Oct-Nov-21-p2-CIE-4037-22 : Solution

2021 (Oct-Nov) CIE (4037-Additional Mathematics), Paper 2/22 ၏ Question နှင့် Solution များ ဖြစ်ပါသည်။ Question Paper ကို ဒီနေရာမှာ Download ယူနိုင်ပါသည်။

1. (a) On the axes, draw the graphs of $y=5+|3 x-2|$ and $y=11-x$. [4]
   
   (b) Using the graphs, or otherwise, solve the inequality $11-x < 5+|3 x-2|$. [2]






$\begin{aligned} &y=5+|3 x-2| \\\\ &y=3\left|x-\frac{2}{3}\right|+5 \\\\ \therefore\ & \text { vertex }=\left(\frac{2}{3}, 5\right) \\\\ &\text { When } x=0, y=7 \\\\ \therefore\ & y \text {-intercept }=(0,7)\\\\ &y=11-x\\\\ &\text { When } x=0, y=11 \\\\ \therefore\ & y-\text {-intercept } =(0,11)\\\\ &\text { When } y=0, x=11 \\\\ \therefore\ & x-\text {-intercept } =(11,0)\\\\ \end{aligned}$ According to the graph, the solution set to satisfy the inequality $1-x < 5+| 3 x-2|$ is $\{x \mid x <-2 \text { or } x>2\}$.




2. (a) Expand $(2-3x)^4$, evaluating all of the coefficients. [4]
   (b) The sum of the first three terms in ascending powers of $x$ in the expansion of $(2-3 x)^{4}\left(1+\dfrac{a}{x}\right)$ is $\dfrac{32}{x}+b+c x$, where $a, b$ and $c$ are integers. Find the values of each of $a, b$ and $c .$ [4]
   






$\begin{aligned} (2-3 x)^{4} &=2^{4}+4\left(2^{3}\right)(-3 x)+6\left(2^{2}\right)(-3 x)^{2}+4(2)(-3 x)^{3}+(-3 x)^{4} \\\\ &=16-96 x+216 x^{2}-24 x^{3}+81 x^{4} \\\\ \end{aligned}$ $\begin{aligned} (2-3 x)\left(1+\frac{a}{x}\right) &=\frac{32}{x}+b+c x+\cdots \\\\ \left(16-96 x+216 x^{2}-24 x+81 x^{4}\right)\left(1+\frac{a}{x}\right) &=\frac{32}{x}+b+c x+\cdots \\\\ 16+\frac{16 a}{x}-96 x-96 a+216 a x+\cdots &=\frac{32}{x}+b+c x+\cdots \\\\ \frac{16 a}{x}+(16-96 a)+(216 a-96) x+\cdots &=\frac{32}{x}+b+c x+\cdots \\\\ \end{aligned}$ $\begin{aligned} \therefore\ & 16 a =32, a=2 \\\\ & b =16-96 a=-176 \\\\ & c =216 a-96=336 \end{aligned}$




3. (a) Show that $\dfrac{1}{\sec x-1}+\dfrac{1}{\sec x+1}=2 \cot x \operatorname{cosec} x$. [4]
   (b) Hence solve the equation $\dfrac{1}{\sec x-1}+\dfrac{1}{\sec x+1}=3 \sec x$ for $0^{\circ} < x < 360^{\circ}$. [4]






