Oct-Nov-21-p1-CIE-0606/11 : Solution

2021 CIE (Additional Mathematics), Paper 1 ၏ Question နှင့် Solution များ ဖြစ်ပါသည်။ Question Paper ကို ဒီနေရာမှာ Download ယူနိုင်ပါသည်။


  1. The diagram shows the graph of $y=a \sin \dfrac{x}{b}+c$ for $-360^{\circ} \le x \le 360^{\circ}$,
    (a) Write down the period of $a \sin \dfrac{x}{b}+c$. [1]
    (b) Find the value of $a$, of $b$ and of $c$. [3]



  2. $\begin{aligned} \text{(a)}\quad \frac{1}{4} \text{ period }&=270^{\circ}\\\\ 1 \text{ period } &=270^{\circ} \times 4\\\\ &=1080^{\circ}\\\\ \end{aligned}$
    $\begin{aligned} \text{(b)}\quad &-1 \le \sin \frac{x}{b} \le 1 \\\\ &-a \le a \sin \frac{x}{b} \le a \\\\ &-a+c \le a \sin \frac{x}{b}+c \le a+c\\\\ \therefore\ &-a+c=-6 \quad \ldots(1)\\\\ \end{aligned}$
    $\begin{aligned} a+c &=2 \ldots(2) \\\\ \text{By } (1) &+(2), \\\\ 2 c &=-4 \\\\ c &=-2 \\\\ \text{By } (2) &-(1), \\\\ 2 a &=8 \\\\ a &=4\\\\ \text{When } x&=90^{\circ},\\\\ a \sin \frac{x}{b}+c&=0\\\\ 4 \sin \frac{90^{\circ}}{b}-2&=0\\\\ \sin \frac{30^{\circ}}{b}&=\frac{1}{2}\\\\ \frac{90^{\circ}}{b}&=30^{\circ}\\\\ \therefore\ b&=3 \end{aligned}$

  3. Points $A$ and $C$ have coordinates $(-4, 6)$ and $( 2, 18)$ respectively. The point $B$ lies on the line $A C$ such that $\overrightarrow{A B}=\dfrac{2}{3} \overrightarrow{A C}$.
    (a) Find the coordinates of $B$. [2]
    (a) Find the equation of the line $l$, which is perpendicular to $AC$ and passes through B. [2]
    (c) Find the area enclosed by the line $l$ and the coordinate axes. [2]



  4. $\begin{aligned} \text{ (a) } \quad \overrightarrow{O A} &=\left(\begin{array}{c} -4 \\\\ 6 \end{array}\right) \\\\ \overrightarrow{O C} &=\left(\begin{array}{c} 2 \\\\ 18 \end{array}\right) \\\\ \text { Let } \overrightarrow{O B} &=\left(\begin{array}{l} x \\\\ y \end{array}\right) \\\\ \overrightarrow{A B} &=\overrightarrow{O B}-\overrightarrow{O A} \\\\ &=\left(\begin{array}{l} x \\\\ y \end{array}\right)-\left(\begin{array}{c} -4 \\\\ 6 \end{array}\right) \\\\ &=\left(\begin{array}{l} x+4 \\\\ y-6 \end{array}\right)\\\\ \overrightarrow{A C} &=\overrightarrow{O C}-\overrightarrow{O A} \\\\ &=\left(\begin{array}{c} 2 \\\\ 18 \end{array}\right)-\left(\begin{array}{r} -4 \\\\ 6 \end{array}\right) \\\\ &=\left(\begin{array}{c} 6 \\\\ 12 \end{array}\right) \\\\ \overrightarrow{A B} &=\dfrac{2}{3} \overrightarrow{A C} \\\\ \left(\begin{array}{c} x+4 \\\\ y-6 \end{array}\right) &=\dfrac{2}{3}\left(\begin{array}{c} 6 \\\\ 12 \end{array}\right) \\\\ \left(\begin{array}{c} x+4 \\\\ y-6 \end{array}\right) &=\left(\begin{array}{c} 4 \\\\ 8 \end{array}\right)\\\\ \therefore x+4 &=4 \\\\ x &=0 \\\\ y-6 &=8 \\\\ y &=14 \\\\ \end{aligned}$
    $\therefore$ The point $B$ is $(8,14)$.
    $\text{ (a) }$ Let the gradient (slope) of $A C$ be $m_{A C}.$
    $\begin{aligned} &\\ &\therefore m_{A C}=\dfrac{18-6}{2-(-4)}=\dfrac{12}{6}=2 \\\\ &l \perp A C \\\\ &\therefore m_{l}=-\dfrac{1}{2}\\\\ \end{aligned}$
    $\therefore$ Equation of the line l is
    $\begin{aligned} &\\ y-14&=-\dfrac{1}{2}(x-0)\\\\ y&=-\dfrac{1}{2} x+14\\\\ \end{aligned}$

