Question 1
Topic – ALV: 2.2
Solve the equation $\ln(x+5) = 5 + \ln x$. Give your answer correct to 3 decimal places.
▶️Answer/Explanation
Solution :-
Use exponentials or laws for the logarithm of a product, quotient or power
Correctly remove logarithms
Obtain a correct equation in $x$
Obtain 0.034
Question 2
Topic – ALV: 2.1
Find the quotient and remainder when $2x^4 – 27$ is divided by $x^2 + x + 3$.
▶️Answer/Explanation
Solution :-
2. Divide to obtain quotient $2x^{2} \pm 2x + k$ ($k \neq 0$)
Obtain [quotient] $2x^{2} – 2x – 4$
Obtain [remainder] $10x – 15$
Alternative Method for Question 2
Expand $(x^{2} + x + 3)(Ax^{2} + Bx + C) + (Dx + E)$ and reach $A = 2$, $B = \pm 2$, $C = k$
Obtain [quotient] $2x^{2} – 2x – 4$
Obtain [remainder] $10x – 15$
Question 3
Topic – ALV: 3.9
On a sketch of an Argand diagram, shade the region whose points represent complex numbers $z$ satisfying the inequalities $|z-3-i| \leq 3$ and $|z| \geq |z-4i|$.
▶️Answer/Explanation
Solution :-
Show a circle with centre $3+i$
Show a circle with radius 3 and centre not at the origin
Show the line $y=2$
Shade the correct region
Question 4
Topic – ALV: 3.4
4. The parametric equations of a curve are
\[x=\frac{\cos \theta}{2-\sin \theta}\]
\[y=\theta+2\cos \theta.\]
Show that
\[\frac{dy}{dx}=(2-\sin \theta)^{2}.\]
▶️Answer/Explanation
Solution :-
State $\frac{dy}{d\theta}=1-2\sin\theta$
Use correct quotient rule, or product rule if rewrite x as $\cos\theta(2-\sin\theta)^{-1}$
Obtain $\frac{dx}{d\theta}=\frac{-(2-\sin\theta)\sin\theta+\cos^{2}\theta}{(2-sin\theta)^{2}}$ o.e.
Use $\frac{dy}{dx}=\frac{dy}{d\theta}\div\frac{dx}{d\theta}$
Obtain $\frac{dy}{dx}=(2-\sin\theta)^{2}$.
Question 5
Topic – ALV: 1.7
The diagram shows the part of the curve \(y=x^{2}\cos 3x\) for \(0\le x\le\frac{1}{6}\pi,\) and its maximum point M, where
x = a.
(a) Show that a satisfies the equation
\[a=\frac{1}{3}\tan^{-1}\left(\frac{2}{3a}\right)\]
(b) Use an iterative formula based on the equation in (a) to determine a correct to 2 decimal places.
Give the result of each iteration to 4 decimal places.
▶️Answer/Explanation
Solution :-
(a) Use correct product rule
Obtain correct derivative in any form
Equate derivative to zero and obtain $a = \frac{1}{3} \tan^{-1} \left( \frac{2}{3a} \right)$
(b) Use the iterative process $a_{n+1} = \frac{1}{3} \tan^{-1} \left( \frac{2}{3a_{n}} \right)$ correctly at least twice during successive iterations in the numerous iterations
Obtain final answer 0.36
Show sufficient iterations to 4 or more d.p. to justify 0.36 to 2 d.p. or show there is a sign change in the interval (0.355, 0.365)
0.5 0.4 0.3 0.2 0.1 π/6 π/12
0.3091 0.3435 0.3826 0.4264 0.4740 0.3017 0.3989
0.3789 0.3650 0.3499 0.3339 0.3176 0.3820 0.3439
0.3513 0.3566 0.3625 0.3688 0.3754 0.3502 0.3649
0.3619 0.3599 0.3576 0.3552 0.3526 0.3624 0.3567
0.3578 0.3604 0.3614 0.3576
0.3580
Question 6
Topic – ALV: 1.5
(a) Express \(3 \cos x + 2 \cos(x – 60^{\circ})\) in the form \(R \cos(x – \alpha)\), where \(R > 0\) and \(0^{\circ} < \alpha < 90^{\circ}\).
State the exact value of R and give α correct to 2 decimal places.
