Question 1
Topic – ALV: 2.1
Solve the inequality $|5x – 3| < 2|3x – 7|$.
▶️Answer/Explanation
Solution :-
State or imply non-modular inequality $(5x-3)^{2}<2^{2}(3x-7)^{2}$, or corresponding quadratic equation, or pair of linear equations $(5x-3)= \pm2(3x-7)$
Solve a 3-term quadratic, or solve two linear equations for $x$
Obtain critical values $x=\frac{17}{11}$ and $x=11$
State final answer $x<\frac{17}{11}$ $x>11$
Alternative Method for Question 1
Obtain critical value $x=11$ from a graphical method, or by inspection, or by solving a linear equation or an inequality
Obtain critical value $x=\frac{17}{11}$ similarly
State final answer $x<\frac{17}{11}$ $x>11$
Question 2
Topic – ALV: 2.2
Solve the equation $\ln(2x^2 – 3) = 2 \ln x – \ln 2$, giving your answer in an exact form.
▶️Answer/Explanation
Solution :-
1. Use law of the logarithm of a power, quotient or product
2. Remove logarithms and obtain a correct equation in $x$
3. Obtain final answer $x=\sqrt{2}$ only
Question 3
Topic – ALV: 3.9
(a) On an Argand diagram, sketch the locus of points representing complex numbers $z$ satisfying
$|z + 3 – 2i| = 2$.
(b) Find the least value of $|z|$ for points on this locus, giving your answer in an exact form.
▶️Answer/Explanation
Solution :-
(a)
Show a circle with radius 2
(b) Carry out a correct method for finding the least value of |z|
Obtain answer $\sqrt{13}-2$ or $\sqrt{17-4\sqrt{13}}$
Question 4
Topic – ALV: 1.5
Solve the equation $2 \cos x – \cos \frac{1}{2}x = 1$ for $0 \leq x \leq 2\pi$.
▶️Answer/Explanation
Solution :-
4. Use correct double angle formula to obtain an equation in \(cos(\frac{x}{2})\) only
Obtain a 3 term quadratic in \(cos(\frac{x}{2})\) ,
Obtain \(cos(\frac{x}{2})=-\frac{3}{4}\) and \(cos(\frac{x}{2})=1\)
Solve for the original x
Obtain x = 0 and 4.84 and no others in the interval
Alternative Method for Question 4
Use correct double angle formula to obtain an equation in \(cos~x\) only
Obtain a 3 term quadratic in \(cos~x\)
Obtain \(cos~x=\frac{1}{8}\) and \(cos~x=1\)
Solve for x
Obtain answers x = 0 and 4.84 and no others in the interval
Question 5
Topic – ALV: 3.9
The complex number $2 + yi$ is denoted by $a$, where $y$ is a real number and $y < 0$. It is given that
$$f(a) = a^3 – a^2 – 2a.$$
(a) Find a simplified expression for $f(a)$ in terms of $y$.
(b) Given that $Re(f(a)) = -20$, find arg(a).
▶️Answer/Explanation
Solution :-
(a) Substitute \(2+yi\) in \(a^{3}-a^{2}-2a\) and attempt expansions of \(a^{2}\) and \(a^{3}\)
Use \(i^{2}=-1\)
Obtain final answer \(-5y^{2}+(6y-y^{3})i\)
(b) Equate their \(-5y^{2}\) to -20 and solve for y
Obtain \(y=-2\)
Obtain final answer arg a $= -\frac{\pi}{4}$
Question 6
Topic – ALV: 3.6
The equation $\cot x = 3x$ has one root in the interval $0 < x < \pi$, denoted by $a$.
(a) Show by calculation that $a$ lies between 0.5 and 1.
(b) Show that, if a sequence of positive values given by the iterative formula
$$x_{n+1} = \frac{1}{3} \left(x_n + 4 \tan^{-1}\left(\frac{1}{3x_n}\right)\right)$$
converges, then it converges to $a$.
(c) Use this iterative formula to calculate $a$ correct to 2 decimal places. Give the result of each iteration to 4 decimal places.
