Public Examinations in Mathematics – Page 8

December 2, 2021October 24, 2022

202112021444 Solution to 1979-CE-AMATH-I-1

Let $\displaystyle{y=x+\frac{1}{x^2}}$ . Find $\displaystyle{\frac{\mathrm{d}y}{\mathrm{d}x}}$ from first principles.

Solution.

From first principles,

$\begin{aligned} y:=f(x) & =x+\frac{1}{x^2} \\ f(x+\Delta x) & =(x+\Delta x) + \frac{1}{(x+\Delta x)^2} \\ f(x+\Delta x)-f(x) & = \Delta x + \frac{1}{(x+\Delta x)^2} - \frac{1}{x^2} \\ & = \Delta x + \frac{x^2-(x+\Delta x)^2}{x^2(x+\Delta x)^2} \\ & = \Delta x - \frac{2x\Delta x+(\Delta x)^2}{x^2(x+\Delta x)^2} \\ \frac{f(x+\Delta x)-f(x)}{\Delta x} & = 1 - \frac{2x + \Delta x}{x^2(x+\Delta x)^2} \\ \lim_{\Delta x\to 0}\frac{f(x+\Delta x)-f(x)}{\Delta x} & = 1 - \frac{2x + 0}{x^2(x+0)^2} \\ \frac{\mathrm{d}y}{\mathrm{d}x} & = 1 - \frac{2}{x^3} \\ \end{aligned}$

Playground.

To make fun of calculus, do let

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

where $f(x,y)$ is a degree $3$ polynomial in two variables $x$ and $y$ .

The set of solutions to $f(x,y)=0$ is

$\begin{cases} \enspace x \neq 0 \\ \enspace y = \frac{x^3+1}{x^2} \end{cases}$

Computing $\partial_xf$ and $\partial_yf$ as follow:

$\begin{aligned} \partial_xf & = \frac{\partial}{\partial x}f(x,y) \\ & = 3x^2-2xy \\ \end{aligned}$

and

$\begin{aligned} \partial_yf & = \frac{\partial}{\partial y}f(x,y) \\ & = -x^2 \\ \end{aligned}$ ,

we oversee the relation

$\mathrm{d}f = (\partial_xf)(\Delta x)+(\partial_yf)(\Delta y)$ .

Definition. (Singularity and Smoothness)

A point $p=(a,b)$ on a curve $\mathcal{C}=\{ (x,y)\in\mathbb{C}^2:f(x,y)=0\}$ is said to be singular if

$\begin{aligned} \frac{\partial f}{\partial x}(a,b) & = 0 \\ \frac{\partial f}{\partial y}(a,b) & = 0 \\ \end{aligned}$ .

A point that is not singular is called smooth. If there is at least one singular point on $\mathcal{C}$ , then curve $\mathcal{C}$ is called a singular curve. If there are no singular points on $\mathcal{C}$ , the curve $\mathcal{C}$ is called a smooth curve.

_{C.f. Definition 1.9.1, T. Carrity, et al., Algebraic Geometry: A Problem Solving Approach}

That said, in our scenario

$f(x,y)=x^3-x^2y+1$ ,

the set of candidates for singularity

$\{ (a,b)\in\mathbb{C}^2\enspace \mathrm{s.t.}\enspace f(a,b)=0\textrm{ and }\partial_xf(a,b)=\partial_yf(a,b)=0\}$ .

is empty.

$\therefore$ The curve $\mathcal{C}=\{ (x,y)\in\mathbb{C}^2:x^3-x^2y+1=0\}$ is smooth, thus everywhere differentiable.

Wait, the function in question should be the curve

$F(x)=\displaystyle{x+\frac{1}{x^2}}$ ;

still, for better or worse, $F(x)$ is $\textrm{\scriptsize{NOT}}$ a differentiable function because the derivative does $\textrm{\scriptsize{NOT}}$ exist at $x=0$ , for that point is a discontinuity.