(b) Hence solve the equation
\[3 \cos 2\theta + 2 \cos(2\theta – 60^{\circ}) = 2.5\]
for \(0^{\circ} < \theta < 180^{\circ}\).
▶️Answer/Explanation
Solution :-
(a) Expand $\cos(x-60^{\circ})$ correctly and evaluate $3 \cos x + 2 \cos(x-60^{\circ})$ to obtain $4 \cos x + \sqrt{3} \sin x$ or unsimplified coefficients
State $R=\sqrt{19}$ [$R \cos \alpha=4$ $R \sin \alpha=\sqrt{3}$]
Use correct trig formulae for their expansion to find $\alpha$
e.g. $\alpha=\tan^{-1}\frac{\sqrt{3}}{4}$ or $\cos^{-1}\frac{4}{\sqrt{19}}$ or $\sin^{-1}\frac{\sqrt{3}}{\sqrt{19}}$
Obtain $\alpha=23.41^{\circ}$
(b) $\cos^{-1}\left(\frac{2.5}{R}\right)$
Use a correct method to find a value of $2\theta (not x)$ in the interval.
Allow sign error in moving $\alpha$ to right side
Obtain one correct answer e.g. $39.2^{\circ}$
Obtain second correct answer e.g. $164.2^{\circ}$ and no others in the interval
Question 7
Topic – ALV: 1.8
(a) Use the substitution \(u = \cos x\) to show that
\[\int_{0}^{\pi} \sin 2x e^{2 \cos x} dx = \int_{-1}^{1} 2ue^{2u} du.\]
(b) Hence find the exact value of
\[\int_{0}^{\pi} \sin 2x e^{2 \cos x} dx.\]
▶️Answer/Explanation
Solution :-
(a) \[\frac{du}{dx}=-\sin x\]
Use double angle formula and substitute for x and dx throughout the integral
Obtain \[\pm\int2ue^{2u}du\]
Justify new limits and obtain \[\int_{-1}^{1}2ue^{2u}du\] from correct working
(b) Commence integration and reach $au e^{2u} + b\int e^{2u} du$, where $ab \neq 0$, $b < 0$
Complete integration and obtain $ue^{2u} – \frac{1}{2}e^{2u}$
Use correct limits correctly in $c u e^{2u} + d e^{2u}$ having integrated twice
or in $c \cos x e^{2 \cos x} + d e^{2 \cos x}$
Obtain $\frac{1}{2}e^{2} – \frac{3}{2}e^{-2}$
Question 8
Topic – ALV: 3.8
The variables x and y satisfy the differential equation
\[\frac{dy}{dx}=\frac{y^{2}+4}{x(y+4)}\]
for $x > 0$. It is given that $x = 4$ when $y = 2\sqrt{3}$.
Solve the differential equation to obtain the value of x when $y = 2$.
▶️Answer/Explanation
Solution :-
Separate the variables correctly
Obtain $\ln x$
Split the fraction and integrate to obtain $p \ln(y^{2} + 4)$ or $q \tan^{-1}\frac{y}{2}$ correctly
Obtain $\frac{1}{2} \ln(y^{2} + 4)$
Obtain $2 \tan^{-1}\frac{y}{2}$
Use $(4, 2\sqrt{3})$ in an expression containing at least 2 of $a \ln x$, $b \ln(y^{2} + 4)$ and $c \tan^{-1}\frac{y}{2}$ to obtain constant of integration
Correct solution (any form)
e.g. $\frac{1}{2} \ln(y^{2} + 4) + 2 \tan^{-1}\frac{y}{2} = \ln x + \frac{2\pi}{3}$
or $\frac{1}{2} \ln(y^{2} + 4) + 2 \tan^{-1}\frac{y}{2} = \ln x + 2 \tan^{-1}\sqrt{3} + \frac{1}{2} \ln 16 – \ln 4$
Obtain $\sqrt{8}e^{-\frac{1}{6}\pi}$ or 1.68 or more accurate or $2\sqrt{2}e^{-\frac{1}{6}\pi}$ or $\frac{\sqrt{8}}{e^{\frac{1}{6}\pi}}$ or $e^{0.516}$
8. Alternative method for first *M1 A1 A1
\[p \left((y+4) \tan^{-1}\frac{y}{2} – \int \tan^{-1}\frac{y}{2} dy \right)\]
\[(y+4) \frac{1}{2} \tan^{-1}\frac{y}{2} – \frac{y}{2} \tan^{-1}\frac{y}{2} + \frac{1}{2} \int \frac{y}{y^{2}+4} dy\]
Obtain $2 \tan^{-1}\frac{y}{2} + \frac{1}{2} \ln(y^{2} + 4)$
Question 9
Topic – ALV: 3.7
The lines $l$ and $m$ have equations
\[l: \mathbf{r} = \mathbf{a}i + 3\mathbf{j} + b\mathbf{k} + \lambda(\mathbf{c}i – 2\mathbf{j} + 4\mathbf{k}),\]
\[m: \mathbf{r} = \mathbf{i} + 2\mathbf{j} + 3\mathbf{k} + \mu(2\mathbf{i} – 3\mathbf{j} + \mathbf{k}).\]
Relative to the origin $O$, the position vector of the point $P$ is $4\mathbf{i} + 7\mathbf{j} – 2\mathbf{k}$.
