7}x\,\mathrm{d}x}$ .

Warm-up.

$\begin{aligned} f(x) & =\cos^7x \\ f(-x) & = \cos^7(-x) \\ & = \cos^7(x) \\ & = f(x) \\ \end{aligned}$

$\therefore$ $f(x)$ is an even function such that

$\displaystyle{\int_{-a}^{a}f(x)\,\mathrm{d}x = 2\int_{0}^{a}f(x)\,\mathrm{d}x=2\int_{-a}^{0}f(x)\,\mathrm{d}x}$ .

Assume $f(x+T)=f(x)$ is a periodic function for some $T\in (0,2\pi ]$ .

$\begin{aligned} & x\overset{f} \longrightarrow \cos^7x \\ \sim\enspace & x\overset{g}\longrightarrow \cos x\overset{h}\longrightarrow \cos^7x \\ \end{aligned}$

where $g(x)=\cos x$ , $h(x)=x^7$ , and $f=h\circ g$ .

$g:[0,\pi ]\to [-1,1]$ by $x\mapsto \cos x$ is one-to-one and onto.

$h:[-1,1]\to [-1,1]$ by $x\mapsto x^7$ is injective and surjective as well.

Hence $f$ , the composition $h\circ g$ of bijections $g$ and $h$ , is also bijective. Besides $f:[0,\pi ]\to [-1,1]$ is a continuously differentiable function that $f(0)=1$ , $f(\frac{\pi}{2}) =0$ , and $f(\pi )=-1$ .

How do you evaluate the integral below:

$\displaystyle{F(\theta )=\int_{0}^{\theta}f(x)\,\mathrm{d}x}$

where $F$ is an odd function such that

$\begin{aligned} F(0) & = F(2\pi ) = 0 \\ F(\theta ) & = -F(-\theta ) \\ \end{aligned}$

The curve of function $f$ is flat at point(s) whose slope $f'$ is zero, i.e.,

$\begin{aligned} f'(x) & = 0 \\ \frac{\mathrm{d}}{\mathrm{d}x}(\cos^7x) & = 0 \\ -7\cos^6x\sin x & = 0 \\ x & = 0,\frac{\pi}{2}, \pi \\ \end{aligned}$

Roughwork. Show

Integrating by parts,

$\begin{aligned} \int_{0}^{\pi}\cos^7x\,\mathrm{d}x & = \int_{0}^{\pi}\cos^6x\,\mathrm{d}(\sin x) \\ & = \big[\cos^6x\sin x\big]\big|_{0}^{\pi} - \int_{0}^{\pi}\sin x\,\mathrm{d}(\cos^6x)\\ & = 0 + \int_{0}^{\pi}6\sin^2x\cos^5x\,\mathrm{d}x \\ & = \int_{0}^{\pi}6(1-\cos^2x)\cos^5x\,\mathrm{d}x \\ & = \int_{0}^{\pi}6\cos^5x\,\mathrm{d}x - \int_{0}^{\pi}6\cos^7x\,\mathrm{d}x \\ \int_{0}^{\pi}7\cos^7x\,\mathrm{d}x & = \int_{0}^{\pi}6\cos^5x\,\mathrm{d}x \\ \int_{0}^{\pi}\cos^7x\,\mathrm{d}x & = \frac{6}{7}\int_{0}^{\pi}\cos^5x\,\mathrm{d}x \\ & = \frac{6}{7}\int_{0}^{\pi}\cos^4x\,\mathrm{d}(\sin x) \\ &= \frac{6}{7}\bigg\{ \big[\cos^4x\sin x\big]\big|_{0}^{\pi} - \int_{0}^{\pi}\sin x\,\mathrm{d}(\cos^4x)\bigg\} \\ &= \frac{6}{7}\bigg\{ 0 + \int_{0}^{\pi}4\sin^2 x\cos^3x\,\mathrm{d}x\bigg\} \\ & = \frac{24}{7}\int_{0}^{\pi} (1-\cos^2x)\cos^3x\,\mathrm{d}x \\ \frac{30}{7}\int_{0}^{\pi}\cos^5x\,\mathrm{d}x & = \frac{24}{7} \int_{0}^{\pi}\cos^3x\,\mathrm{d}x \\ \int_{0}^{\pi}\cos^5x\,\mathrm{d}x & = \frac{4}{5}\int_{0}^{\pi}\cos^3x\,\mathrm{d}x \\ \end{aligned}$

