May-Jun-21-p1-CIE-0606/11 : Solution

2021 (May-June) CIE (0606-Additional Mathematics), Paper 1 ၏ Question နှင့် Solution များ ဖြစ်ပါသည်။ Question Paper ကို ဒီနေရာမှာ Download ယူနိုင်ပါသည်။

  1. (a) On the axes, sketch the graph of $y = 5(x+1)(3x-2)(x-2)$, stating the intercepts with the coordinate axes. [3]
    (b) Hence find the values of x for which $5(x+1)(3x-2)(x-2)>0$ . [2]



  2. (a) $y=5(x+1)(3 x-2)(x-2)\\\\ $
    When $y=0$,
    $\begin{aligned} &\\ &5(x+1)(3 x-2)(x-2)=0 \\\\ &\therefore\ x=-1 \text { or } x=\dfrac{2}{3} \text { or } x=2\\\\ \end{aligned}$
    $\therefore$ The graph cuts $x$-axis at $(-1,0),\left(\dfrac{2}{3}, 0\right)$ and $(2,0)$.
    When $x=0, y=20\\\\ $.
    The graph cuts $y$ axis at $(0,20)$.
    (b) The set of the values of $x$ to satisfy the inequality $5(x+1)(3 x-2)(x-2)>0$ is $\left\{x \mid-1<x<\dfrac{2}{3} \text { or } x>2\right\}. $

  3. Find $\displaystyle\int_{3}^{5}\left(\dfrac{1}{x-1}-\dfrac{1}{(x-1)^{2}}\right) \mathrm{d} x$, giving your answer in the form $a+\ln b$, where $a$ and $b$ are rational numbers. [5]



  4. $\begin{aligned} \text { let } u&=x-1 \\\\ \therefore\ d u&=d x \\\\ \text { When } x&=3, u=2 \\\\ \text { When } x&=5, u=4\\\\ \end{aligned}$
    $\begin{aligned} & \int_{3}^{5}\left(\frac{1}{x-1}-\frac{1}{(x-1)^{2}}\right) d x \\\\ =& \int_{2}^{4}\left(\frac{1}{u}-\frac{1}{u^{2}}\right) d u \\\\ =& \int_{2}^{4} \frac{1}{u} d u-\int_{2}^{4} \frac{1}{u^{2}} d u \\\\ =& \int_{2}^{4} \frac{1}{u} d u-\int_{2}^{4} u^{-2} d u \\\\ =&\left.\ln u\right]_{2}^{4}+\left.\frac{1}{u}\right]_{2}^{4} \\\\ =&(\ln 4-\ln 2)+\frac{1}{4}-\frac{1}{2} \\\\ =& \ln \frac{4}{2}-\frac{1}{4} \\\\ =&-\frac{1}{4}+\ln 2 \end{aligned}$

  5. The polynomial $p(x)=a x^{3}-9 x^{2}+b x-6$, where $a$ and $b$ are constants, has a factor of $x-2 .$ The polynomial has a remainder of $66$ when divided by $x-3$.
    (a) Find the value of $a$ and of $b$. [4]
    (b) Using your values of $a$ and $b$, show that $p(x)=(x-2)q(x)$ , where $q(x)$ is a quadratic factor to be found. [2]
    (c) Hence show that the equation $p(x) = 0$ has only one real solution. [2]



