# Oct-Nov-21-p1-CIE-4037-12 : Solution

2021 (Oct-Nov) CIE (4037-Additional Mathematics), Paper 1/12 ၏ Question နှင့် Solution များ ဖြစ်ပါသည်။ Question Paper ကို ဒီနေရာမှာ Download ယူနိုင်ပါသည်။

1. The diagram shows the graph of the cubic function $y = f( x )$. The intercepts of the curve with the axes are all integers.
(a) Find the set of values of $x$ for which $f(x) < 0$. [1]
(b) Find an expression for f(x). [3]

2. (a) The set of the values of $x$ for which $f(x)< 0$ is $\{x \mid -3 < x < 1 \text{ or } x>5\}.\\\\$
(b) Let $f(x)=a(x+3)(x-1)(x-5)$ When $x=0, f(x)=15 a\\\\$
According to the diagram
\begin{aligned} &\\ 15 a &=-5 \\\\ a &=-\frac{1}{3}\\\\ \therefore\ f(x) &=-\frac{1}{3}(x+3)(x-1)(x-5) \\\\ &=-\frac{1}{3}(x+3)\left(x^{2}-6 x+5\right) \\\\ &=-\frac{1}{3}\left(x^{3}-6 x^{2}+5 x+3 x^{2}-18 x+15\right) \\\\ &=-\frac{1}{3}\left(x^{3}-3 x^{2}-13 x+5\right) \end{aligned}

3. (a) Given that $\dfrac{\sqrt[3]{x y}(z y)^{2}}{(x z)^{-3} \sqrt{z}}=x^{a} y^{b} z^{c}$, find the exact values of the constants $a, b$ and $c$. [3]
(b) Solve the equation $5\left(2^{2 p+1}\right)-17\left(2^{p}\right)+3=0$. [4]

4. \begin{aligned} \text{ (a) }\quad \frac{\sqrt[3]{x y}(z y)^{2}}{(x z)^{-3} \sqrt{z}} &=\frac{x^{\frac{1}{3}} y^{\frac{1}{3}} z^{2} y^{2}}{x^{-3} z^{-3} z^{\frac{1}{2}}} \\\\ &=x^{\frac{1}{3}+3} y^{\frac{1}{3}+2} z^{2+3-\frac{1}{2}} \\\\ &=x^{\frac{10}{3}} y^{\frac{7}{3}} z^{\frac{9}{2}}\\\\ \frac{\sqrt[3]{x y}(z y)^{2}}{(x z)^{-3} \sqrt{z}} &=x^{a} y^{b} z^{c} \quad \text { (given) } \\\\ x^{\frac{10}{3} \frac{7}{3}} z^{\frac{9}{2}} &=x y^{b} z^{c} \\\\ \therefore \quad a &=\frac{10}{3} \\\\ b &=\frac{7}{3} \\\\ c &=\frac{9}{2}\\\\ \end{aligned}
\begin{aligned} \text{ (b) }\quad 5\left(2^{2 p+1}\right)-17\left(2^{p}\right)+3 &= 0 \\\\ 5\left(2 \times 2^{2 p}\right)-17\left(2^{p}\right)+3 &= 0 \\\\ 10\left(2^{p}\right)^{2}-17\left(2^{p}\right)+3 &= 0 \\\\ \left(5 \times 2^{p}-1\right)\left(2 \times 2^{p}-3\right) &= 0 \\\\ 2^{p} =\frac{1}{5} \text { or } 2^{p}& =\frac{3}{2} \\\\ p=\log_{2} \left(\frac{1}{5}\right) \text { or } p &=\log_{2} \left(\frac{3}{2}\right)\\\\ p=\frac{\ln \frac{1}{5}}{\ln 2} \quad \text { or } p&=\frac{\ln \frac{3}{2}}{\ln 2} \\\\ p=-2.32 \quad \text { or } p&=0.585 \end{aligned}

5. (a) Write $3+2 \lg a-4 \lg b$ as a single logarithm to base $10$ . [4]
(b) Solve the equation $3 \log _{a} 4+2 \log _{4} a=7$. [5]

