May-June-21-CIE-9709-21 : AS and A Level - Problems and Solutions

2021 (May-June) CIE (9709-Pure Mathematics 1), Paper 1/21 ၏ Question နှင့် Solution များ ဖြစ်ပါသည်။ Question Paper ကို ဒီနေရာမှာ Download ယူနိုင်ပါသည်။

1. Solve the inequality $|3x − 7| < |4x + 5|$. [4]






$\begin{aligned} &|3 x-7|< |4 x+5|\\\\ &\sqrt{(3 x-7)^{2}}< \sqrt{(4 x+5)^{2}}\\\\ &(3 x-7)^{2}< (4 x+5)^{2}\\\\ &(3 x-7)^{2}-(4 x+5)^{2}< 0\\\\ &(3 x-7-4 x-5)(3 x-7+4 x+5)< 0\\\\ &(-x-12)(7 x-2)< 0\\\\ &-(x+12)(7 x-2)< 0\\\\ &(x+12)(7 x-2)>0\\\\ &x< -12 \text { or } x>\frac{2}{7} \end{aligned}$




2. By first expanding $\sin(\theta + 30^{\circ})$, solve the equation $\sin(\theta + 30^{\circ})\operatorname{cosec}\theta = 2$ for $0^{\circ} < \theta < 360^{\circ}$. [6]






$\begin{aligned} &\sin \left(\theta+30^{\circ}\right)=\sin \theta \cos 30^{\circ}+\cos \theta \sin 30^{\circ}\\\\ &\sin (\theta+30^{\circ}) \operatorname{cosec} \theta=2\\\\ &\left(\sin \theta \cos 30^{\circ}+\cos \theta \sin 30^{\circ}\right) \frac{1}{\sin \theta}=2\\\\ &\cos 30^{\circ}+\cot \theta \sin 30^{\circ}=2\\\\ &\frac{\sqrt{3}}{2}+\frac{1}{2} \cot \theta=2\\\\ &\cot \theta=4-\sqrt{3}\\\\ &\tan \theta=\frac{1}{4-\sqrt{3}}\\\\ \therefore\ &\theta=23.8^{\circ}\ \text{ or }\ \theta=203.8^{\circ} \end{aligned}$




3. (a) Show that $(\sec x+\cos x)^{2}$ can be expressed as $\sec ^{2} x+a+b \cos 2 x$, where $a$ and $b$ are constants to be determined. [2]
   (b) Hence find the exact value of $\displaystyle\int_{0}^{\frac{1}{4} \pi}(\sec x+\cos x)^{2} dx$. [4]






$\begin{aligned} \text{ (a) } \quad & (\sec x+\cos x)^{2} \\\\ =&\ \sec ^{2} x+2 \sec x \cos x+\cos ^{2} x \\\\ =&\ \sec ^{2} x+\frac{2}{\cos x} \cdot \cos x+\cos ^{2} x \\\\ =&\ \sec ^{2} x+2+\cos ^{2} x \\\\ =&\ \frac{1}{2}\left(2 \sec ^{2} x+4+2 \cos ^{2} x-1+1\right) \\\\ =&\ \frac{1}{2}\left(2 \sec ^{2} x+5+\cos 2 x\right) \\\\ =&\ \sec ^{2} x+\frac{5}{2}+\frac{1}{2} \cos 2 x\\\\ \text{ (b) } \quad & \int_{0}^{\frac{\pi}{4}}(\sec x+\cos x)^{2} d x \\\\ =&\ \int_{0}^{\frac{\pi}{4}}\left(\sec ^{2} x+\frac{5}{2}+\frac{1}{2} \cos x x\right) d x \\\\ =&\ {\left[\tan x+\frac{5}{2} x+\frac{1}{4} \sin 2 x\right]_{0}^{\frac{\pi}{4}} } \\\\ =&\ 1+\frac{5 \pi}{8}+\frac{1}{4} \\\\ =&\ \frac{5}{8}(2+5 \pi) \end{aligned}$




4. A curve has parametric equations $$x=\ln (2 t+6)-\ln t, \quad y=t \ln t .$$
   (a) Find the value of $t$ at the point $P$ on the curve for which $x=\ln 4$. [3]
   
   (b) Find the exact gradient of the curve at $P$. [5]