(a) Given that $l$ is perpendicular to $m$ and that $P$ lies on $l$, find the values of the constants $a$, $b$, and $c$.
(b) The perpendicular from P meets line $m$ at $Q$. The point $R$ lies on $PQ$ extended, with
$PQ$ : $QR$ = 2 : 3.
Find the position vector of $R$.
▶️Answer/Explanation
Solution :-
(a) Perform scalar product of direction vectors and set result equal to zero
Use P to find the value of $\lambda$
Obtain $c = -5$ or $b = 6$
$a = -6$, $b = 6$ and $c = -5$ all correct
(b) Find $\overrightarrow{PQ}$ (or $\overrightarrow{QP}$) for a general point Q on m
\[=\pm((1+2\mu,2-3\mu,3+\mu)-(a+\lambda c,3-2\lambda,b+4\lambda))\]
Equate the scalar product of $\overrightarrow{PQ}$ (or $\overrightarrow{QP}$) and a direction vector for m to zero and obtain an equation in $\mu$
Solve and obtain $\mu=-1$
Obtain $\overrightarrow{OQ}=-i+5j+2k$ or $\overrightarrow{PQ}=-5i-2j+4k$
Must be labelled correctly
Carry out a method to find the position vector of R
Alternative method for DM1
\[\overrightarrow{OR}=(4,7,-2)+t(-5,-2,4) \ \ \ \overrightarrow{OR}=\overrightarrow{OR}-\overrightarrow{OQ}\]
Solve $|\overrightarrow{QR}|^{2}=\frac{9}{4}|\overrightarrow{PQ}|^{2}$ or $|\overrightarrow{QR}|=\frac{3}{2}|\overrightarrow{PQ}| \ \ t=2.5$
Obtain $-\frac{17}{2}i + 2j + 8k$ from correct working
Question 10
Topic – ALV: 2.1
Let $f(x) = \frac{21 – 8x – 2x^{2}}{(1 + 2x)(3 – x)^{2}}$.
(a) Express $f(x)$ in partial fractions.
(b) Hence obtain the expansion of $f(x)$ in ascending powers of $x$, up to and including the term in $x^{2}$.