    $\text{ (c) }$ When $x=0, y=14\\\\ $.
    $\therefore\ l$ cuts $y$-axis at $(0,14).\\\\ $
    When $y=0, x=28\\\\ $
    $\therefore\ l$ cuts $x$-axis at $(28,0).\\\\ $
    The area enclosed by $l$ and coordinate axes is
    $\begin{aligned} &\\ A &=\dfrac{1}{2} \times 28 \times 14 \\\\ &=196 \text { sq-units. } \end{aligned}$

  5. (a) Find the vector which has magnitude $39$ and is in the same direction as $\left(\begin{array}{r}12 \\ -5\end{array}\right)$. [2]
    (b) Given that $\mathbf{a}=\left(\begin{array}{r} 2 \\ -1 \end{array}\right)$ and $\mathbf{b}=\left(\begin{array}{r} -4 \\ 5 \end{array}\right)$, find the constants $\lambda$ and $\mu$ such that $5 \mathbf{a}+\lambda\left(\begin{array}{l} 4 \\ 6 \end{array}\right)=\mu \mathbf{b}$ [4]



  6. $\text {(a) }\quad$ Let $\vec{a}=\left(\begin{array}{c}12 \\\\ -5\end{array}\right).$
    $\begin{aligned} &\\ \therefore\ |\vec{a}| &=\sqrt{12^{2}+(-5)^{2}} \\\\ &=\sqrt{169} \\\\ &=13 \\\\ \therefore\ \quad \hat{a} &=\frac{\vec{a}}{|\vec{a}|} \\\\ &=\frac{1}{13}\left(\begin{array}{l} 12 \\\\ -5 \end{array}\right) \end{aligned}$
    Let the required vector be $\vec{b}$.
    $\begin{aligned} &\\ \therefore\ |\vec{b}|=39 \text { (given) } \end{aligned}$
    Since $\vec{b}$ is in the same direction as $\vec{a}$.
    $\begin{aligned} &\\ \vec{b} &=|\vec{b}| \hat{a} \\\\ &=39 \times \frac{1}{13}\left(\begin{array}{c} 12 \\\\ -5 \end{array}\right) \\\\ &=\left(\begin{array}{c} 36 \\\\ -15 \end{array}\right)\\\\ \end{aligned}$
    $\begin{aligned} \text {(b) } \quad &\vec{a}=\left(\begin{array}{c} 2 \\\\ -1 \end{array}\right) \\\\ &\vec{b}=\left(\begin{array}{c} -4 \\\\ 5 \end{array}\right) \\\\ &5 \vec{a}+\lambda\left(\begin{array}{c} 4 \\\\ 6 \end{array}\right)=\mu b \\\\ &5\left(\begin{array}{c} 2 \\\\ -1 \end{array}\right)+\lambda\left(\begin{array}{c} 4 \\\\ 6 \end{array}\right)=\mu\left(\begin{array}{c} -4 \\\\ 5 \end{array}\right) \\\\ &\left(\begin{array}{c} 10 \\\\ -5 \end{array}\right)+\left(\begin{array}{c} 4 \lambda \\\\ 6 \lambda \end{array}\right)=\left(\begin{array}{c} -4 \mu \\\\ 5 \mu \end{array}\right) \\\\ &\left(\begin{array}{c} 10+4 \lambda \\\\ -5+6 \lambda \end{array}\right)=\left(\begin{array}{c} -4 \mu \\\\ 5 \mu \end{array}\right)\\\\ \end{aligned}$
    $\begin{aligned} \therefore\ 10+4 \lambda &=-4 \mu \\\\ \therefore\ \mu &=\frac{-(5+2 \lambda)}{2} \\\\ -5+6 \lambda &=5 \mu \\\\ -5+6 \lambda &=\frac{-5(5+2 \lambda)}{2} \\\\ -10+12 \lambda &=-25-10 \lambda \\\\ 22 \lambda &=-15 \\\\ \lambda &=-\frac{15}{22} \\\\ \mu &=-\frac{1}{2}\left(5+2\left(-\frac{15}{22}\right)\right) \\\\ &=-\frac{20}{11} \end{aligned}$