$\begin{aligned} \text{ (a) } \quad\quad x &=\ln (2 t+6)-\ln t \\\\ y &=t \ln t \\\\ \text { When } x &=\ln 4 \\\\ \ln 4 &=\ln (2 t+6)-\ln t \\\\ \ln 4 &=\ln \left(\frac{2 t+6}{t}\right) \\\\ \therefore \frac{2 t+6}{t} &=4 \\\\ \therefore 2 t+6 &=4 t \\\\ 2 t &=6 \\\\ t &=3\\\\ \frac{d x}{d t} &=\frac{2}{2 t+6}-\frac{1}{t} \\\\ &=-\frac{6}{t(2 t+6)} \\\\ \frac{d y}{d t} &=1+\ln t \\\\ \frac{d y}{d x} &=\frac{d y / d t}{d x / d t} \\\\ &=-\frac{t(2 t+6)(1+\ln t)}{6}\\\\ \end{aligned}$ $\text{ (b) } \ $ At the point $P$, $\begin{aligned} &\\ \qquad\frac{d y}{d x} &=-\frac{3(12)(1+\ln 3)}{6} \\\\ &=-6(1+\ln 3)\\\\ \end{aligned}$ $\quad\therefore\ $ The gradient of the curve at the point $P$ is $-6(1+\ln 3).$





5. The diagram shows the curve with equation $y=\dfrac{3 x+2}{\ln x}$. The curve has a minimum point $M$.
   (a) Find an expression for $\dfrac{dy}{dx}$ and show that the $x$-coordinate of $M$ satisfies the equation $x=\dfrac{3 x+2}{3 \ln x}$. [3]
   
   (b) Use the equation in part (a) to show by calculation that the $x$-coordinate of $M$ lies between $3$ and $4$. [2]
   
   (c) Use an iterative formula, based on the equation in part (a), to find the $x$-coordinate of $M$ correct to $5$ significant figures. Give the result of each iteration to $7$ significant figures. [3]






$\begin{aligned} \text{ (a) } \quad & y=\dfrac{3 x+2}{\ln x} \\\\ & \dfrac{d y}{d x}=\dfrac{3 \ln x-\dfrac{3 x+2}{x}}{\ln ^{2} x} \\\\ & \dfrac{d y}{d x}=0, \text { Shen } \\\\ & \dfrac{3 \ln x-\dfrac{3 x+2}{x}}{\ln ^{2} x}=0 \\\\ \therefore\ & 3 \ln x-\dfrac{3 x+2}{x}=0 \quad \left(\because \ln ^{2} x \neq 0\right) \\\\ \therefore\ & 3 \ln x=\dfrac{3 x+2}{x}\\\\ \therefore\ x &=\dfrac{3 x+2}{3 \ln x} \\\\ \end{aligned}$ $\begin{aligned} \text{ (b) } \quad x-\dfrac{3 x+2}{3 \ln x} &=0 \quad (\text {by part (a)}) \\\\ \text { Let } f(x) &=x-\dfrac{3 x+2}{3 \ln x} \\\\ f(3) &=3-\dfrac{11}{3 \ln 3} \\\\ &=-0.34 \\\\ f(4) &=4-\dfrac{14}{3 \ln 4} \\\\ &=0.63\\\\ \end{aligned}$ Since $f(3)$ and $f(4)$ has opposite signs, there is one root for $f(x)=0$ between $x=3$ and $x=4.\\\\ $ $\therefore$ The $x$-coordinates of $m$ lies between $x=3$ and $x=4.\\\\ $ As we have $x=\dfrac{3 x+2}{3 \ln x},\\\\ $ $\text{let } x_{n+1}=\dfrac{3 x_{n}+2}{3 \ln x_{n}}.\\\\ $ $\begin{array}{|c|c|c|} \hline n & x_{n} & x_{n+1}=\dfrac{3 x+2}{3 \ln x} \\ \hline 1 & 3.5 & 3.325982 \\ \hline 2 & 3.325982 & 3.322321 \\ \hline 3 & 3.322321 & 3.322319 \\ \hline 4 & 3.322319 & 3.322319 \\ \hline \end{array}$ $\therefore\ $ The $x$ coordinate of $M=3.3223$ (correct to 5 sigmiticant figures)




6. (a) Use the trapezium rule with three intervals to find an approximation to $\displaystyle\int_{1}^{4} \frac{6}{1+\sqrt{x}}dx$. Give your answer correct to $5$ significant figures. [3]
   (b) Find the exact value of $\displaystyle\int_{1}^{4} 2 e^{\frac{1}{2} x-2}dx$. [3]
   
   (c) The diagram shows the curves $y=\dfrac{6}{1+\sqrt{x}}$ and $y=2 e^{\frac{1}{2} x-2}$ which meet at a point with $x$-coordinate $4$ . The shaded region is bounded by the two curves and the line $x=1$.
   
