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Question 1

(a) Topic-3.7 – Complex Numbers

(b) Topic-3.7 – Complex Numbers

The complex number z satisfies $|z|=2$ and $0\le arg~z\le\frac{1}{4}\pi.$

(a) On the Argand diagram below, sketch the locus of the points representing z.

(b) On the same diagram, sketch the locus of the points representing z².

▶️Answer/Explanation

Solution –

(a) For all 4 marks, scales must be approximately equal, dashes can replace numbers.

Arcs don’t have to be perfectly circular, mark intention.

Show an arc of a circle, centre the origin and radius 2.

Only need 2 on Re(z) or 2i on Im(z) or $r=2$ to show correct radius

Show an arc centre the origin for $0\le arg~z\le\frac{1}{4}\pi$ with any radius

Max B1 if sector shaded

(b) Show an arc of a circle, centre the origin and radius 4.

Only need 4 on Re(z) or 4i on Im(z) or $r=4$ to show correct radius

Show an arc centre the origin for $0\le arg~z\le\frac{1}{4}\pi$ with any radius

Max B1 if sector shaded

Question 2

(a) Topic-3.6 – Numerical Methods

(b) Topic-3.6 – Numerical Methods

Let $f(x)=2x^{3}-5x^{2}+4$.

(a) Show that if a sequence of values given by the iterative formula

$x_{n+1}=\sqrt{\frac{4}{5-2x_{n}}}$

converges, then it converges to a root of the equation $f(x)=0$.

(b) The equation has a root close to 1.2.

Use the iterative formula from part (a) and an initial value of 1.2 to determine the root correct to
2 decimal places. Give the result of each iteration to 4 decimal places.

▶️Answer/Explanation

Solution –

(a) State or imply the equation $x=\sqrt{\frac{4}{5-2x}}$ and square the equation

Rearrange this with at least one intermediate step in the form $2x^{3}-5x^{2}+4=0$

Alternative Method 1 for Question 2(a)

Rearrange $2x^{3}-5x^{2}+4=0$ to $x^{2}(5-2x)=4$ (or a different intermediate step)
and to either $x=\frac{4}{5-2x}$ or $x=\sqrt{\frac{4}{5-2x}}$
and then obtain the iterative formula $x_{n+1}=\sqrt{\frac{4}{5-2x_{n}}}$

Alternative Method 2 for Question 2(a)

Rearrange $2x^{3}-5x^{2}+4=0$ to $x^{2}(5-2x)=4$ (or a different intermediate step) and
to $x^{2}=\frac{4}{5-2x}$ and to ${x_{n+1}}^{2}=\frac{4}{5-2x_{n}}$

Obtain the iterative formula $x_{n+1}=\sqrt{\frac{4}{5-2x_{n}}}$

(b) Use the iterative process correctly at least once

Obtain final answer 1.28

Show sufficient iterations to at least 4 dp to justify 1.28 to 2 dp or show that there is
a sign change in the interval (1.275, 1.285)

Question 3

(a) Topic-3.2 – Logarithmic and exponential functions
(b) Topic-3.2 – Logarithmic and exponential functions

The number of bacteria in a population, P, at time t hours is modelled by the equation $P=ae^{kt}$, where
a and k are constants. The graph of $ln~P$ against t, shown in the diagram, has gradient $\frac{1}{20}$ and intersects
the vertical axis at (0,3).

(a) State the value of k and find the value of a correct to 2 significant figures.

(b) Find the time taken for P to double. Give your answer correct to the nearest hour.

▶️Answer/Explanation

Solution –

(a) State or imply that $ln~P=ln~a+kt$ or $ln~P=ln~a+k(ln~e)t$

$ln~P=\frac{1}{20}t+3$ B0 until associated with a and/or k

State $k=\frac{1}{20},$ not from $\frac{dP}{dt}=k$

$ln~a=3 \Rightarrow a=20$ to 2 sf

(b) Form a correct equation in t using a and k, or their a and k where a will cancel (or
are both numerical)

Obtain $t=14[hours]$

Question 4

Topic-3.7 – Complex Numbers

Find the complex number z satisfying the equation

$\frac{z-3i}{z+3i}=\frac{2-9i}{5}$.

Give your answer in the form x + iy, where x and y are real.