  7. (a) Given that $\dfrac{q^{-2} \sqrt{p r}}{\sqrt[3]{r}(p q)^{-3}}=p^{a} q^{b} r^{c}$, find the value of each of the constants $a, b$ and $c$. [3]
    (b) Solve the equation $3 x^{\frac{4}{5}}-8 x^{\frac{2}{5}}+5=0$. [4]



  8. $\begin{aligned} \text{ (a) } \hspace{2.5cm} \frac{q^{-2} \sqrt{p r}}{\sqrt[3]{r}(p q)^{-3}} &=p^{a} q^{b} r^{c} \\\\ \frac{q^{-2} p^{\frac{1}{2}} r^{\frac{1}{2}}}{r^{\frac{1}{3}} p^{-3} q^{-3}} &=p^{a} q^{b} r^{e} \\\\ (p)^{\frac{1}{2}+3} (q)^{-2+3} (r)^{\frac{1}{2}-\frac{1}{3}} &=p^{a} q^{b} r^{e} \\\\ p^{\frac{7}{2}} q r^{\frac{1}{6}} &=p^{a} q^{b} r^{e}\\\\ \therefore a &=\frac{7}{2} \\\\ b &=1 \\\\ c &=\frac{1}{6}\\\\ \end{aligned}$
    $\begin{aligned} \text{ (b) } \hspace{2cm} 3 x^{\frac{4}{5}}-8 x^{\frac{2}{5}}+5 &=0 \\\\ 3\left(x^{2 / 5}\right)^{2}-8 x^{2 / 5}+5&=0 \\\\ \left(3 x^{2 / 5}-5\right)\left(x^{2 / 5}-1\right)&=0 \\\\ x^{2 / 5}=\frac{5}{3} \text { (or) } x^{2 / 5}&=1 \\\\ x=\left(\frac{5}{3}\right)^{5 / 2} \text { (or) } x&=7 \end{aligned}$

  9. The polynomial $\mathrm{p}(x)=a x^{3}+b x^{2}+6 x+4$, where $a$ and $b$ are integers, is divisible by $x-2$. When $\mathrm{p}^{\prime}(x)$ is divided by $x+1$ the remainder is $-7$.
    (a) Find the value of $a$ and of $b$. [5]
    (b) Using your answers to part (a), find the remainder when $\mathrm{p}^{\prime \prime}(x)$ is divided by $x$. [2]