  6. $\begin{aligned} \text{(a) } \quad &p(x)=a x^{3}-9 x^{2}+b x-6 \\\\ & x-2 \text { is a factor of } p(x) . \\\\ \therefore\ & p(2)=0 \\\\ & 8 a-3 b+2 b-6=0 \\\\ & 4 a+b=21 \ldots(1)\\\\ \end{aligned}$
    $\quad $ When $p(x)$ is divided by $(x-3)$,
    $\begin{aligned} &\\ \quad\ \text{ the remainder } &=p(3)\\\\ \therefore\ p(3)&=66\\\\ 27 a-81+3 b-6 &=66 \\\\ 9 a+b &=51 \ldots(2)\\\\ \end{aligned}$
    Solving equation $(1)$ and $(2)$,
    $\begin{aligned} &\\ \quad &a=6 \text { and } b=-3 \\\\ &\therefore p(x)=6 x^{3}-9 x^{2}-3 x-6\\\\ \end{aligned}$
    $\text{(b) } \quad\ $ Let $6 x^{3}-9 x^{2}-3 x-6=(x-2)\left(6 x^{2}+k x+3\right)$.
    $\begin{aligned} &\\ \quad\ \therefore\ 6 x^{3}-9 x^{2}-3 x-6 &=6 x^{3}-12 x^{2}+k x^{2}-2 k x+3 x-6 \\\\ &=6 x^{3}+(k-12) x^{2}+(3-2 k) x-6 \\\\ \therefore\ k-12 &=-9 \\\\ \therefore\ k &=3 \\\\ \therefore\ 6 x^{3}-9 x^{2}-8 x-6 &=(x-2)\left(6 x^{2}+3 x+3\right)\\\\ \therefore\ q(x) &=6 x^{2}+3 x+3 \\\\ \end{aligned}$
    $\begin{aligned} \text{(c) } \quad\ \text { When } 6 x^{3}-9 x^{2}-8 x-6 &=0 \\\\ (x-2)\left(6 x^{2}+3 x+3\right) &=0 \\\\ 6(x-2)\left(x^{2}+\dfrac{1}{2} x+\dfrac{1}{2}\right) &=0 \\\\ (x-2)\left(x^{2}+\dfrac{1}{2} x+\dfrac{1}{16}+\dfrac{7}{16}\right) &=0 \\\\ (x-2)\left(\left(x+\dfrac{1}{4}\right)^{2}+\dfrac{7}{16}\right) &=0 \\\\ \text{Since } \left(x+\dfrac{1}{4}\right)^{2}+\dfrac{7}{16} &>0, \\\\ x-2 &=0 \\\\ \therefore\ x &=2\\\\ \end{aligned}$
    $\therefore$ There is only one real solution for $p(x)=0$.

  7. The first 3 terms in the expansion of $(a+x)^{3}\left(1-\dfrac{x}{3}\right)^{5}$, in ascending powers of $x$, can be written in the form $27+b x+c x^{2}$, where $a, b$ and $c$ are integers. Find the values of $a, b$ and $c$. [8]



  8. $\begin{aligned} (a+x)^{3}\left(1-\frac{x}{3}\right)^{5} &=27+b x+c x^{2}+\cdots \\\\ \left(a^{3}+3 a^{2} x+3 a x^{2}+x^{3}\right)\left(1-5\left(\frac{x}{3}\right)+10\left(\frac{x}{3}\right)^{2}+\cdots\right) &=27+b x+c x^{2}+\cdots \\\\ \left(a^{3}+3 a^{2} x+3 a x^{2}+x^{3}\right)\left(1-\frac{5 x}{3}+\frac{10 x^{2}}{9}+\cdots\right) &=27+b x+c x^{2}+\cdots \\\\ \left(a^{3}-\frac{5 a^{3}}{3} x+\frac{10 a^{3}}{9} x^{2}+3 a^{2} x-5 a^{2} x^{2}+3 a x^{2}+\cdots\right) &=27+b x+c x^{2}+\cdots \\\\ a^{3}+\left(3 a^{2}-\frac{5 a^{3}}{3}\right) x+\left(\frac{10 a^{3}}{9}-5 a^{2}+3 a\right) x^{2}+\cdots &=27+b x+c x^{2}+\cdots\\\\ \end{aligned}$
    $\begin{aligned} \quad\therefore\quad a^{3} &=27 \\\\ a &=3\\\\ b &=3 a^{2}-\frac{5 a^{3}}{3} \\\\ &=3\left(3^{2}\right)-\frac{5(27)}{3} \\\\ &=27-45 \\\\ &=-18 \\\\ c &=\frac{10 a^{3}}{9}-5 a^{2}+3 a \\\\ &=\frac{10(27)}{9}-5(3)^{2}+3(3) \\\\ &=30-45+9 \\\\ &=-6 \end{aligned}$