▶️Answer/Explanation
Solution :-
(a) State or imply the form
\[ \frac{A}{1+2x} + \frac{B}{3-x} + \frac{C}{(3-x)^{2}} \]
Use a correct method to find a constant
Obtain one of \(A=2, B=2\) and \(C=-3\)
Obtain a second value
Obtain the third value
(b) Use a correct method to obtain the first two terms of one of the unsimplified expansions
\[(1+2x)^{-1}, \left(1-\frac{1}{3}x\right)^{-1}, \left(1-\frac{1}{3}x\right)^{-1}(3-x)^{-1}, (3-x)^{-2}\]
Obtain the correct unsimplified expansions up to the term in $x^{2}$ for each partial fraction
If correct, should be working with
\[\frac{2}{1+2x}+\frac{2}{3-x}-\frac{3}{(3-x)^{2}}\]
or
\[\frac{2}{1+2x}+\frac{-2x+3}{(3-x)^{2}}\]
Obtain final answer
\[\frac{7}{3}-4x+\frac{215}{27}x^{2}\]
(b) Alternative Method for Question 10(b)
For the form
\[\frac{A}{1+2x}+\frac{Dx+E}{(3-x)^{2}}\]
Obtain the correct unsimplified expansions up to the term in $x^{2}$ for each partial fraction
Multiply out fully
Obtain final answer
\[\frac{7}{3}-4x+\frac{215}{27}x^{2}\]
(b) Alternative Method for Question 10(b): Maclaurin’s Series
Correct derivatives for $A(1+2x)^{-1}$, $B(3-x)^{-1}$ and $C(3-x)^{-2}$
\[(-1)(2)A(1+2x)^{-2}, (-1)(-1)B(3-x)^{-2}\]
and \[(-2)(-1)C(3-x)^{-3}\]
One of following $(-2)(-1)(2)(1)A(1+2x)^{-3}$, $(-2)(-1)(-1)(-1)B(3-x)^{-3}$ and
\[(-3)(-1)(-2)(-1)C(3-x)^{-4}\]
All correct
Substitute in
\[f(0)+xf'(0)+\frac{x^{2}}{2}f”(0)\]
Obtain final answer
\[\frac{7}{3}-4x+\frac{215}{27}x^{2}\]
Question 11
Topic – ALV: 3.9
The complex number $z$ is defined by
\[z=\frac{5a-2i}{3+ai},\]
where $a$ is an integer. It is given that $\arg z=-\frac{1}{4}\pi$.
(a) Find the value of $a$ and hence express $z$ in the form $x + iy$, where $x$ and $y$ are real.
(b) Express $z^{3}$ in the form $re^{i\theta}$, where $r > 0$ and $-\pi < \theta < \pi$. Give the simplified exact values of $r$ and $\theta$.
▶️Answer/Explanation
Solution :-
(a) Multiply numerator and denominator by $(3-ai)$
Use $i^{2}=-1$ at least once and separate real and imaginary parts
Obtain
\[\frac{13a-i(5a^{2}+6)}{9+a^{2}} \text{ or } \frac{13a-5a^{2}i-6i}{9+a^{2}}\]
Use $\arg z$ to form equation in $a$
\[-\frac{5a^{2}+6}{13a}=\pm \tan(\pm\frac{\pi}{4}) \text{ or } -\frac{13a}{5a^{2}+6}=\pm \tan(\pm\frac{\pi}{4})\]
or
\[\tan^{-1}\left(-\frac{5a^{2}+6}{13a}\right)=\pm\frac{\pi}{4} \text{ or } \tan^{-1}\left(-\frac{13a}{5a^{2}+6}\right)=\pm\frac{\pi}{4}\]
Obtain $a=2$
Obtain $z=2-2i$ only
(a) Alternative Method 1 for the first four marks
$\arg z = \arg(5a-2i) – \arg(3+ai)$
$= \tan^{-1}\left(-\frac{2}{5a}\right) – \tan^{-1}\left(\frac{a}{3}\right)$
$= \tan^{-1}\left(\frac{-\frac{2}{5a} – \frac{a}{3}}{1 + \left(-\frac{2}{5a}\right)\left(\frac{a}{3}\right)}\right)$
$= \tan^{-1}\left(-\frac{5a^{2}+6}{13a}\right) \text{ or } \tan^{-1}\left(-\frac{13a}{5a^{2}+6}\right)$
$\pm\frac{\pi}{4} = \tan^{-1}\left(-\frac{5a^{2}+6}{13a}\right) \text{ or } \tan^{-1}\left(-\frac{13a}{5a^{2}+6}\right)$
Alternative Method 2 for the first four marks
$(x+iy)(3+ai) = 5a-2i$
$3x-ay=5a$ and $ax+3y=-2$
$x=\pm y$ Find x or y in terms of a, e.g. $x=\frac{2}{3-a}$ or $x=\frac{5a}{3+a}$
Substitute in other equation, for example
$3\left(\frac{2}{3-a}\right)+a\left(\frac{2}{3-a}\right)=5a$
(b) State $\arg(z^{3}) = \frac{3\pi}{4}$ or evaluate from $z=b-bi$ or from $-2b^{3}(1+i)$
Complete method to obtain $r$ from their $z$
\[r=16\sqrt{2}\]