6. \begin{aligned} \text{ (a) }\qquad & 3+2 \lg a-4 \lg b \\\\ =&\ \lg 10^{3}+\lg a^{2}-\lg b^{4} \\\\ =&\ \lg \frac{10^{3} a^{2}}{b^{4}}\\\\ \end{aligned}
\begin{aligned} \text{ (b) } \hspace{1.5cm} 3 \log _{a} 4+2 \log a &=7 \\\\ \frac{3}{\log _{4} a}+2 \log a &=7 \\\\ 3+2\left(\log _{4} a\right)^{2} &=7 \log _{4} a \\\\ 2(\log_{4} a)^{2}-7 \log _{4} a+3 &=0 \\\\ (2 \log_{4} a-1)\left(\log _{4} a-3\right) &=0 \\\\ \log _{4} a =\frac{1}{2} \text { or } \log _{4} a &=3 \\\\ a =2 \text { or } a &=64 \end{aligned}

7. Solve the equation $\cot \left(2 x+\dfrac{\pi}{3}\right)-\sqrt{3}=0$, where $-\pi<x<\pi$ radians. Give your answers in terms of $\pi$. [4]

8. \begin{aligned} & \cot \left(2 x+\frac{\pi}{3}\right)-\sqrt{3}=0,-\pi<x<\pi\\\\ & \cot \left(2 x+\frac{\pi}{3}\right)=\sqrt{3}\\\\ & \tan \left(2 x+\frac{\pi}{3}\right)=\frac{1}{\sqrt{3}}\\\\ & 2 x+\frac{\pi}{3}=-\frac{5 \pi}{6} \text { or } 2 x+\frac{\pi}{3}=\frac{\pi}{6} \text { or } 2 x+\frac{\pi}{3}=\frac{7 \pi}{6} \text { or } 2 x+\frac{\pi}{3}=\frac{13 \pi}{6}\\\\ & x=-\frac{7 \pi}{12} \text { or } x=-\frac{\pi}{12} \text { or } x=\frac{5 \pi}{12} \text { or } x=\frac{11 \pi}{12} \end{aligned}

9. Find the possible values of the constant $c$ for which the line $y=c$ is a tangent to the curve $y=5 \sin \dfrac{x}{3}+4$. [3]

10. Curve: $y=5 \sin \frac{x}{3}+4\\\\$
Since $y=c$ is tangent to the curve, $c$ must be either maximum or minimum value.
\begin{aligned} & \\ & -1 \leq \sin \frac{x}{3} \leq 1\\\\ & -5 \leq 5 \sin \frac{x}{3} \leq 5\\\\ & -5+4 \leq 5 \sin \frac{x}{3}+4 \leq 5+4\\\\ & -1 \leq y \leq 9\\\\ \therefore\ & c=-1\ \text{ or }\ c=9. \end{aligned}

11. DO NOT USE A CALCULATOR IN THIS QUESTION.

The polynomial $\mathrm{p}(x)=10 x^{3}+a x^{2}-10 x+b$, where $a$ and $b$ are integers, is divisible by $2 x+1$. When $\mathrm{p}(x)$ is divided by $x+1$, the remainder is $-24$.
(a) Find the value of $a$ and of $b$. [4]
(b) Find an expression for $p(x)$ as the product of three linear factors. [4]
(c) Write down the remainder when $p(x)$ is divided by $x$. [1]