  9. The functions $f$ and $g$ are defined as follows.
    $\begin{aligned} &\\ &f(x)=x^{2}+4 x \text { for } x \in \mathbb{R} \\\\ & g(x)=1+e^{2 x} \text { for } x \in \mathbb{R}\\\\ \end{aligned}$
    (a) Find the range of $f$. [2]
    (b) Write down the range of $g$. [1]
    (c) Find the exact solution of the equation $fg(x) = 21$, giving your answer as a single logarithm. [4]



  10. $\begin{aligned} \text{(a) } \hspace{1.5cm} f(x) &=x^{2}+4 x, x \in R \\\\ f(x) &=x^{2}+4 x+4-4 \\\\ f(x) &=(x+2)^{2}-4 \\\\ (x+2)^{2} & \geq 0 \text { for all } x \in \mathbb{R} \\\\ (x+2)^{2}-4 & \ge-4 \\\\ \therefore \quad f(x) & \ge-4\\\\ \text{(b) }\hspace{1.5cm} g(x)&=1+e^{2 x}, x \in \mathbb{R}\\\\ \end{aligned}$
    $\qquad$ Since $e^{2 x}>0$ for all $x \in \mathbb{R}$,
    $\begin{aligned} &\\ 1+e^{2 x} &>1 \\\\ \therefore\ g(x) &>1\\\\ \text{(c) }\hspace{2cm} f g(x)&=21\\\\ f(g(x)) &=21 \\\\ f\left(1+e^{2 x}\right) &=21 \\\\ \left(1+e^{2 x}+2\right)^{2}-4 &=21 \\\\ \left(e^{2 x}+3\right)^{2} &=25 \\\\ e^{2 x}+3 &=5 \\\\ e^{2 x} &=2 \\\\ 2 x &=\ln 2 \\\\ x &=\frac{1}{2} \ln 2 \\\\ &=\ln \sqrt{2} \end{aligned}$

  11. (a) (i) Find how many different 5-digit numbers can be formed using the digits 1, 3, 5, 6, 8 and 9. No digit may be used more than once in any 5-digit number. [1]
    (ii) How many of these 5-digit numbers are odd? [1]
    (iii) How many of these 5-digit numbers are odd and greater than 60000? [3]
    (b) Given that $45 \times{ }^{n} \mathrm{C}_{4}=(n+1) \times{ }^{n+1} \mathrm{C}_{5}$, find the value of $n$. [4]



  12. $\begin{aligned} \text{(a) (i) } \quad & { }^{6} P_{5}=720\\\\ \quad\text{(ii) }\quad & \text{Total number of 5-digit odd numbers} \\\\ =&\ 4 \times 5 ! \\\\ =&\ 480\\\\ \text{(iii) } \quad &\text{Total number of 5-digit odd numbers start with 6 and 8}\\\\ =&\ 2 \times 4 \times 3 \times 2 \\\\ =&\ 192\\\\ &\text{Total number of 5 -digit odd numbers start with 9 }\\\\ =&\ \times 4 \times 3 \times 3 \\\\ =&\ 72\\\\ \therefore\ &\text{Total number of 5-digit odd numbers } > 6000\\\\ =&\ 192+72 \\\\ =&\ 264\\\\ \end{aligned}$
    $\begin{aligned} \text{(c) } \hspace{1.5cm} 45 \times{ }^{n} C_{4} &=(n+1) \times{ }^{n+1} C_{5} \\\\ 45 \times \frac{n !}{4 !(n-4) !}&=(n+1) \times \frac{(n+1) !}{5 !(n+1-5) !}\\\\ 4 !(n-4) ! &=(n+1) \times \frac{(n+1) n !}{5 \times 4 !(n-4) !} \\\\ 45 &=\frac{(n+1)^{2}}{5} \\\\ (n+1)^{2} &=225\\\\ n+1 &=15 \\\\ n &=14 \end{aligned}$

  13. (a) In this question, all lengths are in metres and time, t, is in seconds.

