Edexcel A Level Pure Math 1: 2020/OCT/9MA0/01 - Solution

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Question 1

(a) Find the first four terms, in ascending powers of $x$, of the binomial expansion of
$\begin{aligned} &\\ (1+8 x)^{\frac{1}{2}}\\\\ \end{aligned}$
giving each term in simplest form. (3)
(b) Explain how you could use $x=\dfrac{1}{32}$ in the expansion to find an approximation for $\sqrt{5}$ There is no need to carry out the calculation. (2)

$\begin{aligned} \text{(a) }\quad & \quad (1+8 x)^{1 / 2}\\\\ & \approx 1+\frac{\left(\frac{1}{2}\right)(8 x)}{1 !}+\frac{\left(\frac{1}{2}\right)\left(\frac{1}{2}-1\right)(8 x)^{2}}{2 !}+\frac{\left(\frac{1}{2}\right)\left(\frac{1}{2}-1\right)\left(\frac{1}{2}-2\right)(8 x)^{3}}{3 !}+\cdots \\\\ &=1+4 x-8 x^{2}+32 x^{3}+\cdots\\\\ \end{aligned}$ $\begin{aligned} \text{(b) }\qquad\text { When } x&=\frac{1}{32}\\\\ \text { LHS } &=\left(1+8 \times \frac{1}{32}\right)^{1 / 2}\\\\ &=\frac{\sqrt{5}}{2}\\\\ \text { RHS }&=1+4\left(\frac{1}{32}\right)-8\left(\frac{1}{32}\right)^{2}+32\left(\frac{1}{32}\right)^{3}\\\\\ &=\frac{1145}{1024}\\\\ \therefore\ \frac{\sqrt{5}}{2}&=\frac{1145}{1024}\\\\ \sqrt{5}&=\frac{1145}{512}\\\\ \end{aligned}$

Question 2

By taking logarithms of both sides, solve the equation

$\begin{aligned} &\\ 4^{3 p-1}=5^{210}\\\\ \end{aligned}$
giving the value of $p$ to one decimal place. (3)

$\begin{aligned} 4^{3 p-1}&=5^{210} \\\\ \ln 4^{3 p-1}&=\ln 5^{210} \\\\ (3 p-1) \ln 4&=210 \ln 5 \\\\ 3 p-1&=\frac{210 \ln 5}{\ln 4} \\\\ 3 p&=244.80 \ldots \\\\ p&=81.6\quad(1 \text{decimal place}) \end{aligned}$

Question 3

Relative to a fixed origin $O$

• point $A$ has position vector $2 \mathbf{i}+5 \mathbf{j}-6 \mathbf{k}$
• point $B$ has position vector $3 \mathbf{i}-3 \mathbf{j}-4 \mathbf{k}$
• point $C$ has position vector $2 \mathbf{i}-16 \mathbf{j}+4 \mathbf{k}$

(a) Find $\overrightarrow{A B}$ (2)
(b) Show that quadrilateral $O A B C$ is a trapezium, giving reasonsfor your answer. (2)

$\begin{aligned} \text{(a) }\quad \overrightarrow{A B} &=\overrightarrow{O B}-\overrightarrow{O A} \\\\ &=(3 \hat{i}-3 \hat{j}-4 \hat{k})-(2 \hat{i}+5 \hat{j}-6 \hat{k}) \\\\ &=\hat{i}-8 \hat{j}+2 \hat{k} \\\\ \text{(b) }\quad \overrightarrow{O C} &=2 \hat{i}-16 \hat{j}+4 \hat{k} \\\\ &=2(\hat{i}-8 \hat{j}+2 \hat{k}) \\\\ &=2 \overrightarrow{A B} \\\\ \overrightarrow{A B} &\parallel \overrightarrow{O C}\quad \text { and } \\\\ O C&=2 A B.\\\\ \therefore \quad & O A B C \text { is a trapezium.} \end{aligned}$

Question 4

The function $\mathrm{f}$ is defined by
$\begin{aligned} &\\ f(x)=\frac{3 x-7}{x-2} \quad x \in \mathbb{R}, x \neq 2 \end{aligned}$
(a) Find $f^{-1}(7)$ (2)
(b) Show that $ff(x)=\dfrac{a x+b}{x-3}$ where $a$ and $b$ are integers to be found. (3)