▶️Answer/Explanation
Solution :-
(a) Calculate the values of a relevant expression or pair of expressions at x = 0.5 and x = 1
Complete the argument correctly with conclusion about change of sign or change of inequalities and with correct calculated values. Can all be in symbols – an explanation in words is not required.
6(b) State $x = \frac{1}{3} \left( x + 4~\tan^{-1} \frac{1}{3x} \right)$
Rearrange to the given equation $\cot \left( \frac{x}{2} \right) = 3x$
Need intermediate step between $\frac{x}{2} = \tan^{-1} \frac{1}{3x}$ and $\cot \left( \frac{x}{2} \right) = 3x$
(c) Use the iterative process correctly at least once
Obtain final answer 0.79
Show sufficient iterations to at least 4 d.p. to justify 0.79 to 2 d.p. or show there is a sign change in the interval (0.785, 0.795)
Question 7
Topic – ALV: 3.4
The equation of a curve is $3x^2 + 4xy + 3y^2 = 5$.
(a) Show that $\frac{dy}{dx} = -\frac{3x + 2y}{2x + 3y}$.
(b) Hence find the exact coordinates of the two points on the curve at which the tangent is parallel to $y + 2x = 0$.
▶️Answer/Explanation
Solution :-
(a) State or imply \(6y\frac{dy}{dx}\) as the derivative of \(3y^{2}\)
State or imply \(4x\frac{dy}{dx}+4y\) as the derivative of \(4xy\)
Equate derivative of LHS to zero and solve for \(\frac{dy}{dx}\)
Obtain \(\frac{dy}{dx}=-\frac{3x+2y}{2x+3y}\)
(b) Equate \(\frac{dy}{dx}\) to -2 and solve for x in terms of y or for y in terms of x
Obtain \(x=-4y\) or \(y=-\frac{x}{4}\)
Substitute their \(x=-4y\) or their \(y=-\frac{x}{4}\) in curve equation
Obtain \(y=\pm\frac{1}{\sqrt{7}}\) or \(x=\pm\frac{4}{\sqrt{7}}\)
Obtain both pairs of values
Question 8
Topic – ALV: 3.8
(a) The variables x and y satisfy the differential equation
\[\frac{dy}{dx}=\frac{4+9y^{2}}{e^{2x+1}}\]
It is given that \(y=0\) when \(x=1\).
Solve the differential equation, obtaining an expression for y in terms of x.
(b) State what happens to the value of y as x tends to infinity. Give your answer in an exact form.
▶️Answer/Explanation
Solution :-
(a) Separate variables correctly
Obtain term $\frac{1}{2}e^{-2x-1}$
Obtain term of the form $a~\tan^{-1}(\frac{3y}{2})$
Obtain term $\frac{1}{6}\tan^{-1}(\frac{3y}{2})$
Use $x=1, y=0$ to evaluate a constant or as limits in a solution containing
or derived from terms of the form $atan^{-1}(by)$ and $ce^{\pm(2x+1)}$
Obtain correct answer in any form
Obtain final answer $y=\frac{2}{3}\tan(3e^{-3}-3e^{-2x-1})$
(b) State that y approaches $\frac{2}{3}\tan(3e^{-3})$
Question 9
Topic – ALV: 2.5
Let $f(x) = \frac{2x^2 + 17x – 17}{(1 + 2x)(2 – x)^2}.$
(a) Express f(x) in partial fractions.
(b) Hence show that $\int_{0}^{1} f(x) dx = \frac{5}{2} – \ln 72$.
▶️Answer/Explanation
Solution :-
9(a) State or imply the form
\[ \frac{A}{1+2x} + \frac{B}{2-x} + \frac{C}{(2-x)^{2}} \]
Use a correct method for finding a coefficient
Obtain one of A = -4, B = -3 and C = 5
Obtain a second value
Obtain the third value
(b) Integrate and obtain terms $-2\ln(1+2x)+3\ln(2-x)+\frac{5}{2-x}$
Substitute limits correctly in an integral with two terms (obtained correctly)
of the form $a\ln(1+2x)+b\ln(2-x)+\frac{c}{2-x}$, where $abc \neq 0$
Obtain answer $\frac{5}{2}-\ln~72$ after full and correct working
Question 10
Topic – ALV: 1.8
The diagram shows the curve $y = (x + 5)\sqrt{3 – 2x}$ and its maximum point M.