$\begin{aligned} \int_{0}^{\pi}\cos^7x\,\mathrm{d}x & = \frac{6}{7}\bigg\{ \frac{4}{5}\int_{0}^{\pi}\cos^3x\,\mathrm{d}x \bigg\} \\ & = \frac{24}{35} \int_{0}^{\pi} \cos^2x\,\mathrm{d}(\sin x) \\ & = \frac{24}{35}\bigg\{ \big[ \cos^2x\sin x\big]\big|_{0}^{\pi} - \int_{0}^{\pi}\sin x\,\mathrm{d}(\cos^2x) \bigg\} \\ & = \frac{24}{35} \bigg\{ 0 + \int_{0}^{\pi}2\sin^2x\cos x\,\mathrm{d}x \bigg\} \\ & = \frac{48}{35} \int_{0}^{\pi}(1-\cos^2x)\cos x\,\mathrm{d}x \\ \bigg(\frac{24}{35}+\frac{48}{35}\bigg) \int_{0}^{\pi} \cos^3x\,\mathrm{d}x & = \frac{48}{35}\int_{0}^{\pi}\cos x\,\mathrm{d}x \\ \int_{0}^{\pi} \cos^3x\,\mathrm{d}x & = \frac{2}{3}\int_{0}^{\pi}\cos x\,\mathrm{d}x \\ \int_{0}^{\pi}\cos^7x\,\mathrm{d}x & = \frac{24}{35}\bigg\{ \frac{2}{3}\int_{0}^{\pi}\cos x\,\mathrm{d}x \bigg\} \\ & = \frac{16}{35}\big[ \sin x\big]\big|_{0}^{\pi} \\ & = 0 \\ \end{aligned}$

Solution.

$\begin{aligned} \int_{0}^{\pi}\cos^7x\,\mathrm{d}x & = \int_{0}^{\frac{\pi}{2}}\cos^7x\,\mathrm{d}x + \int_{\frac{\pi}{2}}^{\pi}\cos^7x\,\mathrm{d}x \\ & = \int_{0}^{\frac{\pi}{2}}\cos^7x\,\mathrm{d}x + \int_{0}^{\frac{\pi}{2}}\cos^7(x-\pi /2)\,\mathrm{d}(x-\pi /2) \\ & = \int_{0}^{\frac{\pi}{2}}\cos^7x\,\mathrm{d}x - \int_{0}^{\frac{\pi}{2}}\sin^7x\,\mathrm{d}x \\ & = \int_{0}^{\frac{\pi}{2}}(\cos^7x-\sin^7x)\,\mathrm{d}x \\ \end{aligned}$

The rest is left as an exercise to the reader.

October 13, 2022October 18, 2022

202210131233 Solution to 1969-CE-AMATH-I-XX

Find $y$ in terms of $x$ , if $\displaystyle{\frac{\mathrm{d}y}{\mathrm{d}x}=8x-1}$ and $y=-1$ when $x=-1$ .

Roughwork.

$\begin{aligned} \frac{\mathrm{d}y}{\mathrm{d}x} & = 8x-1 \\ y(x) & = \int \frac{\mathrm{d}y}{\mathrm{d}x}\,\mathrm{d}x \\ & = \int (8x-1)\,\mathrm{d}x \\ & = 4x^2-x+C\qquad\textrm{for some constant }C \\ \because\enspace -1 & = y(-1) \\ -1 & = 4(-1)^2-(-1)+C \\ \therefore\enspace C & = -6 \\ \end{aligned}$

In conclusion, $y(x) = 4x^2-x-6$ .