$\begin{aligned} \text{(a) }\quad f(x)&=\frac{3 x-7}{x-2}\\\\ f(x) &=\frac{3 x-7}{x-2} \\\\ \text { Let } f^{-1}(7) &=k, \text { then }\\\\ f(k) &=7 \\\\ \frac{3 k-7}{k-2} &=7 \\\\ 3 k-7 &=7 k-14 \\\\ 4 k &=7 \\\\ k &=\frac{7}{4} \\\\ \therefore f^{-1}(7) &=\frac{7}{4}\\\\ \end{aligned}$ $\begin{aligned} \text{(b) }\quad f f(x) &=f\left(f(x)\right)\\\\ f f(x) &=\frac{3\left(\frac{3 x-7}{x-2}\right)-7}{\left(\frac{3 x-7}{x-2}\right)-2} \\\\ &=\frac{3(3 x-7)-7(x-2)}{3 x-7-2(x-2)} \\\\ &=\frac{9 x-21-7 x+4}{3 x-7-2 x+4} \\\\ \therefore f(x) &=\frac{2 x-7}{x-3} \end{aligned}$

Question 5

A car has six forward gears. The fastest speed of the car

• in $1^{\text {st }}$ gear is $28 \mathrm{~km} \mathrm{~h}^{-1}$
• in $6^{\text {th }}$ gear is $115 \mathrm{~km} \mathrm{~h}^{-1}$

Given that the fastest speed of the car in successive gears is modelled by an arithmetic sequence,
(a) find the fastest speed of the car in $3^{\mathrm{rd}}$ gear. (3)
Given that the fastest speed of the car in successive gears is modelled by a geometric sequence,
(b) find the fastest speed of the car in $5^{\text {th }}$ gear. (3)

$\begin{aligned} u_{n}&=a+(n-1) d \\\\ 115&=28+(6-1) d \\\\ \therefore\ d&=17.4 \\\\ u_{3}&=28+(3-1) \times 17.4\\\\ &=62.8 \mathrm{knh}^{-1} \\\\ u_{n}&=a r^{n-1} \\\\ 115&=28 r^{6-1} \\\\ r^{5}&=\frac{115}{28}\\\\ \therefore\ r&=1.326 \ldots \\\\ u_{5}&=28 \times \left(1.326 \ldots\right)^{5-1}\\\\ &=86.7 \mathrm{kmh}^{-1} \end{aligned}$

Question 6

(a) Express $\sin x+2 \cos x$ in the form $R \sin (x+\alpha)$ where $R$ and $\alpha$ are constants, $R>0$ and $0< \alpha<\dfrac{\pi}{2}.$
Give the exact value of $R$ and give the value of $\alpha$ in radians to 3 decimal places. (3)
The temperature, $\theta^{\circ} \mathrm{C}$, inside a room on a given day is modelled by the equation

$\theta=5+\sin \left(\dfrac{\pi t}{12}-3\right)+2 \cos \left(\dfrac{\pi t}{12}-3\right) \quad 0 \leq t<24 $

where $t$ is the number of hours after midnight.
Using the equation of the model and your answer to part (a),
(b) deduce the maximum temperature of the room during this day, (1)
(c) find the time of day when the maximum temperature occurs, giving your answer to the nearest minute. (3)