  10. $\begin{aligned} &p(x)=a x^{3}+b x^{2}+6 x+4 \\\\ &p^{\prime}(x)=3 a x^{2}+2 b x+6 \\\\ &p(x) \text { is divisible by } x-2, \\\\ \therefore\ & p(2)=0 \\\\ &8 a+4 b+12+4=0 \\\\ &2 a+b=-4 \\\\ &b=-2 a-4 \\\\ &\text { When } p^{\prime}(x) \text { is divided by }(x+1),\\\\ & p(x)=a x^{3}+b x^{2}+6 x+4 \\\\ & p^{\prime}(x)=3 a x^{2}+2 b x+6 \\\\ & p(x) \text { ie divisible by } x-2, \\\\ \therefore\ & p(2)=0 \\\\ & 8 a+4 b+12+4=0 \\\\ & 2 a+b=-4 \\\\ & b=-2 a-4 \\\\ & \text { When } p^{\prime}(x) \text { is divided by }(x+1) \\\\ & \text { the remainder is - } 7 . \\\\ \therefore\ & p(-1)=-7 \\\\ & 3 a-2 b+6=-7\\\\ & 3 a-2 b=-13 \\\\ & 3 a-2(-2 a-4)=-13 \\\\ & 3 a+4 a+8=-13 \\\\ & 7 a=-21 \\\\ & a=-3 \\\\ \therefore \ & b=-2(-3)-4 \\\\ & b=2 \\\\ \therefore \ & p^{\prime}(x)=-9 x^{2}+4 x+6 \\\\ \therefore \ & p^{\prime \prime}(x)=-18 x+4\\\\ \end{aligned}$
    When $p^{\prime \prime}(x)$ is divided by $x$,
    $\begin{aligned} &\\ \text{ the remainder } &=p^{\prime \prime}(0)\\\\ &=-18(0)+4 \\\\ &=4 \end{aligned}$

  11. A curve with equation $y=f(x)$ is such that $\dfrac{\mathrm{d}^{2} y}{\mathrm{~d} x^{2}}=6 \mathrm{e}^{3 x}+4 x$. The curve has a gradient of 5 at the point $\left(0, \dfrac{5}{3}\right)$. Find $f(x)$. [7]



  12. $\begin{aligned} y &=f(x) \\\\ \frac{d^{2} y}{d x^{2}} &=6 e^{3 x}+4 x \\\\ \therefore\ \frac{d y}{d x} &=\displaystyle\int\left(6 e^{3 x}+4 x\right) d x \\\\ &=6 \displaystyle\int e^{3 x} d x+4 \displaystyle\int x d x \\\\ &=\frac{6}{3} e^{3 x}+\frac{4}{2} x^{2}+c_{1} \\\\ &=2\left(e^{3 x}+x^{2}\right)+c_{1}\\\\ \end{aligned}$
    The curve has gradient of 5 at the point $\left(0, \dfrac{5}{3}\right)$.
    $\begin{aligned} &\\ \therefore\ {{\left. {\frac{{dy}}{{dx}}} \right|}_{{\left( {0,\frac{5}{3}} \right)}}}&=5 \\\\ 2\left(e^{0}+0\right)+c_{1} &=5 \\\\ \therefore\ c_{1} &=3 \\\\ \therefore\ \frac{d y}{d x} &=2 e^{3 x}+2 x^{2}+3\\\\ \therefore\ y &=\displaystyle\int\left(2 e^{3 x}+2 x^{2}+3\right) d x \\\\ &=2 \displaystyle\int e^{3 x} d x+2 \displaystyle\int x^{2} d x+3 \displaystyle\int d x \\\\ &=\frac{2}{3} e^{3 x}+\frac{2}{3} x^{3}+3 x+c_{2}\\\\ \end{aligned}$
    Since $\left(0, \dfrac{5}{3}\right)$ lies on the curve,
    $\begin{aligned} &\\ \frac{5}{3} &=\frac{2}{3} e^{0}+\frac{2}{3}(0)^{3}+3(0)+c_{2} \\\\ \therefore\ c_{2} &=1 \\\\ \therefore\ f(x) &=\frac{2}{3} e^{3 x}+\frac{2}{3} x^{3}+3 x+7 \end{aligned}$

  13. The first three terms, in ascending powers of $x$, in the expansion of $(2+a x)^{n}$ can be written as $64+b x+c x^{2}$, where $n, a, b$ and $c$ are constants.
    (a) Find the value of $n$. [1]
    (b) Show that $5 b^{2}=768 c$. [4]
    (c) Given that $b = 12$, find the exact value of $a$ and of $c$. [2]