    The diagram shows the displacement–time graph for a runner, for $0\le t\le 40$.
    (i) Find the distance the runner has travelled when $t = 40$.
    [1]
    (ii) On the axes, draw the corresponding velocity–time graph for the runner, for $0\le t\le 40$. [2]

    (b) A particle, $P$, moves in a straight line such that its displacement from a fixed point at time $t$ is $s$. The acceleration of $P$ is given by $(2 t+4)^{-\frac{1}{2}}$, for $t>0$.
    (i) Given that $P$ has a velocity of $9$ when $t=6$, find the velocity of $P$ at time $t$.
    [3]
    (ii) Given that $s=\dfrac{1}{3}$ when $t=6$, find the displacement of $P$ at time $t$. [3]



  14. (i) At $0 \leq t <10$
    distance travelled by runner $=50 \mathrm{~m}$
    At $10 \leq t<40$,
    distance travelled by runner $=60 \mathrm{~m}$
    the distance the runner has travelled when $t=40$ is
    $(50+60)=110 \mathrm{~m}$

    (ii) $0 \leqslant t<10$,
    $\quad v=\dfrac{50}{10}=5 \mathrm{~m} / \mathrm{s}$
    $\quad 10 \leq t \leq 40$,
    $\quad v=\dfrac{-10-50}{40-10}=\dfrac{-60}{30}=-2 \mathrm{~m} / \mathrm{s}$


    $\begin{aligned} \text { (b) } \quad \text { displacement } &=s\\\\ \text { velocity } &=\frac{d s}{d t} \\\\ \text { acceleration } &=\frac{d^{2} s}{d t^{2}} \\\\ \frac{d^{2} s}{d t} &=(2 t+4)^{-1 / 2}, t>0 \\\\ \text { velocity } &=\displaystyle\int(2 t+4)^{-1 / 2} d t\\\\ \text { Let } u &=2 t+4 \\\\ d u &=2 d t \\\\ d t &=\frac{1}{2} d u \\\\ \text { velocity } &=\frac{1}{2} \int u^{-1 / 2} d u \\\\ &=\sqrt{u}+c_{1} \\\\ &=\sqrt{2 t+4}+c_{1} \\\\ \text { When } t &=6, v=9 . \\\\ 9 &=\sqrt{16}+c_{1} \\\\ c_{1} &=5 \\\\ \therefore\ \frac{d s}{d t} &=\sqrt{2 t+4}+5 \text { (at time t) } \\\\ \text { displacement } &=\int(\sqrt{2 t+4}+5) d t \\\\ &=\int \sqrt{2 t+4} d t+5 \int d t \\\\ &=\frac{1}{2}(2 t+4)^{3 / 2}+5 t+c_{2}\\\\ s &=\frac{1}{3} \text { When } t=6 \\\\ \frac{1}{3} &=\frac{1}{3}(12+4)^{3 / 2}+5(6)+c_{2} \\\\ \frac{1}{3} &=\frac{1}{3}(64)+30+c_{2} \\\\ \therefore\ c_{2} &=-51 \\\\ \therefore\ s &=\frac{1}{3}(2 t+4)^{3 / 2}+5 t-51 \end{aligned}$

  15. DO NOT USE A CALCULATOR IN THIS QUESTION.

    A curve has equation $y=(2-\sqrt{3}) x^{2}+x-1$. The $x$-coordinate of a point $A$ on the curve is $\dfrac{\sqrt{3}+1}{2-\sqrt{3}}$.
    (a) Show that the coordinates of $A$ can be written in the form $(p+q \sqrt{3}, r+s \sqrt{3})$, where $p, q, r$ and $s$ are integers. [5]
    (b) Find the $x$-coordinate of the stationary point on the curve, giving your answer in the form $a+b \sqrt{3}$, where $a$ and $b$ are rational numbers. [3]