$\begin{aligned} \text{(a) } \quad\text{Let } \sin x+2 \cos x&=R \sin x \cos \alpha+R \cos x \sin \alpha\\\\ \therefore \quad R \cos \alpha &=1 \\\\ R \sin \alpha &=2 \\\\ \tan \alpha &=2 \\\\ \alpha &=1.107 \\\\ R^{2}\left(\sin ^{2} \alpha+\cos ^{2} \alpha\right) &=5 \\\\ R^{2} &=5 \\\\ R &=\sqrt{5}\\\\ \therefore \quad R \cos \alpha &=1 \\\\ R \sin \alpha &=2 \\\\ \tan \alpha &=2 \\\\ \alpha &=1.107 \\\\ R^{2}\left(\sin ^{2} \alpha+\cos ^{2} \alpha\right) &=5 \\\\ R^{2} &=5 \\\\ R &=\sqrt{5} \\\\ \end{aligned}$ $\therefore \quad \sin x+2 \cos x =\sqrt{5} \sin (x+1.107) \\\\ $ $\begin{aligned} \text{(b) } \quad \theta &=5+\sin \left(\frac{\pi t}{12}-3\right)+2 \cos \left(\frac{\pi t}{12}-3\right) \\\\ &=5+\sqrt{5} \sin \left(\frac{\pi t}{12}-3+1.107\right) \\\\ &=5+\sqrt{5} \sin \left(\frac{\pi t}{12}-1.893\right) \\\\ \text{Since} & -1 \leq \sin \left(\frac{\pi t}{12}-1.893\right) \leq 1, \\\\ \theta_{\max } &=5+\sqrt{5}^{\circ} C \quad\left(\text{when} \sin \left(\frac{\pi t}{12}-1.893\right)=1\right)\\\\ \end{aligned}$ $\begin{aligned} \text{(c) } \quad \theta_{\text{max}} \text{ occurs when }\\\\ \sin \left(\frac{\pi t}{12}-1.893\right) &=1 \\\\ \therefore\ \frac{\pi t}{12}-1.893 &=\frac{\pi}{2} \\\\ \therefore\ t &=13.29 \\\\ &=13+(0.23 \times 60)-12\ \mathrm{pm} \\\\ &=1: 14\ \mathrm{pm} \end{aligned}$

Question 7

Figure 1 shows a sketch of a curve $C$ with equation $y=f(x)$ and a straight line $l$. The curve $C$ meets $l$ at the points $(-2,13)$ and $(0,25)$ as shown. The shaded region $R$ is bounded by $C$ and $l$ as shown in Figure 1 . Given that

• $f(x)$ is a quadratic function in $x$
• $(-2,13)$ is the minimum turning point of $y=f(x)$

use inequalities to define $R$. (3)

$\begin{aligned} C: y&=a(x+2)^{2}+13\\\\ \text{When } x&=0, y=25.\\\\ \therefore\ 25 &=4 a+13 \\\\ a &=3\\\\ \text{The } & \text{ equation of the curve } C \text{ is}\\\\ &y=3(x+2)^{2}+13 . \\\\ &\text { Gradient of } l: m=\frac{25-13}{0+2}=6\\\\ \therefore \ l: y&=6 x+25\\\\ \end{aligned}$ Since $R$ is below the line $l$ and above the curve $C$, $R$ is defined by $3(x+2)^{2}+13<y<6 x+25$.

Question 8

A new smartphone was released by a company.

The company monitored the total number of phones sold, $n$, at time $t$ days after the phone was released.

The company observed that, during this time, the rate of increase of $n$ was proportional to $n$.

Use this information to write down a suitable equation for $n$ in terms of $t$.
(You do not need to evaluate any unknown constants in your equation.) (2)

$ n=f(t)\\\\ $ Since the rate of increase of $n$ was proportional to $n$, $\begin{aligned} &\\ \dfrac{dn}{dt} &\propto n \\\\ \dfrac{dn}{dt} &= \text{constant} \times n\\\\ \end{aligned}$ $\therefore\quad n$ must be an exponetial function. Hence $n$ can be described as $$ n=A e^{k t} $$ where $A$ and $k$ are positive constants.

Question 9

Figure 2 shows a sketch of the curve $C$ with equation $y=\mathrm{f}(x)$ where
$\begin{aligned} &\\ f(x)=4\left(x^{2}-2\right) e^{-2 x} \quad x \in \mathbb{R}\\\\ \end{aligned}$
(a) Show that $f^{\prime}(x)=8\left(2+x-x^{2}\right)e^{-2 x}$. (3)
(b) Hence find, in simplest form, the exact coordinates of the stationary points of $C$. (3)
The function $g$ and the function $h$ are defined by
$\begin{aligned} &\\ g(x)&=2 f(x) \quad x \in \mathbb{R} \\\\ h(x)&=2 f(x)-3 \quad x \geq 0\\\\ \end{aligned}$
(c) Find
(i) the range of $g$.
(ii) the range of $h$. (2)