$\begin{aligned} & \frac{1}{\sec x-1}+\frac{1}{\sec x+1} \\\\ =&\ \frac{\sec x+1+\sec x-1}{(\sec x-1)(\sec x+1)} \\\\ =&\ \frac{2 \sec x}{\sec ^{2} x-1} \\\\ =&\ \frac{2 \sec x}{\tan ^{2} x} \\\\ =&\ 2 \cdot \frac{1}{\cos x} \frac{1}{\tan x} \frac{1}{\tan x}\\\\ =&\ \cdot \frac{1}{\cos x} \cdot \frac{\cos x}{\sin x} \cdot \cot x \\\\ =&\ \frac{2}{\sin x} \cdot \cot x \\\\ =&\ 2\cot x \operatorname{cosec} x\\\\ \end{aligned}$ $\begin{aligned} \frac{1}{\sec x-1}+\frac{1}{\sec x+1} &=3 \sec x, 0^{\circ}<x<360^{\circ} \\\\ 2 \cot x \operatorname{cosec} x &=3 \sec x \\\\ 2 \frac{\cos x}{\sin x} \frac{1}{\sin x} &=\frac{3}{\cos x} \\\\ 2 \cos ^{2} x &=3 \sin ^{2} x \\\\ \tan ^{2} x &=\frac{2}{3} \\\\ \tan x &=\pm \frac{2}{3} \\\\ x &=39.2^{\circ} \text { or } \\\\ x &=140.8^{\circ} \text { or } \\\\ x &=219.2^{\circ} \text { or } \\\\ x &=320.8^{\circ} \end{aligned}$




4. (a) Find the $x$-coordinates of the stationary points on the curve $y=3 \ln x+x^{2}-7 x$, where $x>0$. [5]
   (b) Determine the nature of each of these stationary points. [3]






$\begin{aligned} \text{(a) }\quad &y=3 \ln x+x^{2}-7 x \\\\ &\frac{d y}{d x}=\frac{3}{x}+2 x-7 \\\\ &\frac{d y}{d x}=0 \text { when, } \\\\ &\frac{3}{x}+2 x-7=0 \\\\ &2 x^{2}-7 x+3=0 \\\\ &(2 x-1)(x-3)=0 \\\\ &x=\frac{1}{2} \text { or } x=3\\\\ \end{aligned}$ $\qquad$ Hence, the curve has stationary points where $x=\dfrac{1}{2}$ or $x=3$. $\begin{aligned} &\\ \text{(b) }\quad &\frac{d^{2} y}{d x^{2}}=-\frac{3}{x^{2}}+2\\\\ &\text { When } x=\frac{1}{2}, \frac{d^{2} y}{d x^{2}}=-\frac{3}{\frac{1}{4}}+2=-10<0 \\\\ &\text { When } x=3, \frac{d^{2} y}{d x^{2}}=-\frac{3}{9}+2=\frac{5}{3}>0\\\\ \end{aligned}$ $\therefore\quad$ The curve is maximum where $x=\dfrac{1}{2}$ and is minimum where $x=3.$




5. (a) Solve the following simultaneous equations.
   $\begin{aligned} &\\ \qquad e^{x}+e^{y}&=5 \\\\ \qquad 2 e^{x}-3 e^{y}&=8 \end{aligned}$
   [5]
   (b) Solve the equation $e^{(2 t-1)}=5 e^{(5 t-3)}.$ [4]






$\begin{aligned} \text{ (a) }\hspace{1.5cm} e^{x}+e^{y}&=5 \\\\ e^{y} &=5-e^{x} \\\\ 2 e^{x}-3 e^{y}&=8 \\\\ \therefore\ 2 e^{x}-3\left(5-e^{x}\right) &=8 \\\\ 2 e^{x}-15+3 e^{x} &=8 \\\\ 5 e^{x} &=23 \\\\ e^{x} &=\frac{23}{5} \\\\ x &=\ln \left(\frac{23}{5}\right) \\\\ &=1.53\\\\ e^{y} &=5-\frac{23}{5} \\\\ &=\frac{2}{5} \\\\ y &=\ln \left(\frac{2}{5}\right) \\\\ &=-0.916\\\\ \text{ (b) }\hspace{1.5cm}\frac{e^{(2 t-1)}}{e^{2 t-1}} &=5 e^{(5 t-3)} \\\\ e^{5 t-3} &=5 \\\\ \frac{1}{e^{3 t}-2} &=5 \\\\ e^{3 t-2} &=\frac{1}{5} \\\\ 3 t-2 &=\ln \frac{1}{5} \\\\ t &=\frac{1}{3}\left(2+\ln \frac{1}{5}\right) \\\\ &=0.13 \end{aligned}$