12. $p(x)=10 x^{3}+a x^{2}-10 x+b\\\\$
$p(x)$ is divisible by $2 x+1$.
\begin{aligned} &\\ \therefore\ & p\left(-\frac{1}{2}\right)=0 \\\\ &10\left(-\frac{1}{2}\right)^{3}+a\left(-\frac{1}{2}\right)^{2}-10\left(-\frac{1}{2}\right)+b=0 \\\\ &-\frac{5}{4}+\frac{a}{4}+5+b=0 \\\\ &-5+a+20+4 b=0 \\\\ \therefore\ & a+4 b=-15 \ldots(1)\\\\ \end{aligned}
When $p(x)$ is divided by $(x+1)$, the remainder $=-24$
\begin{aligned} &\\ \therefore\ & p(-1)=-24 \\\\ & 10(-1)^{3}+a(-1)^{2}-10(-1)+b=-24 \\\\ &-10+a+10+b=-24 \\\\ \therefore\ & a+b=-24 \ldots(2)\\\\ \end{aligned}
Subtracting equation (2) from equation (1),
\begin{aligned} &\\ 3 b=9\\\\ b=3\\\\ \end{aligned}
Substituting $b=3$ in equation (2),
\begin{aligned} &\\ & a+3=-24 \\\\ & a=-27 \\\\ \therefore\ & p(x)=10 x^{3}-27 x^{2}-10 x+3 \\\\ & \text { Let } p(x)=(2 x+1)\left(5 x^{2}+c x+3\right) . \\\\ \therefore\ &(2 x+1)\left(5 x^{2}+c x+3\right)=10 x^{3}-27 x^{2}-10 x+3 \\\\ & \text { When } x=1, \quad 3(8+c)=10-27-10+3 \end{aligned}

13. (a)The diagram shows triangle $O A C$, where $\overrightarrow{O A}=\mathbf{a}, \overrightarrow{O B}=\mathbf{b}$ and $\overrightarrow{O C}=\mathbf{c}$. The point $B$ lies on the line $A C$ such that $A B: B C=m: n$, where $m$ and $n$ are constants.

(i) Write down $\overrightarrow{A B}$ in terms of $\mathbf{a}$ and $\mathbf{b}$. [1]
(ii) Write down $\overrightarrow{B C}$ in terms of $\mathbf{b}$ and $\mathbf{c}$. [1]
(iii) Hence show that $n \mathbf{a}+m \mathbf{c}=(m+n) \mathbf{b}$. [2]
(b) Given that $\lambda\left(\begin{array}{l}2 \\ 1\end{array}\right)+(\mu-1)\left(\begin{array}{r}-4 \\ 7\end{array}\right)=(\lambda+1)\left(\begin{array}{r}4 \\ -2\end{array}\right)$, find the value of each of the constants $\lambda$ and $\mu$. [4]

14. \begin{aligned} \text{(a) (i)} \hspace{1.5cm} \overrightarrow{A B} &=\overrightarrow{O B}-\overrightarrow{O A} \\\\ &=\vec{b}-\vec{a} \\\\ \text{(ii)} \hspace{1.5cm} \overrightarrow{B C} &=\overrightarrow{O C}-\overrightarrow{O B} \\\\ &=\vec{c}-\vec{b} \\\\ \text{(iii)} \quad AB: B C &=m: n \\\\ \therefore\ n \overrightarrow{A B} &=m \overrightarrow{B C} \\\\ n(\vec{b}-\vec{a}) &=m(\vec{c}-\vec{b}) \\\\ n \vec{b}-n \vec{a} &=m \vec{c}-m \vec{b} \\\\ \therefore\ \vec{b}+m \vec{b} &=n \vec{a}+m \vec{c} \\\\ n \vec{a}+m \vec{c} &=(m+n) \vec{b}\\\\ \end{aligned}
\begin{aligned} \text{(b)} \quad \lambda\left(\begin{array}{l} 2 \\\\ 1 \end{array}\right)+(\mu-1)\left(\begin{array}{r} -4 \\\\ 7 \end{array}\right) &=(\lambda+1)\left(\begin{array}{r} 4 \\\\ -2 \end{array}\right) \\\\ (\mu-1)\left(\begin{array}{r} -4 \\\\ 7 \end{array}\right) &=(\lambda+1)\left(\begin{array}{r} 4 \\\\ -2 \end{array}\right)-\lambda\left(\begin{array}{l} 2 \\\\ 1 \end{array}\right) \\\\ \left(\begin{array}{l} 4-4 \mu \\\\ 7 \mu-7 \end{array}\right) &=\left(\begin{array}{r} 4 \lambda+4 \\\\ -2 \lambda-2 \end{array}\right)-\left(\begin{array}{l} 2 \lambda \\\\ \lambda \end{array}\right) \\\\ \left(\begin{array}{r} 4-4 \mu \\\\ 7 \mu-7 \end{array}\right) &=\left(\begin{array}{r} 2 \lambda+4 \\\\ -3 \lambda-2 \end{array}\right) \\\\ 4-4 \mu &=2 \lambda+4 \\\\ \lambda &=-2 \mu\\\\ 7 \mu-7 &=-3 \lambda-2 \\\\ 7 \mu-7 &=-3(-2 \mu)-2 \\\\ \mu &=5 \\\\ \therefore\ \lambda &=-10 \end{aligned}