  14. $\begin{aligned} &(2+a x)^{n}=64+b x+c x^{2}+\cdots \\\\ &2^{n}+n 2^{n-1}(a x)+\frac{n(n-1)}{2} 2^{n-2}(a x)^{2}=64+b x+c x^{2}+\cdots \\\\ \therefore \quad &2^{n}=64 \\\\ &2^{n}=2^{6} \\\\ &n=6 \\\\ \therefore \quad & 6\left(2^{5}\right)(a x)=b x \\\\ \therefore \quad & b=192 a \\\\ &a=\frac{b}{192}\\\\ \end{aligned}$
    $\begin{aligned} \frac{6 \times 5}{2} \times 2^{4} a^{2} x^{2} &=c x^{2} \\\\ 240 a^{2} x^{2} &=c x^{2} \\\\ c &=240 a^{2} \\\\ \therefore \quad c &=240\left(\frac{b}{192}\right)^{2} \\\\ \therefore \quad 5 b^{2} &=768 c\\\\ b &=12\quad \text { (given) } \\\\ \therefore a &=\frac{12}{192} \\\\ &=\frac{1}{16} \\\\ c &=\frac{5 \times 12^{2}}{768} \\\\ &=\frac{15}{16} \end{aligned}$


  15. The diagram shows a circle, centre $O$, radius $5$ cm. The lines $AOB$ and $COD$ are diameters of this circle. The line $AC$ has length $6$ cm.
    (a) Show that angle $AOC = 1.287$ radians, correct to $3$ decimal places. [2]
    (b) Find the perimeter of the shaded region. [2]
    (c) Find the area of the shaded region. [3]



  16. By cosine rule,
    $\begin{aligned} &\\ \cos (\angle A O C) &=\frac{5^{2}+5^{2}-6^{2}}{2 \times 5 \times 5} \\\\ &=\frac{7}{25} \\\\ \angle A O C &=\cos ^{-1}\left(\frac{7}{25}\right) \\\\ \therefore\ \angle B O D &=1.287\quad \mathrm{rad}\\\\ \end{aligned}$
    length of arc $B D=O B\times\angle B O D$
    $\begin{aligned} &\\ &=5(1.287) \\\\ &=6.44\\\\ \end{aligned}$
    $\begin{aligned} &\quad \text{perimeter of shaded regron}\\\\ &=A C+\operatorname{arc} B D+A B+C D \\\\ &=6+6.44+10+10 \\\\ &=32.44 \\\\ &=32.4 \text { (3~sff) }\\\\ \end{aligned}$
    $\begin{aligned} &\quad \text {area of } \triangle A O C\\\\ &=\frac{1}{2} O C \sin (\angle A O C) \times O A \\\\ &=\frac{1}{2}(5) \sin (1.287) \times 5 \\\\ &=12 \\\\ &\quad \text {area of sector } BOD \\\\ &=\frac{1}{2} \times O B^{2}\times\angle B O D \\\\ &=\frac{1}{2}\left(5^{2}\right)(1.287) \\\\ &=16.1 \\\\ &\quad \text {area of shaded region } \\\\ &=12+16.1 \\\\ &=28.1 \end{aligned}$

  17. (a) Find the coordinates of the stationary points on the curve $y=(2 x+1)(x-3)^{2}$. Give your answers in exact form. [4]
    (b) On the axes below, sketch the graph of $y=\left|(2 x+1)(x-3)^{2}\right|,$ stating the coordinates of the points where the curve meets the axes. [4]

    (c) Hence write down the value of the constant $k$ such that $\left|(2 x+1)(x-3)^{2}\right|=k $ has exactly 3 distinct solutions. [1]