$\begin{aligned} \text{(a) }\quad f(x)=& 4\left(x^{2}-2\right) e^{-2 x} \quad x \in \mathbb{R} \\\\ f^{\prime}(x) &=4\left[\left(x^{2}-2\right) \frac{d}{d x} e^{-2 x}+e^{-2 x} \frac{d}{d x}\left(x^{2}-2\right)\right] \\\\ &=4\left[-2\left(x^{2}-2\right) e^{-2 x}+2 x e^{-2 x}\right] \\\\ &=4 \times 2 e^{-2 x}\left[-\left(x^{2}-2\right)+x\right] \\\\ &=8\left(2+x-x^{2}\right) e^{-2 x}\\\\ \end{aligned}$ $\begin{aligned} \text{(b) }\quad \text{When } \qquad f^{\prime}(x)&=0,\\\\ 8\left(2+x-x^{2}\right) e^{-2 x}&=0\\\\ \therefore\ 2+x-x^{2}&=0 \quad\left(\because e^{-2 x} \neq 0\right)\\\\ \end{aligned}$ $\begin{aligned} \therefore \quad &(1+x) (2-x)=0 \\\\ x&=-1 \text { or } x=2 \\\\ f(-1) &=4(1-2) e^{2} \\\\ &=-4 e^{2}\quad (\text{minimum}) \\\\ f(2) &=4(4-2) e^{-2} \\\\ &=8 e^{-2}\quad (\text{maximum} )\\\\ \end{aligned}$ $\therefore$ The stationary points are $\left(-1,-4 e^{2}\right)$ and $\left(2,8 e^{-2}\right)$. $\begin{aligned} &\\ \text{(c) (i) }\quad g(x)& =2 f(x), x \in \mathbb{R}\\\\ \therefore\ &\text{ The minimum point of } g \text{ is } \left(-1,-8 e^{2}\right).\\\\ \therefore\ &\text{ The range of } g=\left[-8 e^{2}, \infty\right)\\\\ h(x)&=2 f(x)-3, x \geq 0 \\\\ f(0)&=-8 \\\\ h(0)&=2(-8)-3=-19\\\\ \text{For } & x \geq 0, \text{ the maximum value of } h \text{ is } 16 e^{-2}-3.\\\\ \therefore\ & \text{ The range of } h=\left[-19,16 e^{-2}-3\right]. \end{aligned}$

Question 10

(a) (a) Use the substitution $x=u^{2}+1$ to show that

$\displaystyle\int_{5}^{10} \frac{3 ~dx}{(x-1)(3+2 \sqrt{x-1})}=\int_{p}^{q} \frac{6 ~du}{u(3+2 u)}$

where $p$ and $q$ are positive constants to be found. (4)
(b) Hence, using algebraic integration, show that

$\displaystyle\int_{5}^{10} \frac{3 ~dx}{(x-1)(3+2 \sqrt{x-1})}=\ln a$

where $a$ is a rational constant to be found. (6)