November 25, 2021October 24, 2022

202111251031 Solution to 1995-IB-MATH-HL-I-3

Find the value of $x$ such that

$\begin{array}{|cc|} x & 6 \\ x & x \end{array} = \begin{array}{|ccc|} 2 & 1 & 3 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{array}$ .

Working.

$\begin{aligned} \textrm{LHS} & = x^2 - 6x \\ \textrm{RHS} & = 2 \cdot \begin{array}{|cc|} 2 & 3 \\ 1 & 1 \end{array} - 1 \cdot \begin{array}{|cc|} 1 & 3 \\ 3 & 1 \end{array} + 3 \cdot \begin{array}{|cc|} 1 & 2 \\ 3 & 1 \end{array} \\ & = 2\big( (2)(1)-(1)(3) \big) - 1\big( (1)(1)-(3)(3) \big) + 3\big( (1)(1)-(3)(2) \big) \\ & = 2(-1) - 1(-8) + 3(-5) \\ & = -9 \\ \end{aligned}$

$\begin{aligned} \textrm{LHS} & = \textrm{RHS} \\ x^2-6x & = -9 \\ x^2-6x+9 & = 0 \\ (x-3)^2 & = 0 \\ x & = 3 \qquad\textrm{(double roots)} \\ \end{aligned}$

Answers.

$x=3$ .

November 25, 2021October 24, 2022

202111250959 Solution to 1995-IB-MATH-HL-I-2

Calculate $x$ and $y$ exactly, as rational numbers, if

$\begin{aligned} 5^{3x}\times 25^{y}& =\frac{1}{5} \\ \textrm{and} \qquad\qquad 7^{x}\times 49^{2y} & = 1 \\ \end{aligned}$ .

Working.

From the first equation,

$\begin{aligned} 5^{3x}\times 25^{y} & = \frac{1}{5} \\ 5^{3x} \times 5^{2y} & = 5^{-1} \\ 5^{3x+2y} & = 5^{-1} \\ 3x+2y & = -1 \\ \end{aligned}$ .

From the second equation,

$\begin{aligned} 7^x\times 49^{2y} & = 1 \\ 7^x\times 7^{4y} & = 7^{0} \\ 7^{x+4y} & = 7^0 \\ x+4y & = 0 \\ \end{aligned}$ .

We have two linear equations with two unknowns:

$\begin{cases} 3x+2y = -1 \\ x+4y = 0 \\ \end{cases}$

Substituting $-4y$ for $x$ ,

$\begin{aligned} 3(-4y)+2y & =-1 \\ -12y+2y & = -1 \\ -10 y & = -1 \\ y & = \frac{1}{10} \\ \end{aligned}$

Then,

$x=-4y=\displaystyle{-4\bigg(\frac{1}{10}\bigg)=-\frac{2}{5}}$ .

Answers.

$\begin{cases} \enspace x = - \displaystyle{\frac{2}{5}} \\ \enspace y = \displaystyle{\frac{1}{10}} \\ \end{cases}$

November 25, 2021October 24, 2022

202111250947 Solution to 1995-IB-MATH-HL-I-1

Given the complex numbers $z_1=3+\mathrm{i}$ and $z_2=2-\mathrm{i}$ , find

(a) $z_1z_2$ :

(b) $\displaystyle{\frac{z_1}{z_2}}$ .

Working.

(a)

$\begin{aligned} z_1z_2 & = (3+\mathrm{i})(2-\mathrm{i}) \\ & = 6-3\mathrm{i}+2\mathrm{i}-\mathrm{i}^2 \\ & = 6-\mathrm{i}-(-1) \\ & = 7-\mathrm{i} \\ \end{aligned}$

(b)

$\begin{aligned} \frac{z_1}{z_2} & = \frac{3+\mathrm{i}}{2-\mathrm{i}} \\ & = \frac{(3+\mathrm{i})(2+\mathrm{i})}{(2-\mathrm{i})(2+\mathrm{i})} \\ & = \frac{6+3\mathrm{i}+2\mathrm{i}+\mathrm{i}^2}{(2)^2-(\mathrm{i})^2} \\ & = \frac{5+5\mathrm{i}}{5} \\ & = 1 + \mathrm{i} \\ \end{aligned}$

Answers.