6. DO NOT USE A CALCULATOR IN THIS QUESTION.
   
   All lengths in this question are in centimetres.
   You may use the following trigonometrical ratios.
   $\begin{aligned} &\\ \qquad &\sin 60^{\circ}=\dfrac{\sqrt{3}}{2} \\\\ \qquad &\cos 60^{\circ}=\dfrac{1}{2} \\\\ \qquad &\tan 60^{\circ}=\sqrt{3}\\\\ \end{aligned}$
   The diagram shows triangle $A B C$ with $A C=\sqrt{6}-\sqrt{2}, A B=\sqrt{6}+\sqrt{2}$ and angle $C A B=60^{\circ}$.
   
   (a) Find the exact length of $B C$. [3]
   (b) Show that $\sin A C B=\dfrac{\sqrt{6}+\sqrt{2}}{4}$. [2]
   (c) Show that the perpendicular distance from $A$ to the line $B C$ is $1$. [2]






$\text { (a) }$ By cosine Rule, $\begin{aligned} &\\ \qquad BC^{2} &=A B^{2}+A C^{2}-2(A B)(A C) \cos A \\\\ &=(\sqrt{6}+\sqrt{2})^{2}+(\sqrt{6}-\sqrt{2})^{2}-2(\sqrt{6}+\sqrt{2})(\sqrt{6}-\sqrt{2}) \cos 60^{\circ} \\\\ &=6+2 \sqrt{12}+2+6-2 \sqrt{12}+2-2(6-2) \frac{1}{2} \\\\ &=12 \\\\ B C &=\sqrt{12} \\\\ B C &=2 \sqrt{3}\\\\ \end{aligned}$ $\text { (b) }$ By Sine Rule, $\begin{aligned} &\\ \qquad \frac{\sin (\angle A C B)}{A B} &=\frac{\sin (\angle B A C)}{B C} \\\\ \sin (\angle A C B) &=\frac{A B}{B C} \sin (\angle B A C) \\\\ &=\frac{\sqrt{6}+\sqrt{2}}{2 \sqrt{3}} \times \sin 60^{\circ} \\\\ &=\frac{\sqrt{6}+\sqrt{2}}{2 \sqrt{3}} \times \frac{\sqrt{3}}{2} \\\\ &=\frac{\sqrt{6}+\sqrt{2}}{4}\\\\ \end{aligned}$ $\text { (c) }$ Let $h$ be the perpendicular distance from $A$ to $B C$. $\begin{aligned} &\\ \qquad \therefore\ h &=A C \sin (\angle A C B) \\\\ &=(\sqrt{6}-\sqrt{2}) \frac{\sqrt{6}+\sqrt{2}}{4} \\\\ &=\frac{6-2}{4} \\\\ &=1 \end{aligned}$




7. It is given that $\dfrac{d^{2} y}{dx^{2}}=e^{2 x}+\dfrac{1}{(x+1)^{2}}$ for $x>-1$.
   (a) Find an expression for $\dfrac{dy}{dx}$ given that $\dfrac{d y}{dx}=2$ when $x=0$. [3]
   (b) Find an expression for $y$ given that $y=4$ when $x=0$. [3]
   