$\displaystyle\int_{5}^{10} \dfrac{3 ~d x}{(x-1)(3+2 \sqrt{x-1})}$ $\begin{aligned} &\\ \text{ Let } x&=u^{2}+1, \text{ then }\\\\ x &=5, u=2 \\\\ x &=10, u=3 \\\\ \therefore \quad p &=2 \text { and } q=3 . \\\\ x &=u^{2}+1 \\\\ \therefore \quad \dfrac{d x}{d x} &=2 u \\\\ \therefore \quad d x &=2u ~d u\\\\ \end{aligned}$ Substituting $x=u^{2}+1,$ $\begin{aligned} &\\ &(x-1)(3+2 \sqrt{x-1}) \\\\ =&\ \left(u^{2}+1-1\right)\left(3+2 \sqrt{u^{2}+1-1}\right) \\\\ =&\ u^{2}\left(3+2 \sqrt{u^{2}}\right) \\\\ =&\ u^{2}(3+2 u) \\\\ \therefore\quad & \int_{5}^{10} \dfrac{3 ~d x}{(x-1)(3+2 \sqrt{x-1})} \\\\ =& \ \int_{2}^{3} \dfrac{3(2 u ~d u)}{u^{2}(3+2 u)} \\\\ =&\ \int_{2}^{3} \dfrac{6 ~d u}{u(3+2 u)}\\\\ \end{aligned}$ Let $\dfrac{6}{u(3+2 u)}=\dfrac{A}{u}+\dfrac{B}{(3+2 u)}\\\\ $ When $\quad u=1, \dfrac{6}{5}=A+\dfrac{B}{5}\\\\ $ $\therefore\quad 5 A+B=6 \ldots(1)\\\\ $ When $\quad u=-1,-6=-A+B\\\\ $ $\therefore\quad A-B=6 \ldots(2)\\\\ $ Solving equations (1) and (2), $A=2 \text { and } B=-4\\\\ $ $\begin{aligned} &\therefore \int_{5}^{10} \frac{3 d x}{(x-1)(3+2 \sqrt{x-1})} \\\\ &=\int_{2}^{3} \frac{6 d u}{u(3+2 u)} \frac{3}{\left(x=u^{2}+1\right)} \\\\ &=\int_{2}^{3} \frac{2}{u} d u \frac{4}{3+2 u} d u \\\\ &=2 \int_{2}^{3} \frac{1}{u} d u-\frac{4}{2} \int_{3}^{3} \frac{1}{3+2 u} d(3+2 u) \\\\ &=2\left[\ln u-\ln (3+2 u)\right]_{2}^{3}\\\\ &=2[(\ln 3-\ln 9)-(\ln 2-\ln 7)] \\\\ &=2\left[\ln \dfrac{1}{3}-\ln \dfrac{2}{7}\right] \\\\ &=2\left[\ln \dfrac{7}{6}\right] \\\\ &=\ln \dfrac{49}{36} \end{aligned}$

Question 11

Circle $C_{1}$ has equation $x^{2}+y^{2}=100$.
Circle $C_{2}$ has equation $(x-15)^{2}+y^{2}=40$.
The circles meet at points $A$ and $B$ as shown in Figure 3 .
(a) Show that angle $A O B=0.635$ radians to 3 significant figures, where $O$ is the origin. (4)
The region shown shaded in Figure 3 is bounded by $C_{1}$ and $C_{2}$
(b) Find the perimeter of the shaded region, giving your answer to one decimal place. (4)

$\begin{aligned} \text{(a) }\quad C_{1}: & x^{2}+y^{2}=100 \\\\ \therefore\ & \text { center: }(0,0) \\\\ & \text { radius }=10 \\\\ &\text { Rearranging, we have } \\\\ &y^{2}=100-x^{2} \\\\ C_{2}:& (x-15)^{2}+y^{2}=40 \\\\ \therefore\ & \text { center: }(15,0) \\\\ & \text { radius }=2 \sqrt{10}\\\\ & \text { Rearranging, we have } \\\\ & y^{2}=40-(x-15)^{2}\\\\ \end{aligned}$ At the points of intersection of $C_{1}$ and $C_{2}$, $\begin{aligned} &\\ 100-x^{2} &=40-(x-15)^{2} \\\\ 100-x^{2} &=40-x^{2}+30 x+225 \\\\ 30 x &=285 \\\\ x &=\dfrac{19}{2}\\\\ \therefore \quad y^{2} &=100-\left(\dfrac{19}{2}\right)^{2} \\\\ &=\dfrac{39}{4} \\\\ \therefore y &=\pm \dfrac{\sqrt{39}}{2}\\\\ \end{aligned}$ $\therefore$ The coordinates of the points $A$ and $B$ are $\left(\dfrac{19}{2}, \dfrac{\sqrt{39}}{2}\right)$ and $\left(\dfrac{19}{2},-\dfrac{\sqrt{39}}{2}\right)$ respectively. $\begin{aligned} &\\ &\therefore A B=\sqrt{39}\\\\ \end{aligned}$ $\begin{aligned} \therefore \angle A O B &=\cos ^{-1}\left[\dfrac{10^{2}+10^{2}-39}{2(10)(10)}\right] \\\\ &=0.635 \text { radians }\\\\ \end{aligned}$ $\text{(b) }\quad$ Let the centre of $C_{2}$ be $P$. $\begin{aligned} &\\ \quad \therefore \angle A P B &=\cos ^{-1}\left[\dfrac{40+40-39}{2(40)}\right] \\\\ &=1.033 \text { radians }\\\\ \end{aligned}$ $\begin{aligned} \quad &\quad \text{perimeter of shaded region}\\\\ \quad &=10(2 \pi-0.635)+2 \sqrt{10}(2 \pi-1.03)\\\\ &=89.7 \quad \text{ (one decimal place)} \end{aligned}$