(a) $7-\mathrm{i}$ ; (b) $1 + \mathrm{i}$ .

November 24, 2021October 24, 2022

202111241152 Solution to 1971-CE-AMATH-3

Prove by mathematical induction that $n(n^2+2)$ is divisible by $3$ for all positive integral values of $n$ .

Solution.

Let $P(n)$ be the statement that

$P(n):\enspace n(n^2+2)\textrm{ is divisible by }3\textrm{ for }n\in\mathbb{Z}^{+}$

First, we wish to show $P(1)$ holds true.

$\begin{aligned} &\quad (1)\big( (1)^2 + 2\big) \\ & = 1(1+2) \\ & = 3\qquad (\textrm{divisible by }3)\\ \end{aligned}$

$\therefore$ $P(1)$ is true.

Once we assume that $P(n)$ is true for some positive integral values, we wish to show $P(n+1)$ is true.

$\begin{aligned} &\quad (n+1)\big( (n+1)^2 + 2\big) \\ & = (n+1)(n^2+2n+3) \\ & = n^3 + 3n^2 + 5n +3 \\ \end{aligned}$

Let $f(x)=x^3+3x^2+5x+3$ .

Suppose we can divide $f(x)$ by $3$ , i.e.,

$f(x)=3\cdot g(x)$

for some $g(x)=Ax^3+Bx^2+Cx+D$ where $A,B,C,D\in\mathbb{R}$ .

Then, let’s see

$\begin{aligned} \frac{f(x)}{g(x)} & = 3 \\ \frac{\mathrm{d}}{\mathrm{d}x}\bigg( \frac{f(x)}{g(x)}\bigg) & = \frac{\mathrm{d}}{\mathrm{d}x}(3) \\ \frac{g(x)f'(x)-f(x)g'(x)}{[g(x)]^2} & = 0 \\ g(x)f'(x)-f(x)g'(x) & = 0 \\ \end{aligned}$

Plugging in $f(x)$ , $f'(x)$ , $g(x)$ , and $g'(x)$ , we have

$\begin{aligned} &\quad (Ax^3+Bx^2+Cx+D)(3x^2+6x+5) \\ & = (x^3+3x^2+5x+3)(3Ax^2+2Bx+C) \\ \end{aligned}$

$\begin{aligned} &\quad \enspace \textrm{LHS} \\ & = (3A)x^5 + (6A+3B)x^4 + (5A+6B+3C)x^3 \\ &\qquad\quad + (5B+6C+3D)x^2 + (5C+6D)x + (5D) \\ \end{aligned}$

$\begin{aligned} &\quad \enspace \textrm{RHS} \\ & = (3A)x^5 + (9A+2B)x^4 + (15A+6B+C)x^3 \\ &\qquad\quad + (9A+10B+3C)x^2 + (6B+5C)x + (3C) \\ \end{aligned}$

We have a set of five equations below:

$\begin{aligned} 6A + 3B & = 9A + 2B \\ 5A+6B+3C & = 15A+6B+C \\ 5B+6C+3D & = 9A+10B+3C \\ 5C +6D & = 6B+5C \\ 5D & = 3C \\ \end{aligned}$

which can be summarized to

$\displaystyle{A=\frac{B}{3}=\frac{C}{5}=\frac{D}{3}}$ ,

that is equivalent to saying

$g(x)=\displaystyle{\frac{f(x)}{3}=\frac{x^3}{3}+x^2+\frac{5x}{3}+1}$ ,

viz. I have been wasting time over a circular reasoning of no use.

I should have tried to attest otherwise that $g(x)$ is of some positive integral values, even though this is $\textrm{\scriptsize{NOT}}$ proving the statement $P(n)$ by mathematical induction firsthand.