  18. $\begin{aligned} \text{(a) }\quad y &=(2 x+1)(x-3)^{2} \\\\ \dfrac{d y}{d x} &=(2 x+1) \dfrac{d}{d x}(x-3)^{2}+(x-3)^{2} \dfrac{d}{d x}(2 x+1) \\\\ &=2(2 x+1)(x-3)+2(x-3)^{2} \\\\ &=2(x-3)(2 x+1+x-3) \\\\ &=2(x-3)(3 x-2) \\\\ \dfrac{d y}{d x} &=0 \text { when }\\\\ \end{aligned}$
    $\begin{aligned} \quad &2(x-3)(3 x-2)=0 \\\\ \quad &\therefore x=3 \text { or } x=\dfrac{2}{3} \\\\ \quad &\text { When } x=\dfrac{2}{3}, y=\dfrac{343}{27} \\\\ \quad &\text { When } x=3, y=0 \\\\ \end{aligned}$
    $\therefore$ The stationary points are $\left(\dfrac{2}{3}, \dfrac{343}{27}\right)$ and $(3,0).$
    $\begin{aligned} &\\ \text{(b) } \hspace{1.5cm} \dfrac{d^{2} y}{d x^{2}} &=2\left[(x-3) \dfrac{d}{d x}(3 x-2)+(3 x-2) \dfrac{d}{d x}(x-3)\right] \\\\ \quad &=2[3 x-9+3 x-2] \\\\ \quad &=2(6 x-11) \\\\ \quad \left.\dfrac{d^{2} y}{d x^{2}}\right|_{\left(\dfrac{2}{3}, 0\right)} &=2\left(6 \times \dfrac{2}{3}-11\right)\\\\ \quad &=-14<0\\\\ \end{aligned}$
    $\quad \therefore \quad \left(\dfrac{2}{3}, 0\right)$ is a minimum turning point.
    $\begin{aligned} &\\ \quad \left.\dfrac{d^{2} y}{d x^{2}}\right|_{(3,0)}&=2(6 \times 3-11)\\\\ \quad &=14>0 \end{aligned}$
    $\quad \therefore\quad (3,0)$ is maximum turning point.
    $\begin{aligned} &\\ \quad \text{When } x=0, y=9\\\\ \quad \therefore\ y-\text{intercept } =(0,9)\\\\ \quad \text{When } y=0, x=-\dfrac{1}{2} \text{ or } x=3\\\\ \quad \therefore\ x-\text{intercept } =\left(-\frac{1}{2}, 0\right) \text{ and } (3,0)\\\\ \end{aligned}$

    $\text{(c) } \quad \left|(2 x+1)(x-3)^{2}\right|=k$ has three distinct solutions.

    This means that the graph of $y=k$ meets the graph $y=\left|(2 x+1)(x-3)^{2}\right|$ at three distinct points.

    By the diagram in part $(b)$, this condition is only possible when $k=\dfrac{343}{27}$.


  19. (a) Jess runs on 5 days each week to prepare for a race.
    In week 1 , every run is $2$ km.
    In week 2, every run is $2.5 $ km.
    In week 3 , every run is $3 $ km.
    Jess increases the distance of the run by $0.5 $ km every week.
    (i) Find the week in which Jess runs $16 $ km on each of the $5$ days. [3]
    (ii) Find the total distance Jess will have run by the end of week $8$.
    [3] (b) Kyle also runs on 5 days each week to prepare for a race.
    In week 1 , every run is $2 $ km.
    In week 2, every run is $2.5 $ km.
    In week 3, every run is $3.125 $ km.
    The distances he runs each week form a geometric progression.
    (i) Find the common ratio of the geometric progression. [3]
    (ii) Find the first week in which Kyle will run more than $16$ km on each of the $5$ days. [3]
    (iii) Find the total distance Kyle will have run by the end of week $8$. [3]



  20. (a) Jeck runs on 5 days according to the following patterns.

    $\begin{array}{|c|c|c|c|} \hline \text{ week 1 } & \text{ week 2 } & \text{ week 3 } & \ldots \\ \hline 2 \mathrm{~km} \text{ a day } & 2.5 \mathrm{~km} \text{ a day } & 3 \mathrm{~km} \text{ a day } & \ldots \\ \hline \end{array}$

    This is an arithmetic progression with the first term $a=2$ and the common difference $d=0.5$.

    (i) Assume that Jeck runs $16 \mathrm{~km}$ in week $n$.

    $\begin{aligned} &\\ \therefore \quad a+(n-1) d &=16 \\\\ 2+(n-1) 0.5 &=16 \\\\ (n-1) 0.5 &=14 \\\\ n &=29\\\\ \end{aligned}$
    Therefore, Jeck runs $16 \mathrm{~km}$ in week $29$ .