Question 12

In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable.
(a) Show that

$\operatorname{cosec} \theta-\sin \theta \equiv \cos \theta \cot \theta \quad \theta \neq(180 n)^{\circ} \quad n \in \mathbb{Z}$

(3) $\quad$ (b) Hence, or otherwise, solve for $0<x<180^{\circ}$

$\operatorname{cosec} x-\sin x=\cos x \cot \left(3 x-50^{\circ}\right)$

(5)

$\begin{aligned} \text{(a) }\quad &\quad\operatorname{cosec} \theta-\sin \theta\\\\ \quad &=\frac{1}{\sin \theta}-\sin \theta \\\\ &=\frac{1-\sin ^{2} \theta}{\sin \theta} \\\\ &=\frac{\cos ^{2} \theta}{\sin \theta} \\\\ &=\cos \theta \times \frac{\cos \theta}{\sin \theta} \\\\ &=\cos \theta \cdot \cot \theta\\\\ \end{aligned}$ $\begin{aligned} \text{(b) }\qquad &\operatorname{cosec} x-\sin x=\cos x \cot(3 x-50) \\\\ &\cos x \cot x=\cos x \cot \left(3 x-50^{\circ}\right) \\\\ \therefore\quad &\cos x\left[\cot x-\cot \left(3 x-50^{\circ}\right)\right] =0 \\\\ \therefore\quad &\cos x=0^{\circ} \text { or } \cot x =\cot \left(3 x-50^{\circ}\right) \\\\ \therefore \quad & x=0^{\circ}\text { or } x=3 x-50^{\circ} \text { or } 180^{\circ}+x =3 x-50^{\circ} \\\\ &x=0^{\circ} \text { or } x =25^{\circ} \text { or } x=115^{\circ} \end{aligned}$

Question 13

A sequence of numbers $a_{1}, a_{2}, a_{3}, \ldots$ is defined by
$\begin{aligned} &\\ a_{n+1}=\frac{k\left(a_{n}+2\right)}{a_{n}} \quad n \in \mathbb{N} \end{aligned}$
where $k$ is a constant.
Given that

• the sequence is a periodic sequence of order $3$.
• $a_{1}=2$

(a) show that
$\begin{aligned} &\\ k^{2}+k-2=0\\\\ \end{aligned}$ (3)
(b) For this sequence explain why $k \neq 1$. (2)
(c) Find the value of
$\begin{aligned} &\\ \displaystyle\sum_{r=1}^{80} a_{r} \end{aligned}$ (3)

$\begin{aligned} \text{(a) }\quad a_{n+1} &=\frac{k\left(a_{n}+2\right)}{a_{n}}, n \in N \\\\ a_{1} &=2 \\\\ a_{2} &=\frac{k\left(a_{1}+2\right)}{a_{1}} \\\\ &=\frac{k(2+2)}{2} \\\\ &=2 k \\\\ a_{3} &=\frac{k\left(a_{2}+2\right)}{a_{2}} \\\\ &=\frac{k(2 k+2)}{2 k} \\\\ &=k+1\\\\ a_{4} &=\frac{k\left(a_{3}+2\right)}{a_{3}} \\\\ &=\frac{k(k+1+2)}{k+1} \\\\ &=\frac{k^{2}+3 k}{k+1}\\\\ \end{aligned}$ Since the sequence is a periodic sequence of order $3$, $\begin{aligned} &\\ a_1 &=a_4 \\\\ \frac{k^{2}+3 k}{k+1} &=2\\\\ \therefore\ & k^{2}+3 k=2 k+2 \\\\ &k^{2}+k-2=0 \\\\ \text{(b) }\quad &(k+2)(k-1)=0 \\\\ &k=-2 \text { or } k=1 \\\\ &\text{When } k=1 \\\\ &a_{1}=2 \\\\ &a_{2}=2(1)=2 \\\\ &a_{3}=1+1=2 \\\\ &a_{4}=\frac{1^{2}+3(1)}{1+1}=2\\\\ \end{aligned}$ $\therefore$ The sequence wouldn't have a period or order $3$ when $k=1\\\\ $. $\therefore k=1$ is not a solution. $\begin{aligned} &\\ \text{When } k &=-2,\\\\ &a_{1}=2 \\\\ &a_{2}=2(-2)=-4 \\\\ &a_{3}=-2+1=-1 \\\\ &a_{4}=a_{1}=2\\\\ \text{(c) }\quad \sum_{r=1}^{80} a_{r} &=a_{1}+a_{2}+a_{3}+a_{4}+\cdots+a_{80} \\\\ &=(2-4-1)+(2-4-1)+(2-4-1)+\cdots\text { to } 80^{\text {th }} \text { term } \\\\ &=26(2-4-1)+2-4 \\\\ &=-80 \end{aligned}$