Reformulation.

To show that the following statement $Q(x)$ holds true:

$Q(x):\quad \displaystyle{\frac{x^3}{3}+x^2+\frac{5x}{3}+1}\textrm{ is an integer for any }x\in\mathbb{Z}^{+}$ .

$Q(1)$ is true because

$\displaystyle{\frac{(1)^3}{3}+(1)^2+\frac{5(1)}{3}+1}=4$

is an integer.

Assume $Q(x)$ is true for some $x\in\mathbb{Z}^{+}$ .

Then,

$\begin{aligned} &\qquad \frac{(x+1)^3}{3}+(x+1)^2+\frac{5(x+1)}{3}+1 \\ & = \frac{(x^3+3x^2+3x+1)}{3} + (x^2+2x+1)+\frac{(5x+5)}{3} + 1 \\ & = \bigg( \frac{x^3}{3}+x^2+\frac{5x}{3}+1\bigg) + \bigg( \frac{3x^2+3x+1}{3}+2x+1+\frac{5}{3}+1 \bigg) \\ \end{aligned}$

The first term is an integer for $Q(x)$ is assumed to be true. We then need to check whether the second term is of integral value(s).

The second term, after simplification, is

$x^2+3x+4$ ,

obviously an integer.

$\therefore$ $Q(x+1)$ thus holds.

Because $Q(1)$ is true, by mathematical induction, $Q(x)$ is true for all positive integers $x$ .

Escape to the original statement.

$3|n^3+3n^2+5n+3\Longleftrightarrow P(n+1)\textrm{ holds true.}$

Because $P(1)$ is true, by mathematical induction, $P(n)$ is true for any positive integers $n$ .

November 24, 2021October 24, 2022

202111241104 Solution to 1969-CE-AMATH-3

Prove by mathematical induction that if $n$ is a positive integer, $f(n)\equiv 3^{2n}-1$ is divisible by $8$ .

Solution.

We wish to show that the statement is true for $n=1$ , and also true for $n+1$ .

When $n=1$ , $f(1)=3^{2(1)}-1=8$ is divisible by $8$ .

Assume $f(n)$ is true for some positive integer $n$ , i.e.,

$\{f(n):3^{2n}-1=8m\, |\, \exists \, m, n\in\mathbb{Z}^{+}\}$ .

For $f(n+1)=3^{2(n+1)}-1$ , we have

$\begin{aligned} & \quad 3^{2(n+1)}-1 \\ & = 3^{2n+2} - 1 \\ & = 3^{2n}\cdot 3^2 - 1 \\ & = 9(3^{2n})-1 \\ & = 3^{2n} -1 + 8(3^{2n}) \\ \end{aligned}$

As is assumed $3^{2n}-1$ divisible by $8$ , the statement is thus also true for $n+1$ .

We have proven by mathematical induction that for any positive integer $n\enspace (\in\mathbb{Z}^+)$ , $f(n)$ is divisible by $8$ .

Afterword.

Try to prove $\textrm{\scriptsize{NOT}}$ by mathematical induction $\textrm{\scriptsize{BUT}}$ by direct proof.

$\begin{aligned} &\quad \frac{3^{2n}-1}{8} \\ & = \frac{3^{2n}-1}{9-1} \\ & = \frac{3^{2n}-1}{3^2-1} \\ & = \dots \end{aligned}$

Let $x=3^2$ , then

$\begin{aligned} &\quad \dots \\ & = \frac{x^n-1}{x-1} \\ & = 1+x+x^2+\cdots + x^{n-2} + x^{n-1} \enspace (\in\mathbb{Z}^{+}) \\ \end{aligned}$

$\therefore$ $f(n)\equiv 3^{2n}-1$ is divisible by $8$ .

November 22, 2021October 24, 2022

202111221509 Solution to 1971-CE-AMATH-1

When the functions $x^2-bx+3$ and $2b-x$ are divided by $(x-a)$ , the remainders are 1 and 4 respectively. Find $a$ and $b$ .