$\begin{aligned} \frac{d^{2} y}{d x^{2}}=e^{2 x} &+\frac{1}{(x+1)^{2}}, x>-1 . \\\\ \therefore \frac{d y}{d x} &=\displaystyle\int\left(e^{2 x}+\frac{1}{(x+1)^{2}}\right) d x \\\\ &=\frac{1}{2} \displaystyle\int e^{2 x} d(2 x)+\displaystyle\int(x+1)^{-2} d x \\\\ &=\frac{1}{2} e^{2 x}-\frac{1}{x+1}+c_{1} \\\\ \frac{d y}{d x} &=2 \text{ when }x=0\\\\ \frac{1}{2}-1+e_{1} &=2 \\\\ c_{1} &=\frac{5}{2} \\\\ \therefore\ \frac{d y}{d x} &=\frac{1}{2} e^{2 x}-\frac{1}{x+1}+\frac{5}{2}\\\\ y &=\displaystyle\int\left[\frac{1}{2} e^{2 x}-\frac{1}{x+1}+\frac{5}{2}\right] d x \\\\ &=\displaystyle\int \frac{1}{2} e^{2 x} d x-\displaystyle\int \frac{1}{x+1} d x+\frac{5}{2} \displaystyle\int d x \\\\ &=\frac{1}{4} e^{2 x}-\ln (x+1)+\frac{5 x}{2}+c_{2} \\\\ y &=4 \text { when } x=0 \\\\ \therefore\ \frac{1}{4}+c_{2} &=4 \\\\ \therefore\ c_{2} &=\frac{15}{4}\\\\ \therefore\ y&=\frac{1}{4} e^{2 x}-\ln (x+1)+\frac{5 x}{2}+\frac{15}{4} \end{aligned}$




8. Variables $x$ and $y$ are such that when $\sqrt{y}$ is plotted against $\log _{2}(x+1)$, where $x>-1$, a straight line is obtained which passes through $(2,10.4)$ and $(4,15.4)$.
   (a) Find $\sqrt{y}$ in terms of $\log _{2}(x+1)$. [4]
   (b) Find the value of $y$ when $x=15$. [1]
   (c) Find the value of $x$ when $y = 25$. [3]






$\text{ (a) }\quad$ The straight line passes through the points $(2,10.4)$ and $(4,15.4)$. $\begin{aligned} &\\ \qquad\therefore \text{ gradient } &=\frac{15 \cdot 4-10 \cdot 4}{4-2}\\\\ &=\frac{5}{2}\\\\ \end{aligned}$ $\quad \therefore\ $ Equation of straight line is $\begin{aligned} &\\ \quad \sqrt{y}-10.4 &=\frac{5}{2}\left(\log _{2}(x+1)-2\right) \\\\ \sqrt{y} &=\frac{5}{2} \log _{2}(x+1)+5-4\\\\ \text{ (b) }\quad \text{ When } x&=15,\\\\ \sqrt{y} &=\frac{5}{2} \log _{2}(16)+5.4 \\\\ &=\frac{5}{2}(4)+5.4 \\\\ &=15.4 \\\\ y &=237.16\\\\ \text{ (c) }\quad \text{ When } y&=25,\\\\ \sqrt{25} &=\frac{5}{2} \log _{2}(x+1)+5.4 \\\\ \log (x+1) &=-0.16 \\\\ x+1 &=2^{-0.16} \\\\ x &=2^{-0.16}-1 \\\\ &=-0.105 \end{aligned}$




9. (a) Find the equation of the normal to the curve $y=x^{3}+x^{2}-4 x+6$ at the point $(1,4)$. [5]
   (b) DO NOT USE A CALCULATOR IN THIS QUESTION.
   Find the exact $x$-coordinate of each of the two points where the normal cuts the curve again. [5]