  16. $\begin{aligned} \text{ (a) } \quad \text{ Curve } : y&=(2-\sqrt{3}) x^{2}+x-1\\\\ \frac{\sqrt{3}+1}{2-\sqrt{3}} &=\frac{\sqrt{3}+1}{2-\sqrt{3}} \times \frac{2+\sqrt{3}}{2+\sqrt{3}} \\\\ &=\frac{2 \sqrt{3}+3+2+\sqrt{3}}{4-3} \\\\ &=5+3 \sqrt{3}\\\\ \text{ When } x&=5+3 \sqrt{3},\\\\ y &=(2-\sqrt{3})(5+3 \sqrt{3})^{2}+5+3 \sqrt{3}-1 \\\\ &=(2-\sqrt{3})(25+30 \sqrt{3}+27)+4+3 \sqrt{3}\\\\ y &=(2-\sqrt{3})(52+30 \sqrt{3})+4+3 \sqrt{3} \\\\ y &=104-52 \sqrt{3}+60 \sqrt{3}-90+4+3 \sqrt{3} \\\\ &=18+11 \sqrt{3}\\\\ \end{aligned}$
    $\qquad \therefore$ The coordinates of the point $A$ is $(5+3 \sqrt{3}, 18+11 \sqrt{3}).$
    $\begin{aligned} &\\ \text{ (b) } \quad \frac{d y}{d x}&=2(2-\sqrt{3}) x+1 \end{aligned}\\\\ $
    $\qquad$ At stationary point,
    $\begin{aligned} &\\ \frac{d y}{d x} &=0 . \\\\ \qquad 2(2-\sqrt{3}) x+1 &=0 \\\\ x &=\frac{1}{2 \sqrt{3}-4}\\\\ x &=\frac{1}{2 \sqrt{3}-4} \times \frac{2 \sqrt{3}+4}{2 \sqrt{3}+4} \\\\ &=\frac{2 \sqrt{3}+4}{12-16} \\\\ &=\frac{2 \sqrt{3}+4}{-4} \\\\ &=-1-\frac{\sqrt{3}}{2}\\\\ \end{aligned}$

  17. (a) (i) Write $6 x y+3 y+4 x+2$ in the form $(a x+b)(c y+d)$, where $a, b, c$ and $d$ are positive integers. [1]
    (ii) Hence solve the equation $6 \sin \theta \cos \theta+3 \cos \theta+4 \sin \theta+2=0$ for $0^{\circ} <\theta<360^{\circ}$.
    [4]
    (b) Solve the equation $\dfrac{1}{2} \sec \left(2 \phi+\dfrac{\pi}{4}\right)=\dfrac{1}{\sqrt{3}}$ for $-\pi<\phi<\pi$, where $\phi$ is in radians. Give your answers in terms of $\pi$. [5]