   Use your answers to parts (a) and (b) to find an approximation to the area of the shaded region. Give your answer correct to $3$ significant figures. [2]
   
   (d) State, with a reason, whether your answer to part (c) is an over-estimate or under-estimate of the exact area of the shaded region. [1]






Let $y=\dfrac{6}{1+\sqrt{x}}\\\\ $ For $\displaystyle\int_{1}^{4} y d x$, $\begin{aligned} &\\ a &=1, b=4, n=3 \text { (given) } \\\\ \therefore \quad h &=\frac{b-a}{n} \\\\ &=\frac{4-1}{3} \\\\ &=1\\\\ \end{aligned}$ $\begin{array}{|c||c|c|c|c|} \hline \quad x\quad & \quad 1\quad & \quad 2\quad & \quad 3\quad & \quad 4\quad \\ \hline \quad y\quad & \quad 3 \quad & \quad 2.4853\quad & \quad 2.1962\quad & \quad 2\quad \\ \hline \end{array}$ $\begin{aligned} & \displaystyle\int_{1}^{4} \frac{6}{1+\sqrt{x}} d x \\\\ =&\ \frac{1}{2}[(3+2)+2(2.4853+2.1962)] \\\\ =&\ 7.1815 \end{aligned}$ $\begin{aligned} \text{(b)}\quad & \int_{1}^{4} 2 e^{\frac{1}{2} x-2} d x \\\\ =&\ 2 \int_{1}^{4} e^{\frac{1}{2} x-2} 2 \cdot d\left(\frac{1}{2} x-2\right) \\\\ =&\ 4 \int_{1}^{4} e^{\frac{1}{2} x-2} d\left(\frac{1}{2} x+2\right) \\\\ =&\ 4\left[e^{\frac{1}{2} x-2}\right]^{4} \\\\ =&\ 4\left[e^{0}-e^{-3 / 2}\right] \\\\ =&\ 4\left[1-e^{-3 / 2}\right]\\\\ \end{aligned}$ $\text{(c)}\quad$ Let $y_{1}=\frac{6}{1+\sqrt{x}}$ and $y_{2}=2 e^{\frac{1}{2} x-2}\\\\ $. For a given interval, Let the area under $y_{1}$ be $A_{1}$ and that under $y_{2}$ be $A_{2}\\\\ $. By part (a), $A_{1}=7.1815\\\\ $ By part (b), $A_{2}=4\left[1-e^{-3 / 2}\right]\\\\ $ The approximate area of shaded region $\begin{aligned} &\\ &=A_{1}-A_{2} \\\\ &=7.1815-4\left[1-e^{-3 / 2}\right] \\\\ &=4.07\\\\ \end{aligned}$ $\text{(d)}\quad$ It can be seen from the diagram that the approximate area of the shade region is an over-estimate since the top edges of the strips of $A_1$ all lie above the curve.




7. The polynomial $\mathrm{p}(x)$ is defined by $p(x)=a x^{3}-11 x^{2}-19 x-a,$ where $a$ is a constant. It is given that $(x-3)$ is a factor of $p(x)$.
   
   (a) Find the value of $a$. [2]
   
   (b) When $a$ has this value, factorise $p(x)$ completely. [3]
   
   (c) Hence find the exact values of $y$ that satisfy the equation $p\left(e^{y}+e^{-y}\right)=0$. [4]






$\begin{aligned} & p(x)=a x^{3}-11 x^{2}-19 x-a \\\\ &(x-3) \text { is a factor of } p(x) \\\\ \therefore\ & p(3)=0 \\\\ & 27 a-99-57-a=0 \\\\ & 26 a=156 \\\\ & a=6 \\\\ \therefore\quad & p(x)=6 x^{3}-11 x^{2}-19 x-6\\\\ &\text{Let } 6 x^{3}-11 x^{2}-19 x-6=(x-3)\left(6 x^{2}+k x+2\right)\\\\ \therefore \quad & k-18 =-11 \\\\ \therefore\quad &k =7 \\\\ \therefore \quad &p(x) =(x-3)\left(6 x^{2}+7 x+2\right) \\\\ \therefore \quad & p(x)=(x-3)(3 x+2)(2 x+1)\\\\ &p\left(e^{y}+e^{-y}\right)=0 \\\\ &\left(e^{y}+e^{-y}-3\right)\left(3\left(e^{y}+e^{-y}\right)+2\right)\left(2\left(e^{y}+e^{-y}\right)+1\right)=0 \\\\ &e^{y}+e^{-y}=3 \text { or } e^{y}+e^{-y}=-\frac{2}{3} \text { or } e^{y}+e^{-y}=-\frac{1}{2} \\\\ &\text {Since } e^{y}+e^{-y}>0, \\\\ &e^{y}+e^{-y}=3 \text { is only a unique solution. } \\\\ &e^{2 y}+1=3 e^{y} \\\\ &\left(e^{y}\right)^{2}-3 e^{y}+1=0\\\\ &\left(e^{y}\right)^{2}-3 e^{y}+\frac{9}{4}=\frac{5}{4} \\\\ &\left(e^{y}-\frac{3}{2}\right)^{2}=\frac{5}{4} \\\\ \therefore \quad & e^{y}=\frac{3 \pm \sqrt{5}}{2} \\\\ \therefore \quad & y=\ln \frac{3+\sqrt{5}}{2} \text { or } y=\ln \frac{3-\sqrt{5}}{2} \end{aligned}$

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