Question 14

A large spherical balloon is deflating.
At time $t$ seconds the balloon has radius $r \mathrm{~cm}$ and volume $V \mathrm{~cm}^{3}$
The volume of the balloon is modelled as decreasing at a constant rate. (a) Using this model, show that
$\begin{aligned} &\\ \frac{dr}{dt}=-\frac{k}{r^{2}}\\\\ \end{aligned}$
where $k$ is a positive constant. (3)
Given that

• the initial radius of the balloon is $40 \mathrm{~cm}$.
• after 5 seconds the radius of the balloon is $20 \mathrm{~cm}$.
• the volume of the balloon continues to decrease at a constant rate until the balloon is empty.

(b) solve the differential equation to find a complete equation linking $r$ and $t$. (2)
(c) Find the limitation on the values of $t$ for which the equation in part (b) is valid. (2)

$\text{(a) }\quad $ A large spherical balloon is deflating. Volume of balloon: $V=f(r)=\dfrac{4}{3} \pi r^{3}\\\\ $ radius of balloon: $r=g(t)\\\\ $ The volume of the balloon is modelled as decreasing at a constant rate. $\begin{aligned} &\\ \text{i.e, }\quad &\frac{d V}{d t}=-c \quad (c= \text{ constant} )\\\\ &\frac{d V}{d r} \cdot \frac{d r}{d t}=-c \\\\ &4 \pi r^{2} \cdot \frac{d r}{d t}=-c \\\\ \therefore\quad & \frac{d r}{d t}=-\frac{c}{4 \pi r^{2}}\\\\ \end{aligned}$ Taking the constant $k=\dfrac{c}{4 \pi}$, $\begin{aligned} &\\ \text{(b) }\quad\frac{d r}{d t} &=-\frac{k}{r^{2}} \\\\ \therefore r^{2} d r &=-k d t\\\\ \end{aligned}$ Integrating both sides, $\begin{aligned} &\\ \int r^{2} d r &=-k \int d t \\\\ \frac{r^{3}}{3} &=-k t+c_{1}\\\\ \end{aligned}$ $\begin{aligned} \text{When } t&=0, r=40 \mathrm{~cm}.\\\\ \therefore\ \frac{(40)^{3}}{3}&=c_{1}\\\\ c_{1}&=\frac{64000}{3}\\\\ \therefore\ \frac{r^{3}}{3}&=-k t+\frac{64000}{3}\\\\ \text{When } t &=58, r=20 \mathrm{~cm}\\\\ \therefore\ \frac{(20)^{3}}{3}&=-5 k+\frac{64000}{3}\\\\ \therefore\ 5 k &=\frac{64000-8000}{3} \\\\ k &=\frac{11200}{3} \\\\ \therefore \ r^{3} &=-11200 t+64000\\\\ \end{aligned}$ $\begin{aligned} \text{(c) }\quad &\text{Since } r^{3} \geq 0\\\\ &-11200 t+64000 \geq 0 \\\\ &t \leq \frac{64000}{11200} \\\\ &t \leq \frac{40}{7} \mathrm{~s} \\\\ \therefore\ & 0 \leq t \leq\frac{40}{7} \mathrm{~s} . \end{aligned}$