▶️Answer/Explanation

Solution –

Substitute $z=x+iy$ and obtain a horizontal equation

Do not allow if this would lead to an equation containing xy terms which do not
cancel

Use $i^{2}=-1$ anywhere

Obtain e.g. $5x+5(y-3)i=(2x+9y+27)+i(2y+6-9x)$
or e.g. $3x^{2}+3y^{2}-12y-63+(9x^{2}+9y^{2}-30x+54y+81)i=0$

Obtain simultaneous equations by equating real and imaginary parts

Obtain $[z=]3-2i$ only

Alternative Method for Question 4:

Obtain a horizontal equation in z

Do not allow if it would lead to an equation containing z² where the xy terms do not
cancel

Use $i^{2}=-1$ anywhere

Obtain $z=\frac{9+7i}{1+3i}$

Multiply top and bottom by 1-3i or equivalent for their z

Obtain $[z=]3-2i$ only

Question 5

(a) Topic-3.2 – Trigonometric Functions

(b) Topic-3.2 – Trigonometric Functions

(a) Show that $cos^{4}\theta-sin^{4}\theta-4~sin^{2}\theta~cos^{2}\theta\equiv cos^{2}2\theta+cos~2\theta-1.$

(b) Solve the equation $cos^{4}\alpha-sin^{4}\alpha=4~sin^{2}\alpha~cos^{2}\alpha$ for $0^{\circ}\le\alpha\le180^{\circ}$.

▶️Answer/Explanation

Solution :-

(a) $cos^{4}\theta~as~(\frac{1+cos~2\theta}{2})^{2}$ or $sin^{4}\theta~as~(\frac{1-cos~2\theta}{2})^{2}$

$4~sin^{2}\theta~cos^{2}\theta~as~sin^{2}2\theta$

Obtain $(\frac{1+cos~2\theta}{2})^{2}-(\frac{1-cos~2\theta}{2})^{2}-sin^{2}2\theta$

Expand to
$\frac{1}{4}+\frac{1}{2}cos~2\theta+\frac{1}{4}cos^{2}2\theta-(\frac{1}{4}-\frac{1}{2}cos~2\theta+\frac{1}{4}cos^{2}2\theta)-(1-cos^{2}2\theta)$

Simplify to obtain

$cos^{2}2\theta+cos~2\theta-1$

Alternative Method 1 for Question 5(a):

Express

$cos^{4}\theta-sin^{4}\theta~as~(cos^{2}\theta+sin^{2}\theta)(cos^{2}\theta-sin^{2}\theta)$

Or rewrite

$4~sin^{2}\theta~cos^{2}\theta~as~sin^{2}2\theta$

Simplify to $cos~2\theta-sin^{2}2\theta$

Use $sin^{2}2\theta=1-cos^{2}2\theta$ to obtain $cos^{2}2\theta+cos~2\theta-1$

Alternative Method 2 for Question 5(a):

Use correct double angle formulae once e.g. replace $cos~2\theta$ with $cos^{2}\theta-sin^{2}\theta$

$(cos^{2}\theta-sin^{2}\theta)^{2}+(cos^{2}\theta-sin^{2}\theta)-1$

Expand to obtain

$cos^{4}\theta-2~sin^{2}\theta~cos^{2}\theta+sin^{4}\theta+cos^{2}\theta-sin^{2}\theta-1$

Leading to

$cos^{4}\theta-2~sin^{2}\theta~cos^{2}\theta+sin^{4}\theta-2~sin^{2}\theta$ leading to
$cos^{4}\theta-2~sin^{2}\theta~cos^{2}\theta+sin^{4}\theta-2~sin^{2}\theta(cos^{2}\theta+sin^{2}\theta)$

Rewrite as

$cos^{4}\theta-2~sin^{2}\theta~cos^{2}\theta-sin^{4}\theta+2~sin^{4}\theta-2~sin^{2}\theta$ leading to
$cos^{4}\theta-2~sin^{2}\theta~cos^{2}\theta-sin^{4}\theta+2~sin^{2}\theta(sin^{2}\theta-1)$ leading to
$cos^{4}\theta-4~sin^{2}\theta~cos^{2}\theta-sin^{4}\theta$

$cos^{4}\theta-2~sin^{2}\theta~cos^{2}\theta+sin^{4}\theta-2~sin^{2}\theta~cos^{2}\theta-2~sin^{4}\theta$
$cos^{4}\theta-4~sin^{2}\theta~cos^{2}\theta-sin^{4}\theta$

(b) State a quadratic equation in $cos~2\alpha$ and solve for $\alpha$
$(cos^{2}2\alpha+cos~2\alpha-1=0)$

Obtain $\alpha=25.9^{\circ}$ or $\alpha=154.1^{\circ}$

Obtain $\alpha=25.9^{\circ}$ and $\alpha=154.1^{\circ}$ and no others in range

Question 6

(a) Topic-3.8 – Vectors

(b) Topic-3.8 – Vectors

The lines $l$ and $m$ have vector equations

$l: \mathbf{r}=2\mathbf{i}+\mathbf{j}-3\mathbf{k}+\lambda(-\mathbf{i}+2\mathbf{k})$ and
$m: \mathbf{r}=2\mathbf{i}+\mathbf{j}-3\mathbf{k}+\mu(2\mathbf{i}-\mathbf{j}+5\mathbf{k}).$

Lines $l$ and $m$ intersect at the point P.