$\begin{aligned} \text{ (a) } \hspace{1.5cm}\text{ Curve } : y&=x^{3}+x^{2}-4 x+6\\\\ \frac{d y}{d x} &=3 x^{2}+2 x-4 \\\\ \left.\frac{d y}{d x}\right|_{(1,4)} &=3+2-4 \\\\ &=1\\\\ \therefore\ \text{ gradient of normal } &=-1\\\\ \end{aligned}$ $\qquad$ The equation of normal line at $(1,4)$ is $\begin{aligned} &\\ \quad y-4 &=-1(x-1) \\\\ y &=5-x\\\\ \end{aligned}$ $\text{ (b) } \quad$ At the point of intersection of normal and curve, $\begin{aligned} &\\ \qquad x^{3}+x^{2}-4 x+6&=5-x \\\\ x^{3}+x^{2}-3 x+1&=0\\\\ \end{aligned}$ $\begin{aligned} \qquad \text{ Let } f(x)&=x^{3}+x^{2}-3 x+1\\\\ f(1) &=1+1-3+1 \\\\ &=0\\\\ \end{aligned}$ $\therefore(x-1)$ is a factor of $f(x)$. $\begin{aligned} &\\ \qquad &\text{ Let } f(x)=(x-1)\left(x^{2}+k x-1\right)\\\\ &\therefore\ x^{3}+x^{2}-3 x+1=(x-1)\left(x^{2}+k x-1\right)\\\\ \end{aligned}$ $\begin{aligned} \qquad \text{ When } x =2,\quad &\\\\ 8+4-6+1 &=(2-1)(4+2 k-1)\\\\ 7&=3+2 k \\\\ k&=2\\\\ \end{aligned}$ $\begin{aligned} \qquad \therefore\ (x-1)\left(x^{2}+2 x-1\right)&=0 \\\\ (x-1)\left(x^{2}+2 x+1-2\right)&=0 \\\\ (x-1)\left((x+1)^{2}-2\right)&=0 \\\\ \therefore x-1=0 \text { or }(x+1)^{2}-2&=0 \\\\ \therefore x=1 \text { or } x=-1 \pm &\sqrt{2}\\\\ \end{aligned}$ $\quad$ The normal cuts the curve again where $x=-1-\sqrt{2}$ and $x=-1+\sqrt{2}$




10. (a) The first three terms of an arithmetic progression are $x, 5x-4$ and $8x+2$. Find $x$ and the common difference. [4]
    (b) The first three terms of a geometric progression are $y, 5y-4$ and $8y+2$.
    (i)) Find the two possible values of $y$. [4]
    (ii)) For each of these values of $y$, find the corresponding value of the common ratio. [2]






$\text{ (a) }\quad x, 5 x-4$ and $8 x+2$ ane in an A.P. $\begin{aligned} &\\ \quad\therefore \quad 5 x-4-x &=8 x+2-5 x+4 \\\\ 4 x-4 &=3 x+6 \\\\ \therefore \quad x &=10\\\\ \end{aligned}$ $\qquad$Let the common difference be $d$. $\begin{aligned} &\\ \quad\therefore\quad d &=5 x-4-x \\\\ &=4 x-4 \\\\ &=4(x-1) \\\\ &=4(10-1) \\\\ &=36\\\\ \end{aligned}$ $\text{ (b) }\quad y, 5 y-4$ and $8 y+2$ are in G.P. $\begin{aligned} &\\ \text{ (i) }\hspace{1.5cm} \therefore \frac{5 y-4}{y} &=\frac{8 y+2}{5 y-4} \\\\ \quad 25 y^{2}-40 y+16 &=8 y^{2}+2 y \\\\ 17 y^{2}-42 y+16 &=0 \\\\ (17 y-8)(y-2) &=0 \\\\ \therefore y=\frac{8}{17} \text { or } y &=2\\\\ \end{aligned}$ $\text{ (ii) }\quad$ Let the common ratio be $r$, $\begin{aligned} &\\ \qquad\text{ When } y&=\frac{8}{17},\\\\ r &=\frac{5 y-4}{y} \\\\ &=\frac{5\left(\frac{8}{17}\right)-4}{\frac{8}{17}} \\\\ &=-\frac{7}{2}\\\\ \text{ When } y&=2,\\\\ r &=\frac{5 y-4}{y} \\\\ &=\frac{10-4}{2} \\\\ &=3 \end{aligned}$

စာဖတ်သူ၏ အမြင်ကို လေးစားစွာစောင့်မျှော်လျက်!