Solution.

Translating mathematically, we have

$\begin{aligned} x^2-bx+3 & = (x-a)F(x)+1 \\ 2b-x & = (x-a)G(x)+4\\ \end{aligned}$

Plugging in $x=a$ , we have

$\begin{aligned} (a)^2-b(a)+3 & = \big( (a)-a\big) F(a)+1 \\ 2b-(a) & = \big( (a)-a\big) G(a)+4 \\ \end{aligned}$

After simplifying it, we have

$\begin{aligned} a^2-ab & = -2\\ -a+2b & = 4\\ \end{aligned}$

Solving two equations with two unknowns,

$\begin{cases} & a =2\\ & b =3 \\ \end{cases}$

is the solution.

October 25, 2021October 24, 2022

202110251056 Solution to 1970-CE-AMATH-1

Given that $F(x)=(b-c)x^3-(a+2b-3c)x^2-(b+2c-3a)x-2(a-b)$ . By using the remainder theorem, prove that $(x-1)(x-2)$ is a factor of $F(x)$ . Find also the remaining factor of $F(x)$ .

Solution.

$\begin{aligned} F(1) & = (b-c)(1)^3 - (a+2b-3c)(1)^2 - (b+2c-3a)(1) - 2(a-b) = 0 \\ F(2) & = (b-c)(2)^3 - (a+2b-3c)(2)^2 - (b+2c-3a)(2) - 2(a-b) = 0 \\ \end{aligned}$

$\therefore$ $(x-1)(x-2)$ is a factor of $F(x)$ .

By trial-and-error,

$\begin{aligned} F(3) & = (b-c)(3)^3-(a+2b-3c)(3)^2-(b+2c-3a)(3)-2(a-b) \\ & = -2a+8b-6c \\ & \neq 0 \\ F(4) & = (b-c)(4)^3-(a+2b-3c)(4)^2-(b+2c-3a)(4)-2(a-b) \\ & = -6a+30b-24c \\ & \neq 0 \\ F(5) & = (b-c)(5)^3-(a+2b-3c)(5)^2-(b+2c-3a)(5)-2(a-b) \\ & = -12a+112b-60c \\ & \neq 0 \\ F(6) & = (b-c)(6)^3-(a+2b-3c)(6)^2-(b+2c-3a)(6)-2(a-b) \\ & = -20a+140b-120c \\ & \neq 0 \\ F(7) & = (b-c)(7)^3-(a+2b-3c)(7)^2-(b+2c-3a)(7)-2(a-b) \\ & = -30a+240b-210c \\ & \neq 0 \\ F(8) & = (b-c)(8)^3-(a+2b-3c)(8)^2-(b+2c-3a)(8)-2(a-b) \\ & = -42a+378b-336c \\ & \neq 0 \\ F(9) & = (b-c)(9)^3-(a+2b-3c)(9)^2-(b+2c-3a)(9)-2(a-b) \\ & = -56a + 560b - 504c \\ & \neq 0 \end{aligned}$

It may be feasible to work in this manner, however endlessly, to find a zero. But it is high time for me to work in another way round.

By long division, the remaining factor is

$(b-c)x+(b-a)$ .

One should check that

$\begin{aligned} &\quad (x-1)(x-2)\big( (b-c)x+(b-a)\big) \\ & = (x^2-3x+2)\big( (b-c)x+(b-a)\big) \\ & = (b-c)x^3 - (a+2b-3c)x^2 - (b+2c-3a)x - 2(a-b) \\ & = F(x) \\ \end{aligned}$

Remark.

This remaining zero, i.e., $\displaystyle{x=\frac{b-a}{b-c}}$ , is hard to find by trial-and-error, so long division is necessary.

Roughwork.