July 7, 2022October 24, 2022

202207071711 Solution to 1976-CE-AMATH-I-XX

July 6, 2022October 24, 2022

202207061152 Solution to 1971-CE-AMATH-I-XX

If $k$ is an integer, simplify

$\displaystyle{\cos\bigg[ \frac{1}{2}(4k+1)\pi +\theta \bigg]\tan\bigg[\frac{1}{2}(4k+3)\pi -\theta \bigg]}$ .

Roughwork.

First,

$\begin{aligned} & \quad \cos\bigg[ \frac{1}{2}(4k+1)\pi +\theta \bigg]\tan\bigg[\frac{1}{2}(4k+3)\pi -\theta \bigg] \\ & = \cos\bigg( 2k\pi+\frac{\pi}{2}+\theta \bigg) \tan \bigg( 2k\pi+\frac{3\pi}{2}-\theta \bigg) \\ \dots\textrm{as } &\textrm{are}\cos (\angle ),\tan (\angle )\textrm{ shifted by full periods, \textit{i.e.}, }k\cdot 2\pi\,\dots \\ & = \cos\bigg(\frac{\pi}{2}+\theta\bigg) \tan \bigg(\frac{3\pi}{2}-\theta \bigg) \\ \end{aligned}$

Second,

recall that

$\begin{aligned} \textrm{(in radians)} & \quad\textrm{(in degrees)} \\ \pi & = 180^\circ \\ \frac{\pi}{2} & = 90^\circ \\ \frac{3\pi}{2} & = 270^\circ\textrm{ \scriptsize{OR} }-90^\circ = -\frac{\pi}{2}\\ \end{aligned}$

Hence,

$\begin{aligned} & \quad \cos\bigg(\frac{\pi}{2}+\theta\bigg) \tan \bigg(\frac{3\pi}{2}-\theta \bigg) \\ & = \cos\bigg(\frac{\pi}{2}+\theta\bigg) \tan \bigg(-\frac{\pi}{2}-\theta \bigg) \\ & = \cos\bigg(\frac{\pi}{2}+\theta\bigg) \tan \bigg[ -\bigg(\frac{\pi}{2}+\theta \bigg)\bigg] \\ \dots &\textrm{ as }\tan(-\theta )=-\tan\theta \enspace\dots \\ & = -\cos\bigg(\frac{\pi}{2}+\theta\bigg) \tan \bigg(\frac{\pi}{2}+\theta\bigg) \\ \end{aligned}$ .

Recall that

$\begin{aligned} \because \enspace & \tan\theta = \frac{\sin\theta}{\cos\theta} \\ \therefore \enspace & \cos\theta\tan\theta = \sin\theta \\ \end{aligned}$ .

Hence,

$\begin{aligned} &\quad -\cos\bigg( \frac{\pi}{2}+\theta\bigg)\tan\bigg( \frac{\pi}{2}+\theta\bigg) \\ & = -\sin\bigg( \frac{\pi}{2}+\theta\bigg) \\ \end{aligned}$

Third,

recall that

$\displaystyle{\sin \bigg(\theta\pm\frac{\pi}{2}\bigg)}=\pm\cos\theta$ .

$\begin{aligned} &\quad -\sin \bigg( \frac{\pi}{2}+\theta\bigg) \\ & = -\cos\theta \\ \end{aligned}$

Answer.

$\displaystyle{\cos\bigg[ \frac{1}{2}(4k+1)\pi +\theta \bigg]\tan\bigg[\frac{1}{2}(4k+3)\pi -\theta \bigg]}=-\cos\theta$ .

July 5, 2022October 24, 2022

202207051339 Solution to 1974-CE-AMATH-I-XX

(a) In the figure below, $ABCDEF$ is a regular hexagon. Which one of the $6$ vectors $\overrightarrow{AD}$ , $\overrightarrow{DA}$ , $\overrightarrow{FC}$ , $\overrightarrow{CF}$ , $\overrightarrow{EB}$ , $\overrightarrow{BE}$ is equal to $\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{DC}$ ?