(a) State the coordinates of P.

(b) Find the exact value of the cosine of the acute angle between $l$ and $m$.

Find the exact area of triangle APB.

▶️Answer/Explanation

Solution –

6(a) P'(2, 1, -3)

(b) Use the correct method to find the scalar product of the direction vectors

Divide the scalar product by the product of the moduli to obtain ±cos θ using
consistent vectors throughout

Obtain $cos~\theta=\frac{8}{5\sqrt{6}}$

(b) Alternative Method for Question 6(b):

Use of cosine rule: e.g. sides of √5, √30 and √19 found

e.g. $cos~\theta=\frac{5+30-19}{2\sqrt{5}\sqrt{30}}$

Obtain $cos~\theta=\frac{8}{5\sqrt{6}}$

$Area=\frac{1}{2}\times2\sqrt{5}\times\sqrt{30}\times\sqrt{1-\frac{64}{150}}.$

$=\sqrt{86}$

Alternative Method for Question 6(c)

$\vec{PA}\times\vec{PB}=-4i-18j-2k$

$Area=\frac{1}{2}|\vec{PA}\times\vec{PB}|=\frac{1}{2}\sqrt{16+324+4}$

$=\sqrt{86}$

Question 7

(a) Topic-3.4 – Differentiation

(b) Topic-3.4 – Differentiation

The parametric equations of a curve are

$x=3~sin~2t, \quad y=tan~t+cot~t,$

for $0<t<\frac{1}{2}\pi.$

(a) Show that
$\frac{dy}{dx}=-\frac{2}{3~sin^{2}2t}.$

(b) Find the equation of the normal to the curve at the point where $t=\frac{1}{4}\pi$. Give your answer in the
form $py+qx+r=0$ where p, q and r are integers.

▶️Answer/Explanation

Solution –

(a) Obtain $\frac{dx}{dt}=6~cos~2t$

Obtain $\frac{dy}{dt}=sec^{2}t-cosec^{2}t$

Use $\frac{dy}{dx}=\frac{dy}{dt}\div\frac{dx}{dt}$

Express as a single fraction

With $\frac{dy}{dx}$ correctly simplified as a single fraction in terms of sin t and cos t
Allow with 6cos 2t expressed as $\frac{1}{6~cos~2t}$ outside bracket

Obtain $(\frac{-4~cos~2t}{6~cos~2t\times sin^{2}2t}=)\frac{-2}{3~sin^{2}2t}$ from full and correct working
Numerator and denominator must have identical terms before cancelling, both cos2t
or both + and – $(sin^{2}t-cos^{2}t)$

7(a) Alternative Method for Question 7(a):

$y=\frac{6}{x}$

$\frac{dy}{dx}=-6x^{-2}$

$\frac{dy}{dx}=-\frac{6}{9~sin^{2}2t}$

Obtain $\frac{dy}{dx}=\frac{-2}{3~sin^{2}2t}$ from full and correct working

(b) Gradient of normal $=\frac{3}{2}$

Use correct method to find the equation of the normal

Obtain $2y-3x+5=0$

Question 8

(a) Topic-3.1 – Algebra

(b) Topic-3.1 – Algebra

Let $f(x)=\frac{7a^{2}}{(a-2x)(3a+x)}, \text{ where } a \text{ is a positive constant.}$

(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 x².

▶️Answer/Explanation

Solution –

(a) State or imply the form $\frac{A}{a-2x}+\frac{B}{3a+x}$ and use a correct method to find a constant

Obtain A = 2a or B = a

Obtain A = 2a and B = a

(b) Use a correct method to obtain the first two terms of the expansion of $(a-2x)^{-1}$ or
$(1-\frac{2x}{a})^{-1}$ or $(3a+x)^{-1}$ or $(1+\frac{x}{3a})^{-1}$

Obtain $2(1+\frac{2x}{a}+\frac{4x^{2}}{a^{2}}+…)$

Obtain $\frac{1}{3}(1-\frac{x}{3a}+\frac{x^{2}}{9a^{2}}+…)$

Obtain $\frac{7}{3}+\frac{35x}{9a}+\frac{217x^{2}}{27a^{2}}$

(b) Alternative Method for Question 8(h)

Expanding $7a^{2}(a-2x)^{-1}(3a+x)^{-1}$ from the original question.