  18. $\begin{aligned} \text { (a) (i) }\quad & 6 x y+3 y+4 x+2 \\\\ =& 3 y(2 x+1)+2(2 x+1) \\\\ =&(2 x+1)(3 y+2)\\\\ \end{aligned}$
    $\begin{aligned} \quad\text{ (ii) } \quad 6 \sin \theta \cos \theta+3 \cos \theta+4 \sin \theta+2&=0\\\\ \therefore(2 \sin \theta+1)(8 \cos \theta+2) &=0 \\\\ \sin \theta=-\frac{1}{2} \text { or } \cos \theta &=-\frac{2}{3}\\\\ \end{aligned}$
    $\begin{aligned} \qquad\text { For } \sin \theta &=-\frac{1}{2} \\\\ \theta &=210^{\circ} \text { or } \\\\ \theta &=330^{\circ} \\\\ \text { For } \cos \theta &=-\frac{2}{3} \\\\ \theta &=180-48.2 \text { or } \\\\ \theta &=180^{\circ}+48.2^{\circ} \\\\ \therefore \theta &=131.8^{\circ} \text { or } \\\\ \theta &=228.2^{\circ}\\\\ \end{aligned}$
    $\begin{aligned} \text { (a) (i) } &\quad \frac{1}{2} \sec \left(2 \phi+\frac{\pi}{4}\right)=\frac{1}{\sqrt{3}},-\pi<\phi<\pi \\\\ &\sec \left(2 \phi+\frac{\pi}{4}\right)=\frac{2}{\sqrt{3}} \\\\ &\cos \left(2 \phi+\frac{\pi}{4}\right)=\frac{\sqrt{3}}{2}\\\\ \end{aligned}$
    $\qquad \therefore$ basic acute angle (reference angle) $=\dfrac{\pi}{6}$
    $\begin{aligned} &\\ \qquad \cos \left(2 \phi+\frac{\pi}{4}\right)&>0 \\\\ \end{aligned}$
    $\qquad \therefore 2 \phi+\dfrac{\pi}{4}$ lies either in the $1^{\text {st }}$ or $4^{\text {th }}$ quadrant.
    $\begin{aligned} &\\ \qquad 2 \phi+\frac{\pi}{4} &=\frac{\pi}{6} \text { or } \\\\ 2 \phi+\frac{\pi}{4}&=-\frac{\pi}{6} \text { or }\\\\ 2 \phi+\frac{\pi}{4}&=\frac{13 \pi}{6} \text { or }\\\\ 2 \phi+\frac{\pi}{4}&=\frac{11 \pi}{6} \\\\ \qquad \therefore \phi &=-\frac{\pi}{24} \text { or }\\\\ \phi&=-\frac{5 \pi}{24} \text { or } \\\\ \phi&=\frac{23 \pi}{24} \text { or } \\\\ \phi&=\frac{19 \pi}{24} \end{aligned}$

  19. In this question all lengths are in centimetres.
    The diagram shows a shaded shape. The arc $AB$ is the major arc of a circle, centre $O$, radius $10$. The line $AB$ is of length $15$, the line $OC$ is of length $25$ and the lengths of $AC$ and $BC$ are equal.
    (a) Show that the angle $AOB$ is $1.70$ radians correct to $2$ decimal places. [2]
    (b) Find the perimeter of the shaded shape. [4]
    (c) Find the area of the shaded shape. [5]



  20. $\text{ (a) } \quad$ By cosine rule,
    $\begin{aligned} &\\ \qquad \cos (\angle A O B) &=\frac{10^{2}+10^{2}-15^{2}}{2(10)(10)} \\\\ &=-\frac{1}{8} \\\\ \angle A O B &=1.70 \text{ rad}\\\\ \end{aligned}$
    $\begin{aligned} \text{ (b) } \quad &\text{ The perpendicular distance from } O \text{ to } A B\\\\ &=\sqrt{10^{2}-\left(\frac{15}{2}\right)^{2}} \\\\ &=6.61\\\\ \therefore\ & \text{ The perpendicular distance from } C \text{ to } A B\\\\ &=25-6.61 \\\\ &=18.39\\\\ \end{aligned}$
    $\begin{aligned} & A C=B C=\sqrt{18.32^{2}+7.5^{2}} \\\\ &=19.86 \\\\ &\quad \text { length of major sector }\\\\ &=(2 \pi-1.7) \times 10 \\\\ &=45.83 \\\\ \therefore & \text { perimeter of shadld region } \\\\ =&\ 2(19.8 b)+45.83 \\\\ =&\ 85.55\\\\ \end{aligned}$
    $\begin{aligned} &\quad \text{ area of major sector }\\\\ &=\frac{1}{2} \times 10^{2}(2 \pi-1.7) \\\\ &=229.15 \\\\ &\quad \text { area of kite AOBC } \\\\ &= \frac{1}{2} \times 25 \times 15 \\\\ &=187.5 \\\\ &=229.15+187.55 \\\\ &=416.65 \end{aligned}$

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