(b) $\mathbf{u}$ and $\mathbf{v}$ are unit vectors making an angle $60^\circ$ with each other as shown in the figure below. $AB$ is a vector making an angle $30^\circ$ with $\mathbf{u}$ and $|\overrightarrow{AB}|=2$ . Express $\overrightarrow{AB}$ in terms of $\mathbf{u}$ and $\mathbf{v}$ .

(c) In the figure below, $\angle B=90^\circ$ and $m(\overrightarrow{AB})=10$ . Calculate $\overrightarrow{AB}\cdot\overrightarrow{AC}$ .

(a)

$\begin{aligned} \mathbf{AD} & = -\mathbf{DA} \\ |\mathbf{AD}| & = |\mathbf{DA}| \\ \mathbf{FC} & = -\mathbf{CF} \\ |\mathbf{FC}| & = |\mathbf{CF}| \\ \mathbf{EB} & = -\mathbf{BE} \\ |\mathbf{EB}| & = |\mathbf{BE}| \\ \end{aligned}$

Ans. $\mathbf{FC}$ $\textrm{\scriptsize{OR}}$ $\overrightarrow{FC}$ by inspection.

Working.

$\begin{aligned} &\quad \mathbf{AB} + \mathbf{BC} + \mathbf{DC} \\ & = (\mathbf{AB} + \mathbf{BC}) + \mathbf{DC} \\ & = \mathbf{AC} + \mathbf{DC} \\ \dots & \textrm{ as }\mathbf{AC}=\mathbf{FD} \enspace\dots \\ & = \mathbf{FD} + \mathbf{DC} \\ & = \mathbf{FC} \\ \end{aligned}$

(b)

Write, in Cartesian components of unit vectors $\hat{\mathbf{i}}$ and $\hat{\mathbf{j}}$ , the following:

$\begin{aligned} \mathbf{AB} & = |\mathbf{AB}|\cos 30^\circ\,\hat{\mathbf{i}} + |\mathbf{AB}|\sin 30^\circ\,\hat{\mathbf{j}} \\ & = 2\cos 30^\circ\,\hat{\mathbf{i}} + 2\sin 30^\circ\,\hat{\mathbf{j}} \\ & = \sqrt{3}\,\hat{\mathbf{i}} + 1\,\hat{\mathbf{j}} \\ \end{aligned}$

and similarly,

$\begin{aligned} \mathbf{u} & = |\mathbf{u}|\,\hat{\mathbf{i}} \\ \mathbf{v} & = |\mathbf{v}|\cos 60^\circ\,\hat{\mathbf{i}} + |\mathbf{v}|\sin 60^\circ\,\hat{\mathbf{j}}\\ & \\ \because\enspace & \mathbf{u}, \mathbf{v}\textrm{ are unit vectors} \\ \therefore\enspace & |\mathbf{u}|=|\mathbf{v}|=1 \\ & \\ \mathbf{u} & = \hat{\mathbf{i}} \\ \mathbf{v} & = \frac{1}{2}\,\hat{\mathbf{i}} + \frac{\sqrt{3}}{2} \hat{\mathbf{j}} \\ \end{aligned}$

$\begin{aligned} \begin{bmatrix} \mathbf{u} \\ \mathbf{v} \end{bmatrix} & = \begin{bmatrix} 1 & 0 \\ \frac{1}{2} & \frac{\sqrt{3}}{2} \end{bmatrix} \begin{bmatrix} \hat{\mathbf{i}} \\ \hat{\mathbf{j}} \end{bmatrix} \\ \begin{bmatrix} \hat{\mathbf{i}} \\ \hat{\mathbf{j}} \end{bmatrix} & = \begin{bmatrix} 1 & 0 \\ \frac{1}{2} & \frac{\sqrt{3}}{2} \end{bmatrix}^{-1}\begin{bmatrix} \mathbf{u} \\ \mathbf{v} \end{bmatrix} \\ \end{aligned}$