(a) Find the exact coordinates of M.
(b) Using the substitution $u = 3 – 2x$, find by integration the area of the shaded region bounded by the curve and the x-axis. Give your answer in the form $a\sqrt{13}$, where $a$ is a rational number.
▶️Answer/Explanation
Solution :-
(a) Use the product rule correctly to obtain \(p(x+5)(3-2x)^{\frac{1}{2}} + q(3-2x)^{\frac{1}{2}}\)
Obtain correct derivative in any form
Equate derivative to zero and obtain a linear equation
Obtain a correct linear equation.
Obtain answer
\[ \left( -\frac{2}{3}, \frac{13\sqrt{39}}{9} \right) \]
Alternative Method for Question 10(a)
Obtain \(y^{2}\) and differentiate
Obtain correct derivative in any form
Equate derivative to zero and solve for x
Obtain \(-\frac{2}{3}\)
Obtain answer
\[ \left( -\frac{2}{3}, \frac{13\sqrt{39}}{9} \right) \]
only
(b) Use the given substitution and reach
\[ \int \left(\frac{13}{2}-\frac{u}{2}\right)^{\frac{1}{2}} du \]
Obtain correct integral
\[ -\frac{1}{2}\int \left(\frac{13}{2}-\frac{u}{2}\right)^{\frac{1}{2}} du \]
x = -5 and $\frac{3}{2}$
Use correct limits the right way round in an integral of the form
\[ \int_{a}^{b} \left(\frac{26}{3}-u-\frac{2}{5}u^{2}\right)^{\frac{1}{2}} du \]
Obtain answer $\frac{169}{15}\sqrt{13}$ or a = $\frac{169}{15}$
Question 11
Topic – ALV: 3.7
11. The points A and B have position vectors $\textbf{i} + 2\textbf{j} – 2\textbf{k}$ and $2\textbf{i} – \textbf{j} + \textbf{k}$ respectively. The line $l$ has equation
$\textbf{r} = \textbf{i} – \textbf{j} + 3\textbf{k} + \mu(2\textbf{i} – 3\textbf{j} + 4\textbf{k})$.
(a) Show that $l$ does not intersect the line passing through A and B.
(b) Find the position vector of the foot of the perpendicular from A to $l$.
▶️Answer/Explanation
Solution :-
(a) Carry out correct method for finding a vector equation for AB
Obtain \(\mathbf{r}= \mathbf{i}+2\mathbf{j}-2\mathbf{k}+\lambda(\mathbf{i}-3\mathbf{j}+3\mathbf{k})\)
Equate two pairs of components of general points on their AB and l
and evaluate λ or μ
Obtain correct answer for λ or μ, e.g. \(\lambda=-1, \mu=-2\)
Verify that all three equations are not satisfied and the lines fail to intersect
(\(\neq\) is sufficient justification e.g. 0≠-3).
11(b) Find $\overline{AP}$ for a general point P on l, e.g. \(-3\mathbf{j}+5\mathbf{k}+\mu(2\mathbf{i}-3\mathbf{j}+4\mathbf{k})\)
Calculate scalar product of their \overline{AP} and a direction vector for l and equate
the result to zero
Obtain \(\mu=-1\)
Obtain answer \(-\mathbf{i}+2\mathbf{j}-\mathbf{k}\)
Alternative Method for Question 11(b)
Find $\overline{AP}$ for a general point P on l, e.g. \(-3\mathbf{j}+5\mathbf{k}+\mu(2\mathbf{i}-3\mathbf{j}+4\mathbf{k})\)
Use Pythagoras and differentiate with respect to \(\mu\) to obtain value of \(\mu\)
corresponding to minimum distance. (No need to prove it is a minimum)
Obtain \(\mu=-1\)
Obtain answer \(-\mathbf{i}+2\mathbf{j}-\mathbf{k}\)