    (ii) The total distance that Jecle will run by the end of week $8$ is
    $\begin{aligned} &\\ & \frac{8}{2} \times\{2(2)+7(0.5)\} \times 5 \\\\ =& 150 \text{ km}\\\\ \end{aligned}$
    (b) Kyle runs on 5 days aecording to the following patterns.

    $\begin{array}{|c|c|c|c|} \hline \text{ week 1 } & \text{ week 2 } & \text{ week 3 } & \ldots \\ \hline 2 \mathrm{~km} \text{ a day } & 2.5 \mathrm{~km} \text{ a day } & 3.125 \mathrm{~km} \text{ a day } & \ldots \\ \hline \end{array}$

    The run of kyle on each week in in the form of geometric progression.
    $\begin{aligned} &\\ \text{(i) }\quad \therefore\ a &=2\\\\ r &=\frac{2 \cdot 5}{2} \\\\ &=1 \cdot 25\\\\ \end{aligned}$
    (ii) Assume that Kyle runs $16 \mathrm{~km}$ in week $n$.
    $\begin{aligned} &\\ a r^{n-1} &>16 \\\\ 2(1.25)^{n-1} &>16 \\\\ (1.25)^{n-1} &>8 \\\\ n-1 &>\frac{\ln 8}{\ln 1.25} \\\\ n-1 &>9.32 \\\\ n &>10.32 \\\\ \therefore\ n=11 \\\\ \end{aligned}$
    Therefore The first week that Kyle runs more than $16 \mathrm{~km}$ is week 11.

    (iii) The total distance Kyle will run by the end of week $8$ is
    $\begin{aligned} &\\ &\frac{2\left(1.25^{8}-1\right)}{1.25-1} \times 5 \\\\ &=198.42 \mathrm{~km} \end{aligned}$

  21. (a) Solve the equation $3 \operatorname{cosec}^{2} \theta-5=5 \cot \theta$ for $0^{\circ} \le \theta \le180^{\circ}$. [4]
    (b) Solve the equation $\sin \left(\phi+\dfrac{\pi}{3}\right)=-\dfrac{1}{2}$, where $\phi$ is in radians and $-\pi \le \phi \le \pi$. Give your answers in terms of $\pi$. [4]



  22. $\begin{aligned} \text{(a) } \quad 3 \operatorname{cosec}^{2} \theta-5&=5 \cot \theta, \quad 0^{\circ} \leq \theta \leq 180^{\circ}\\\\ 3\left(1+\cot ^{2} \theta\right)-5 &=5 \cot \theta \\\\ 3+3 \cot ^{2} \theta-5 &=5 \cot \theta \\\\ 3 \cot ^{2} \theta-5 \cot \theta-2 &=0 \\\\ (3 \cot \theta+1)(\cot \theta-2) &=0 \\\\ \cot \theta =-\dfrac{1}{3} \text { or } \cot \theta&=2 \\\\ \theta=\left(180^{3}-4^{2} .56\right) \text { or } \theta &=26.56 \\\\ \theta=108.44^{\circ} \text { or } \theta &=26.56\\\\ \end{aligned}$
    $\text{(b) } \quad \sin \left(\phi+\dfrac{\pi}{3}\right)=-\dfrac{1}{2},-\pi \leq \phi \leq \pi\\\\ $
    basic acute angle (reference angle) $=\dfrac{\pi}{6}\\\\ $
    $\sin \left(\phi+\dfrac{\pi}{3}\right)<0, \phi+\dfrac{\pi}{3}$ lies either $3^{r d}$ or $4^{\text {th }}$ quadrant.
    $\begin{aligned} &\\ \therefore \phi+\dfrac{\pi}{3}=\pi+\dfrac{\pi}{6} \quad \text { or } \quad \phi+\dfrac{\pi}{3} &=-\dfrac{\pi}{6} \\\\ \phi =\dfrac{5 \pi}{6} \quad \text { or } \quad \phi &=-\dfrac{\pi}{2} \end{aligned}$

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