Question 15

The curve $C$ has equation
$\begin{aligned} &\\ x^{2} \tan y=9 \quad 0<y<\frac{\pi}{2}\\\\ \end{aligned}$
(a) Show that
$\begin{aligned} &\\ \frac{dy}{dx}=\frac{-18 x}{x^{4}+81}\\\\ \end{aligned}$ (4)
(b) Prove that $C$ has a point of inflection at $x=\sqrt[4]{27}$ (3)

$\begin{aligned} \text{(a) }\quad C: \quad x^{2} \tan y&=9 \quad, 0<y<\frac{\pi}{2}\\\\ \therefore\ \tan y&=\frac{9}{x^{2}}\\\\ \tan ^{2} y&=\frac{81}{x^{4}}\\\\ \sec ^{2} y-1&=\frac{81}{x^{4}}\\\\ \therefore\ \sec ^{2} y&=\frac{x^{4}+81}{x^{4}}\\\\ \end{aligned}$ $x^{2} \tan y=9\\\\ $ Differentiale with respect to $x$ $\begin{aligned} &\\ &x^{2} \sec ^{2} y \frac{d y}{d x}+2 x \text { tany }=0 \\\\ &x^{2} \sec ^{2} y \frac{d y}{d x}=-2 x \tan y \\\\ \therefore\ &\frac{d y}{d x}=-\frac{2 x \tan y}{x^{2} \sec ^{2} y}\\\\ \end{aligned}$ $\begin{aligned} \text{(b) }\quad \therefore\ \frac{d y}{d x} &=-\frac{2 x\left(\frac{9}{x^{2}}\right)}{x^{2}\left(\frac{x^{4}+81}{x^{4}}\right)} \\\\ &=-\frac{18}{x} \times \frac{x^{2}}{x^{4}+81} \\\\ &=-\frac{18 x}{x^{4}+81}\\\\ \end{aligned}$ $\begin{aligned} \therefore\ \frac{d^{2} y}{d x^{2}} &=-18\left[\frac{x^{4}+81-x\left(4 x^{3}\right)}{\left(x^{4}+81\right)^{2}}\right] \\\\ &=-\frac{18\left(81-3 x^{4}\right)}{\left(x^{4}+81\right)^{2}} \\\\ \left.\frac{d^{2} y}{d x^{2}}\right|_{x=\sqrt[4]{27}} &=-\frac{18(81-3(27))}{(27+81)^{2}} \\\\ &=0\\\\ \end{aligned}$ $\therefore C$ has a stationary point at $x=\sqrt[4]{27}$ $\begin{aligned} &\\ &x=\sqrt[4]{27} \approx 2.3 \\\\ &\left.\frac{d^{2} y}{d x^{2}}\right|_{x=2}=-\frac{18(81-48)}{(1.6+81)^{2}}>0(\text { concave up }) \\\\ &\left.\frac{d^{2} y}{d x^{2}}\right|_{x=3}=-\frac{18(81-3(81))}{(81+81)}>0 \text { (concave down) }\\\\ \end{aligned}$ $\therefore$ The stationary point of $x=\sqrt[4]{27}$ is a point of inflection.

Question 16

Prove by contradiction that there are no positive integers $p$ and $q$ such that
$\begin{aligned} &\\ 4 p^{2}-q^{2}=25 \end{aligned}$ (4)

Assume that $p$ and $q$ are positive integers such that $\begin{aligned} &\\ & 4 p^{2}-q^{2}=25 \\\\ \therefore\ &(2 p+q)(2 p-q)=25\\\\ \therefore\ & \text {(i) } 2 p+q=25 \text { and } 2 p-q=1 \text { or }\\\\ &\text {(ii) } 2 p+q=1 \text { and } 2 p-q=25 \text { or }\\\\ & \text {(iii) } 2 p+q=5 \text { and } 2 p-q=5\\\\ &\text {(i) } p=\frac{13}{2}, q=12 \quad \text { (false)}\\\\ &\text {(ii) } p=\frac{13}{2}, q=-12 \quad \text { (false)}\\\\ &\text {(iii) } p=\frac{5}{2}, g=0 \quad \text { (false)}\\\\ \end{aligned}$ Hence the assumption is false and the original statement is true.

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