$\begin{aligned} & (b-c)x & +(-a+b) & & \\\cline{2-5} x^2-3x+2\quad \Big)& (b-c)x^3 & +(-a-2b+3c)x^2 & +(3a-b-2c)x & +(-2a+2b) \\ & (b-c)x^3 & +(-3b+3c)x^2 & + (2b-2c)x & \\\cline{2-5} & & (-a+b)x^2 & +(3a-3b)x & + (-2a+2b) \\ & & (-a+b)x^2 & +(3a-3b)x & + (-2a+2b) \\\cline{3-5} \end{aligned}$

October 25, 2021October 24, 2022

202110251042 Solution to 1967-CE-AMATH-1

When $14x^4-27x^3+ax^2+12x+9$ is divided by $(x-2)$ the remainder is $5$ . What is the value of $a$ ?

Solution.

$\begin{aligned} f(x) & \stackrel{\textrm{def}}{=} 14x^4-27x^3+ax^2+12x+9 \\ f(x) & = (x-2)g(x) + 5 \\ f(2) & = 5 \\ 5 & = 14(2)^4-27(2)^3+a(2)^2+12(2)+9 \\ a & = 9 \\ \end{aligned}$

$\therefore$ The value of $a$ is $9$ .

April 8, 2021October 31, 2022

202104080551 Solution to 1973-AL-AMATH-I-5

A particle of mass $m$ moving in a straight line in a medium with a velocity $v$ is subjected to a force $mk(v^3+a^2v)$ in the direction opposite to the velocity, where $k$ and $a$ are constants. Show that, at any subsequent time $t$ , the velocity $v$ is related to the initial velocity $v_0$ by

$\displaystyle{\frac{v}{(v^2+a^2)^{1/2}}=\frac{v_0}{(v_0^2+a^2)^{1/2}}e^{-a^2kt}}$ .

Show that for any value of the initial velocity, no matter how large, the particle will never travel a distance greater than $\pi /(2ka)$ .

Background.

Newton's second law:

$\begin{aligned} F = -mk(v^3+a^2v) & = ma \\ -mk(\dot{x}^3+a^2\dot{x}) & = m\ddot{x} \\ \frac{\mathrm{d}\dot{x}}{\mathrm{d}t} & = -k(\dot{x}^3+a^2\dot{x}) \\ \frac{\mathrm{d}\dot{x}}{\dot{x}^3+a^2\dot{x}} & = -k\,\mathrm{d}t \\ \dots \textrm{integrating from }& (v=v_0,\, t=0)\textrm{ to }(v=v,\, t=t) \\ \int_{v_0}^{v} \frac{1}{\dot{x}(\dot{x}^2+a^2)}\,\mathrm{d}\dot{x} & = \int_{0}^{t}-k\,\mathrm{d}t \end{aligned}$

(to be continued)

Roughwork.

Let . Then,

$\begin{aligned} 1 & \equiv A(u^2+a^2)+Bu^2+Cu \\ 1 & \equiv (A+B)u^2+Cu+Aa^2 \\ A+B & = 0 \\ C & = 0 \\ A & = \frac{1}{a^2} \\ \Rightarrow B & = -\frac{1}{a^2} \end{aligned}$
Thus,
$\displaystyle{\int \frac{1}{u(u^2+a^2)}\,\mathrm{d}u = \int \bigg(\frac{1}{a^2u} - \frac{u}{a^2(u^2+a^2)}\bigg) \,\mathrm{d}u}$

(continue)

$\begin{aligned} & \quad \int_{v_0}^{v} \frac{1}{\dot{x}(\dot{x}^2+a^2)}\,\mathrm{d}\dot{x} \\ & = \int_{v_0}^{v} \bigg(\frac{1}{a^2\dot{x}} - \frac{\dot{x}}{a^2(\dot{x}^2+a^2)}\bigg) \,\mathrm{d}\dot{x} \\ & = \int_{v_0}^v\frac{1}{a^2\dot{x}}\,\mathrm{d}\dot{x} - \int_{v_0}^v \frac{\dot{x}}{a^2(\dot{x}^2+a^2)}\,\mathrm{d}\dot{x} \\ \end{aligned}$