15. (a) A $5$-digit number is made using the digits $0, 1, 4, 5, 6, 7$ and $9$. No digit may be used more than once in any $5$-digit number. Find how many such $5$-digit numbers are even and greater than $50, 000$. [3]
(b) The number of combinations of n objects taken $4$ at a time is equal to $6$ times the number of combinations of $n$ objects taken $2$ at a time. Calculate the value of $n$. [5]

16. $\text{ (a) }\$ For $5$-digit even numbers greater than $50, 000$,
$\qquad$ If the leading digit is $5,7$ or $9$, the unit's digit must be $0, 4$ or $6$.
$\quad\therefore\$ Number of ways $=3 \times{ }^{5} P_{3} \times 3=540$ ways.
$\qquad$ If the leading digit is $6$ , the units digit mult be $0$ or $4$.
$\quad\therefore\$ Number of ways $=1 \times{ }^{5}P_{3} \times 2=120$ ways.
$\quad\therefore\$ Total ways $=540+120=660$ ways.
\begin{aligned} &\\ \text{ (b) } \hspace{2.5cm} {}^nC_{4} &= 6 \times {}^nC_{2} \\\\ \frac{n !}{4 !(n-4) !} &=6 \times \frac{n !}{2 !(n-2) !} \\\\ \frac{1}{4 \times 3 \times 2 !(n-4) !} &=\frac{6}{2 !(n-2)(n-3)(n-4) !} \\\\ \frac{1}{12} &=\frac{6}{(n-2)(n-3)} \\\\ (n-2)(n-3) &=72 \\\\ (n-2)(n-3) &=9 \times 8 \\\\ (n-2)(n-3) &=(11-2)(11-3) \\\\ \therefore\ n &=11 \end{aligned}

17. The diagram shows a circle, centre $O$, radius $12 \mathrm{~cm}$, and a rectangle $A B C D$. The diagonals $A C$ and $B D$ intersect at $O$. The sides $A B$ and $A D$ of the rectangle have lengths $6 \mathrm{~cm}$ and $4 \mathrm{~cm}$ respectively. The points $M$ and $N$ lie on the circumference of the circle such that $M A C$ and $N D B$ are straight lines.

(a) Show that angle $A O D$ is $1.176$ radians correct to 3 decimal places. [2]
(b) Find the perimeter of the shaded region. [4]
(c) Find the area of the shaded region. [3]

18. \begin{aligned} \text { radius } &=12 \mathrm{~cm} \\\\ \therefore O M&=O N =12 \mathrm{~cm} \\\\ A C &=\sqrt{6^{2}+4^{2}} \\\\ &=\sqrt{36+16} \\\\ &=2 \sqrt{13} \mathrm{~cm} \\\\ \therefore O A&=O B=O C =O D=\sqrt{13} \mathrm{cu} \\\\ &=\cos ^{-1} \frac{O A^{2}+O D^{2}-A D^{2}}{2.0 A \cdot O D} \\\\ &=\cos ^{-1} \frac{13+13-16}{2(\sqrt{13})(\sqrt{13})} \\\\ &=1.176 \text{rad}\\\\ \end{aligned}
$\quad$ arc length of major arc $MN$
\begin{aligned} &\\ s&=12 \times(2 \pi-1.176) \mathrm{cm} \\\\ M A&=N D=12-\sqrt{13} \mathrm{~cm}\\\\ \end{aligned}
$\therefore$ perimeter of shaded region
\begin{aligned} &\\ &=m A+N D+A B+C D+B C+S \\\\ &=12-\sqrt{13}+12-\sqrt{13}+6+6+4+12(2 \pi-1.176) \\\\ &=94.1 \mathrm{~cm}\\\\ \end{aligned}
\begin{aligned} &\quad\text { area of major sector } MN \\\\ &=\frac{1}{2} r s \\\\ &=\frac{1}{2} \times 12 \times 12(2 \pi-1.176) \\\\ &=367.72 \mathrm{~cm}^{2} \\\\ &\quad\text { area of rectangle } ABCD \\\\ &=6 \times 4 \\\\ &=24 \mathrm{~cm}^{2}\\\\ &\quad\text { area of } \triangle A O D\\\\ &=\frac{1}{2} \times O A \times O D \times \sin (\angle A O D) \\\\ &=\frac{1}{2} \times 13 \sin (1.176) \\\\ &=6\\\\ \end{aligned}
\begin{aligned} &\quad\text{area of shaded region }\\\\ &=\text{area of major sector } - \text{ area of rectangle } + \text{area of triangle }\\\\ &=367.72-24+6 \\\\ &=349.72 \mathrm{~cm}^{2} \end{aligned}