Lemma. (Inversion of 2-by-2 matrices)

For $2\times 2$ matrices, inversion can be done as follows:

$\begin{aligned} \mathbf{A}^{-1} & = \begin{bmatrix} a & b \\ c & d \end{bmatrix}^{-1} \\ & = \frac{1}{\mathrm{det}\mathbf{A}}\begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \\ & = \frac{1}{ad-bc}\begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \\ \end{aligned}$

_{Wikipedia on Invertible matrix}

Then,

$\begin{bmatrix} 1 & 0 \\ \frac{1}{2} & \frac{\sqrt{3}}{2}\end{bmatrix}^{-1} = \begin{bmatrix} 1 & 0 \\ -\frac{1}{\sqrt{3}} & \frac{2}{\sqrt{3}} \end{bmatrix}$

so that

$\begin{bmatrix} \hat{\mathbf{i}} \\ \hat{\mathbf{j}} \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ -\frac{1}{\sqrt{3}} & \frac{2}{\sqrt{3}} \end{bmatrix}\begin{bmatrix}\mathbf{u} \\ \mathbf{v}\end{bmatrix}$

Thus

$\begin{aligned} \mathbf{AB} & =\begin{pmatrix} \sqrt{3} & 1\end{pmatrix}\begin{bmatrix}\hat{\mathbf{i}} \\ \hat{\mathbf{j}} \end{bmatrix} \\ & = \begin{pmatrix} \sqrt{3} & 1\end{pmatrix} \begin{bmatrix} 1 & 0 \\ -\frac{1}{\sqrt{3}} & \frac{2}{\sqrt{3}} \end{bmatrix} \begin{bmatrix}\mathbf{u} \\ \mathbf{v}\end{bmatrix} \\ & = \begin{bmatrix} \frac{2}{\sqrt{3}} & \frac{2}{\sqrt{3}} \end{bmatrix}\begin{bmatrix}\mathbf{u} \\ \mathbf{v}\end{bmatrix} \\ &=\frac{2}{\sqrt{3}}\,\mathbf{u}+\frac{2}{\sqrt{3}}\,\mathbf{v} \\ \end{aligned}$

There might be some mistake if conceptually.

(c)

I don’t know how. Skip it.

March 7, 2022October 24, 2022

202203071350 Solution to 1968-CE-AMATH-I-XX

Find the indefinite integrals:

i. $\displaystyle{\int\frac{\mathrm{d}x}{x^4}}$

ii. $\displaystyle{\int\frac{\mathrm{d}x}{(2x-3)^4}}$

Solution.

i.

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

ii.

$\begin{aligned} &\quad \int \frac{\mathrm{d}x}{(2x-3)^4} \\ & \cdots\enspace\cdots\enspace\cdots \\ & \textrm{let }y=2x-3\\ & \textrm{then }\frac{\mathrm{d}y}{\mathrm{d}x}=2 \\ & \textrm{and }\mathrm{d}x=\frac{\mathrm{d}y}{2} \\ & \cdots\enspace\cdots\enspace\cdots \\ & = \int\frac{\frac{\mathrm{d}y}{2}}{y^4} \\ & = \frac{1}{2}\int\frac{\mathrm{d}y}{y^4} \\ & \stackrel{\textrm{(i)}}{=} \frac{1}{2}\bigg( -\frac{1}{3x^3}+C\bigg) \\ & = -\frac{1}{6x^3}+C'\textrm{ for some arbitrary constant }C' \\ \end{aligned}$

March 6, 2022October 24, 2022

202203061946 Solution to 1967-CE-AMATH-I-XX

Find the indefinite integral $\int x\sqrt{x^2-2}\,\mathrm{d}x$ .

Solution.

