CIE 0606/2021/m/12 Question

1. Find the exact solutions of the equation $3(\ln 5x)^2+2\ln 5x-1=0.$ [4] 

2. The diagram shows the graph of $y=a\sin bx+c$ where $x$ is in radians and $-2\pi\le x\le 2\pi$, where $a, b$ and $c$ are positive constants. Find the value of each of $a, b$ and $c.$ [3]

3. The line $AB$ is such that the points $A$ and $B$ have coordinates $(-4, 6)$ and $(2, 14)$ respectively. 

(a) The point $C$, with coordinates $(7, a)$ lies on the perpendicular bisector of $AB$. Find the value of $a.$ [4]

(b) Given that the point $D$ also lies on the perpendicular bisector of $AB,$ find the coordinates of $D$ such that the line $AB$ bisects the line $CD.$ [2]

4 (a) Show that $2x^2+5x-3$ can be written in the form $a(x+b)^2+c$,  where $a, b$ and $c$ are constants. [3]

(b) Hence write down the coordinates of the stationary point on the curve with equation $y=2x^2+5x-3.$ [2]

(c) On the axes below, sketch the graph of $y=|2x^2+5x-3|,$ stating the coordinates of the intercepts with the axes.[3]

(d) Write down the value of $k$ for which the equation $|2x^2+5x-3|=k$ has exactly 3 distinct solutions.  [1] 

$\def\cvec#1#2{\left(\begin{array}{c}#1\\#2\end{array}\right)}\def\ABvec#1{\overrightarrow{#1}}$

5. In this question all lengths are in kilometres and time is in hours. Boat A sails, with constant velocity, from a point $O$ with position vector $\cvec{0}{0}$. After 3 hours $A$ is at the point with position vector $\cvec{-12}{9.}$

(a) Find the position vector, $\ABvec{OP},$ of $A$ at time $t$. [1]

At the same time as $A$ sails from $O$, boat $B$ sails from a point with position vector $\cvec{12}{6},$ with constant velocity $\cvec{-5}{8}$.

(b) Find the position vector, $\ABvec{OQ},$ of $B$ at time $t.$ [1] 

(c) Show that at time $t, |\ABvec{PQ}|^2=26t^2+36t+180.$  [3] 

(d) Hence show that $A$ and $B$ do not collide. [2]

6(a) A geometric progression has first term 10 and sum to infinity 6. 

(i) Find the common ratio of this progression.[2] 

(ii) Hence find the sum of the first 7 terms, giving your answer correct to 2 decimal places. [2] ©

(b) The first three terms of an arithmetic progression are $\log_x3 , \log_x (3^2),\log_x(3^3).$

(i) Find the common difference of this progression.[1] 

(ii) Find, in terms of $n$ and $\log_x3,$ the sum to $n$ terms of this progression. Simplify your answer. [2]

(iii) Given that the sum to $n$ terms is $3081 \log_x 3,$ find the value of $n.$ [2] 

(iv) Hence, given that the sum to $n$ terms is also equal to 1027, find the value of $x.$ [2]

7. DO NOT USE A CALCULATOR IN THIS QUESTION 

In this question all lengths are in centimetres.

The diagram shows a trapezium $ABCDE$ such that $AB$ is parallel to $EC$ and $ABCD$ is a rectangle. It is given that $BC=\sqrt{17}+1, ED=\sqrt{17}-1$ and $DC=\sqrt{17}+4$.

(a) Find the perimeter of the trapezium, giving your answer in the form $a+b\sqrt{17},$ where $a$ and $b$ are  integers. [3]

(b) Find the area of the trapezium, giving your answer in the form $c+d\sqrt{17},$ where $c$ and $d$ are  integers.

(c) Find $\tan AED,$ giving your answer in the form $\dfrac{e+f\sqrt{17}}{8},$ where $e$ and $f$ are integers. [2] 

(d) Hence show that $\sec^2 AED = \dfrac{81+9\sqrt{17}}{32}.$  [2]

8 (a) (i) Show that $\sin x\tan x+\cos x=\sec x$. [3] 

(ii) Hence solve the equation $\sin\dfrac{\theta}{2}\tan\dfrac{\theta}{2}+\cos\dfrac{\theta}{2}=4,$ for $0\le \theta\le4\pi$,  where $\theta$ is in radians. [4] 

(b) Solve the equation $\cot(y+38^{\circ})=\sqrt{3}$ for $0\le y\le 360^{\circ}.$ [3]

9. The polynomial $\mathrm{p}(x)=2 x^{3}-3 x^{2}-x+1$ has a factor $2 x-1$.

(a) Find $\mathrm{p}(x)$ in the form $(2 x-1) \mathrm{q}(x)$, where $\mathrm{q}(x)$ is a quadratic factor.

The diagram shows the graph of $y=\dfrac{1}{x}$ for $x>0$, and the graph of $y=-2 x^{2}+3 x+1$. The curves intersect at the points $A$ and $B$.

(b) Using your answer to part (a), find the exact $x$ -coordinate of $A$ and of $B$.

(c) Find the exact area of the shaded region.$[6]$

10 A curve has equation $y=\dfrac{\left(2 x^{2}+10\right)^{\dfrac{3}{2}}}{x-1}$ for $x>1$.

(a) Show that $\dfrac{\mathrm{d} y}{\mathrm{~d} x}$ can be written in the form $\dfrac{\left(2 x^{2}+10\right)^{\dfrac{1}{2}}}{(x-1)^{2}}\left(A x^{2}+B x+C\right)$, where $A, B$ and $C$ are integers.

(b) Show that, for $x>1$, the curve has exactly one stationary point. Find the value of $x$ at this stationary point. $\quad[4]$

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*********math Answer************* Answers 1 $x=e^{\frac 13}/5$ or $x=1/(5e)$  2 $a=3,b=.5,c=4$ 3 a) $a=4$ b) $(-9,16)$ 4 a) $2\left(x+\frac 54\right)^2-\frac{49}{8}$ b) $\left(-\frac{5}{4},-\frac{49}{8}\right)$ c) fig d) 49/8 5) a) $\cvec{-4}{3}t$ b) $\cvec{12-5t}{6+8t}$ c)d)  6) $r=-2/3,S_7=6.35$ bi) $\log_x3,$ bii) $S_n=\frac{n}{2}(n+1)\log_x3$ biii) $n=78$ biv) $x=27$ 7 a)$4\sqrt{17}+14$ b) $29+5\sqrt{17}$ c) $\dfrac{9+\sqrt{17}}{8}$ 8aii) $2.64,9.93$ b) $172,352$ 9a) $(2x-1)(x^2-x-1)$ b) $x=.5,x=\frac{1+\sqrt{5}}{2}$ c) $\ln(1+\sqrt{5})+19/24$ 10a) $(4x^2-6x-10)$ b) $x=2.5$ **********end math solution********************