19. The diagram shows the graph of the curve $y=\dfrac{1}{(x+2)^{2}}+\dfrac{3}{(x+2)}$ for $x>-2$. The points $A$ and $B$ lie on the curve such that the $x$-coordinates of $A$ and of $B$ are $-1$ and 2 respectively.

(a) Find the exact $y$-coordinates of $A$ and of $B$. [2]
(b) Find the area of the shaded region enclosed by the line $A B$ and the curve, giving your answer in the form $\dfrac{p}{q}-\ln r$, where $p, q$ and $r$ are integers. [6]

20. \begin{aligned} \text{ Curve } : y&=\dfrac{1}{(x+2)^{2}}+\dfrac{3}{(x+2)}\\\\ \text{ When } x&=-1,\\\\ y &=\dfrac{1}{(-1+2)^{2}}+\dfrac{3}{(-1+2)} \\\\ &=4\\\\ \text{ When } x&=2,\\\\ y &=\dfrac{1}{(2+2)^{2}}+\dfrac{3}{(2+2)} \\\\ &=\dfrac{13}{16}\\\\ \end{aligned}
$\therefore$ The points $A$ and $B$ respectively are $(-1,4)$ and $\left(2, \dfrac{13}{16}\right).\\\\$
The equation of the line passing through $A$ and $B$ is
\begin{aligned} &\\ y-4 &=\dfrac{\dfrac{13}{16}-4}{2-(-1)}(x+1) \\\\ y-4 &=-\dfrac{17}{16}(x+1) \\\\ y &=-\dfrac{17}{16}(x+1)+4\\\\ &\qquad\text{ area of shaded region }\\\\ &=\int_{-1}^{2}\left[-\dfrac{17}{16}(x+1)+4-\dfrac{1}{(x+2)^{2}}-\dfrac{3}{x+2}\right] d x \\\\ &=\int_{-1}^{2}\left[-\dfrac{17}{16} x-\dfrac{17}{16}+4-(x+2)^{-2}-\dfrac{3}{x+2}\right] d x \\\\ &=\int_{-1}^{2}\left[-\dfrac{17}{16} x+\dfrac{47}{16}-(x+2)^{-2}-\dfrac{3}{x+2}\right] d x \\\\ &=\left[-\dfrac{17}{32} x^{2}+\dfrac{47}{16} x+\dfrac{1}{x+2}-3 \ln (x+2)\right]_{-1}^{2}\\\\ &=\left(-\dfrac{17}{32} \times 4+\dfrac{47}{16} \times 2+\dfrac{1}{4}-3 \ln 4\right)-\left(-\dfrac{17}{32}-\dfrac{47}{16}+1-3 \ln 1\right) \\\\ &=-\dfrac{17}{8}+\dfrac{47}{8}+\dfrac{2}{8}-3 \ln 4+\dfrac{17}{32}+\dfrac{47}{16}+1+3 \ln 1 \\\\ &=\dfrac{207}{32}-3(\ln 4-\ln 1) \\\\ &=\dfrac{207}{32}-3 \ln \dfrac{4}{1}\\\\ &=\dfrac{207}{32}-\ln 64\\\\ \end{aligned}

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