$\begin{aligned} &\quad \int x\sqrt{x^2-2}\,\mathrm{d}x \\ &\enspace \cdots \enspace \cdots \enspace\cdots \\ \qquad & \textrm{let }y=x^2-2 \\ \qquad & \textrm{then }\mathrm{d}y=2x\,\mathrm{d}x \\ &\enspace \cdots \enspace\cdots \enspace\cdots \\ & = \int (x)(\sqrt{y})\bigg(\frac{\mathrm{d}y}{2x}\bigg) \\ & = \frac{1}{2}\int y^{\frac{1}{2}}\,\mathrm{d}y \\ & = \frac{y^{\frac{3}{2}}}{3} + C\textrm{ for some arbitrary constant }C\\ & = \frac{\sqrt{(x^2-2)^3}}{3} + C \\ \end{aligned}$

March 6, 2022October 24, 2022

202203061208 Solution to 1973-CE-AMATH-I-XX

A circle passes through the points $(-1,-1)$ , $(3,3)$ , and $(7,-1)$ . Find the equation of this circle.

Solution.

Let the centre of this circle be located at point $(a,b)$ and its radius be $R$ long. The equation of this circle is thus in the form

$(x-a)^2+(y-b)^2=R^2$ ,

where $a$ , $b$ , and $R$ are three unknowns to be obtained in the following three equations:

$\begin{aligned} (-1-a)^2+(-1-b)^2 & = R^2 \\ (3-a)^2+(3-b)^2 & = R^2 \\ (7-a)^2+(-1-b)^2 & = R^2 \\ \end{aligned}$

$\begin{aligned} (a^2+2a+1) + (b^2+2b+1) & = R^2 \\ (a^2-6a+9) + (b^2-6b+9) & = R^2 \\ (a^2-14a+49) + (b^2+2b+1) & = R^2 \\ \end{aligned}$

$\begin{aligned} a & = 3 \\ b & = -1 \\ R & = 4 \\ \end{aligned}$

$\therefore$ The equation of this circle is

$(x-3)^2+(y+1)^2=16$ .

February 7, 2022October 24, 2022

202202071146 Solution to 1969-CE-AMATH-I-XX

Find $\displaystyle{\int\frac{2x+5}{(x^2+5x+1)^3}\,\mathrm{d}x}$ .

Solution.

Set-up.

$\begin{aligned} \textrm{Let }u & =x^2+5x+1 \\ \frac{\mathrm{d}u}{\mathrm{d}x} & = 2x+5 \\ \mathrm{d}x & = \frac{\mathrm{d}u}{2x+5} \\ \end{aligned}$

Then,

$\begin{aligned} &\quad \int\frac{2x+5}{(x^2+5x+1)^3}\,\mathrm{d}x \\ & = \int \frac{2x+5}{u^3}\bigg( \frac{\mathrm{d}u}{2x+5}\bigg) \\ & = \int \frac{1}{u^3} \,\mathrm{d}u \\ & = -\frac{1}{2u^2}+C\textrm{ for some constant }C \\ & = -\frac{1}{2(x^2+5x+1)^2}+C\\ \end{aligned}$

February 6, 2022October 24, 2022

202202070744 Solution to 1997-CE-AMATH-II-2

Find $\displaystyle{\int x\sqrt{x-1}\,\mathrm{d}x}$ .

[Hint: Let $u=x-1$ .]

Solution.

$\begin{aligned} \textrm{Let } & u = x-1 \\ \textrm{then } & x = u+1 \\ \textrm{and }&\mathrm{d}x = \mathrm{d}u\\ \end{aligned}$

$\begin{aligned} &\quad \int x\sqrt{x-1}\,\mathrm{d}x\\ & = \int (u+1)(u^{\frac{1}{2}})\,\mathrm{d}u \\ & = \int (u^{\frac{3}{2}}+u^{\frac{1}{2}})\,\mathrm{d}u \\ & = \frac{2u^{\frac{5}{2}}}{5} + \frac{2u^{\frac{3}{2}}}{3} + C\textrm{ for some constant }C \\ & = \frac{2(x-1)^{\frac{5}{2}}}{5} + \frac{2(x-1)^{\frac{3}{2}}}{3} + C \\ \end{aligned}$