Use a correct method to obtain the first two terms of the expansion of $(a-2x)^{-1}$ or
$(1-\frac{2x}{a})^{-1}$ or $(3a+x)^{-1}$ or $(1+\frac{x}{3a})^{-1}$

Obtain $7a^{2}(1+\frac{2x}{a}+\frac{4x^{2}}{a^{2}}+…)(1-\frac{x}{3a}+\frac{x^{2}}{9a^{2}}+…)$

Obtain $7(1+\frac{2x}{a}+\frac{4x^{2}}{a^{2}}+…)(1-\frac{x}{3a}+\frac{x^{2}}{9a^{2}}+…)$

Obtain $\frac{7}{3}+\frac{35x}{9a}+\frac{217x^{2}}{27a^{2}}$

Question 9

(a) Topic-3.1 -Algebra

(b) Topic-3.5 – Integration

(a) Find the quotient and remainder when $x^{4}+16$ is divided by $x^{2}+4$.

(b) Hence show that $\int_{2}^{2\sqrt{3}}\frac{x^{4}+16}{x^{2}+4}dx=\frac{4}{3}(\pi+4).$

▶️Answer/Explanation

Solution –

(a) Divide to obtain quotient x²+k

Obtain quotient x²-4

Obtain remainder 32

Alternative Method for Question 9(a)

Expands brackets to get B=0

$C=-4$

$D=32$

(b) $\frac{1}{3}x^{3}-4x$

Obtain $p~tan^{-1}qx$ where $q=2$ or $q=\frac{1}{2}$

Obtain $16~tan^{-1}\frac{1}{2}x$

Use limits correctly in an expression containing $p~tan^{-1}qx$ where $q=2$ or $q=\frac{1}{2}$
and $rx^{3}+sx$

Obtain $\frac{4}{3}(\pi+4)$ from full and correct working

Question 10

(a) Topic-3.7 – Differential Equations

(b) Topic-3.7 – Differential Equations

A water tank is in the shape of a cuboid with base area 40000 cm². At time t minutes the depth of water
in the tank is h cm. Water is pumped into the tank at a rate of 50000 cm³ per minute. Water is leaking
out of the tank through a hole in the bottom at a rate of 600h cm³ per minute.

(a) Show that $200 \frac{dh}{dt} = 250-3h.$

(b) It is given that when $t=0, h=50$.

Find the time taken for the depth of water in the tank to reach 80 cm. Give your answer correct to
2 significant figures.

▶️Answer/Explanation

Solution –

(a) State that $\frac{dV}{dt}=50000-600h$

[Use $V=40000h$ to] obtain $\frac{dV}{dh}=40000$ and use this and their $\frac{dV}{dt}$ in the correct
chain rule to obtain $\frac{dh}{dt}$

[Use $V=40000h$ to] obtain $\frac{dV}{dt}=40000\frac{dh}{dt}$ and equate to their $\frac{dV}{dt}$

(a) Obtain $200\frac{dh}{dt}=250-3h$ from full and correct working

(b) Separate variables correctly and integrate one side correctly

Obtain $-\frac{1}{3}ln|250-3h|=\frac{t}{200}(+C)$

Use $t=0, h=50$ in an expression containing $ln(250-3h)$ or $ln|250-3h|$ to find
the constant of integration.

Obtain $C=-\frac{1}{3}ln~100$

$t=150$

Question 11

(a) Topic-3.4 – Differentiation

(b) Topi-3.5 – Integration

The diagram shows the curve $y=2sin~x\sqrt{2+cos~x}$, for $0\le x\le2\pi$, and its minimum point M, where
$x=a$.

(a) Find the value of a correct to 2 decimal places.

(b) Use the substitution $u=2+cos~x$ to find the exact area of the shaded region R.

▶️Answer/Explanation

Solution –

(a) Use of correct product rule and correct chain rule

Obtain $\frac{dy}{dx}=2~cos~x\sqrt{2+cos~x}-\frac{2~sin^{2}x}{2\sqrt{2+cos~x}}$

Equate the derivative to zero and obtain a horizontal 3 term quadratic equation or 4
term quartic equation in cos a

If MO earlier then needs that expression to be such that arrive at 3 term quadratic or
4 term quartic equation in cos x without further trig errors.

The only error in the form of the differential allowed is for $(2+cos~x)^{-\frac{1}{2}}$ to be
$(2+cos~x)^{+\frac{1}{2}}or(2+cos~x)^{-\frac{3}{2}}$

Solve for cos a

Obtain $a=4.93$

(b) State or imply $du=-sin~x~dx$

Substitute throughout for u and du

Obtain $-\int2\sqrt{u}du$

Integrate to obtain $ku^{\frac{3}{2}}(+C)$

Use correct limits correctly in an expression of the form $ku^{\frac{3}{2}}$ or $k(2+cos~x)^{\frac{3}{2}}$

Obtain $\frac{4}{3}(3\sqrt{3}-1)$ or $4\sqrt{3}-\frac{4}{3}$ or $\frac{4}{3}\sqrt{27}-\frac{4}{3}$

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