The first term is

$\begin{aligned} \int_{v_0}^v\frac{1}{a^2\dot{x}}\,\mathrm{d}\dot{x} & = \bigg[ \frac{\ln \dot{x}}{a^2} \bigg]\bigg|_{v_0}^{v} \\ & = \bigg( \frac{\ln v}{a^2} \bigg) - \bigg( \frac{\ln v_0}{a^2} \bigg) \\ & = \frac{1}{a^2}\ln \bigg( \frac{v}{v_0} \bigg) \end{aligned}$

and the second term is

$\begin{aligned} &\quad -\int_{v_0}^v \frac{\dot{x}}{a^2(\dot{x}^2+a^2)}\,\mathrm{d}\dot{x}\\ \dots & \textrm{ let } \dot{x}=a\tan\theta \enspace\dots \\ \dots & \textrm{ let } \theta = \tan^{-1}\bigg(\frac{v}{a}\bigg) \textrm{ and } \theta_0 = \tan^{-1}\bigg(\frac{v_0}{a}\bigg) \\ & = -\int_{\theta_0}^{\theta} \frac{a\tan\theta}{a^2(a^2\tan^2\theta +a^2)}\cdot (a\sec^2\theta)\,\mathrm{d}\theta \\ & = -\int_{\theta_0}^{\theta} \frac{\tan\theta}{a(a^2\sec^2\theta)}\cdot (a\sec^2\theta )\,\mathrm{d}\theta \\ & = -\int_{\theta_0}^{\theta} \frac{\tan\theta}{a^2}\,\mathrm{d}\theta \\ & = -\bigg[ -\frac{\ln |\cos\theta |}{a^2} \bigg]\bigg|_{\theta_0}^{\theta} \\ & = \frac{1}{a^2} \bigg[\ln \bigg| \frac{a}{\sqrt{\dot{x}^2+a^2}} \bigg|\bigg]\bigg|_{v_0}^{v} \\ & = \frac{1}{a^2} \bigg( \ln\bigg| \frac{a}{\sqrt{v^2+a^2}} \bigg| -\ln\bigg| \frac{a}{\sqrt{v_0^2+a^2}} \bigg| \bigg) \\ & = \frac{1}{a^2} \ln \bigg( \frac{\sqrt{v_0^2+a^2}}{\sqrt{v^2+a^2}} \bigg) \\ \end{aligned}$

Back to an earlier line

$\displaystyle{\int_{v_0}^{v} \frac{1}{\dot{x}(\dot{x}^2+a^2)}\,\mathrm{d}\dot{x} = \int_{0}^{t}-k\,\mathrm{d}t}$

we have

$\begin{aligned} \frac{1}{a^2}\ln\bigg( \frac{v}{v_0} \bigg) + \frac{1}{a^2}\ln \bigg( \frac{\sqrt{v_0^2+a^2}}{\sqrt{v^2+a^2}}\bigg) & = -kt \\ \ln\bigg( \frac{v}{v_0} \bigg) + \ln \bigg( \frac{\sqrt{v_0^2+a^2}}{\sqrt{v^2+a^2}}\bigg) & = -a^2kt \\ \ln\bigg( \frac{v}{v_0}\frac{\sqrt{v_0^2+a^2}}{\sqrt{v^2+a^2}} \bigg) & =-a^2kt \\ \frac{v}{v_0}\frac{\sqrt{v_0^2+a^2}}{\sqrt{v^2+a^2}} & = e^{-a^2kt} \\ \end{aligned}$

or,

$\boxed{\displaystyle{\frac{v}{\sqrt{v^2+a^2}}=\frac{v_0}{\sqrt{v_0^2+a^2}}e^{-a^2kt}}}$

as desired.

It remains to be shown that for any value of the initial velocity, no matter how large, the particle will never travel a distance greater than $\pi /(2ka)$ .

(to be continued)