Edexcel A-Level Mathematics AS Unit Fp1 Further Pure Mathematics 1 Questions

Practise FP1 methods across complex numbers, roots, matrices, series, proof and numerical methods, with algebraic and graphical decisions.

Covered syllabus points

FP1.1 - Complex numbers FP1.2 - Roots of quadratic equations FP1.3 - Numerical solution of equations FP1.4 - Coordinate systems FP1.5 - Matrix algebra FP1.6 - Transformations using matrices FP1.7 - Series FP1.8 - Proof

[Maximum number: 8]

In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable.

\mathrm{f}(z)=4 z^{3}+p z^{2}-24 z+108

where p is a constant.
Given that -3 is a root of the equation f(z)=0

(a)

using algebra, solve f(z)=0 completely, giving the roots in simplest form,

[ 4 ]

$4z^3-8z^2-24z+108=(z+3)\left(4z^2+\ldots z+36\right)$ $=(z+3)\left(4z^2-20z+36\right)$ $4z^2-20z+36=0 \Rightarrow z=\frac{20\pm\sqrt{400-4\times4\times36}}{8}=\ldots$
Roots are $-3,\;\frac{5\pm \mathrm{i}\sqrt{11}}{2}$
(4)

(b)

determine the modulus of the complex roots of f(z)=0

[ 2 ]

Product of complex roots is $\frac{36}{4}=9$ , so modulus is $\sqrt{9}$ .
or
$\text{Modulus}=\sqrt{\left(\frac{5}{2}\right)^2+\left(\frac{\sqrt{11}}{2}\right)^2}$
Hence modulus is 3

(c)

show the roots of f(z)=0 on a single Argand diagram.

[ 2 ]

[Maximum number: 9]

The quadratic equation

2 x^{2}-3 x+7=0

has roots $\alpha$ and $\beta$
Without solving the equation,

(a)

write down the value of $(\alpha+\beta)$ and the value of $\alpha \beta$

[ 1 ]

$2 x^{2}-3 x+7=0$ $\alpha+\beta=\frac{3}{2}, \quad \alpha \beta=\frac{7}{2}$
(1)

(b)

determine the value of $\alpha^{2}+\beta^{2}$

[ 2 ]

$\alpha^{2}+\beta^{2}=(\alpha+\beta)^{2}-2 \alpha \beta$ $=\left(\frac{3}{2}\right)^{2}-2\left(\frac{7}{2}\right)=-\frac{19}{4}$
(2)

(c)

find a quadratic equation which has roots

\left(\alpha-\frac{1}{\beta^{2}}\right) \text { and }\left(\beta-\frac{1}{\alpha^{2}}\right)

giving your answer in the form $p x^{2}+q x+r=0$ where p, q and r are integers to be determined.

[ 6 ]

For roots $\alpha-\frac1{\beta^2}$ and $\beta-\frac1{\alpha^2}$ :
$\text{Sum}=\alpha+\beta-\frac{\alpha^2+\beta^2}{\alpha^2\beta^2} =\frac32-\frac{-19/4}{(7/2)^2}=\frac{185}{98}.$ $\text{Product}=\alpha\beta-\frac{\alpha+\beta}{\alpha\beta}+\frac1{\alpha^2\beta^2} =\frac72-\frac37+\frac4{49}=\frac{309}{98}.$ $x^2-\frac{185}{98}x+\frac{309}{98}=0.$ $98x^2-185x+309=0.$

[Maximum number: 6]

(a)

$\mathrm f(x)=x-4-\cos(5\sqrt{x}),\qquad x>0.$

[ 4 ]

$\mathrm{f}(x)=x-4-\cos (5 \sqrt{x}) \quad x>0$

(i)

Show that the equation f(x)=0 has a root $\alpha$ in the interval [2.5, 3.5]

[ 2 ]

$\mathrm{f}(2.5)=-1.44 \ldots, \mathrm{f}(3.5)=0.497 \ldots$
Sign change (negative, positive) and f(x) is continuous therefore (a root) $\alpha$
is between x=2.5 and x=3.5
(2)

(ii)

Use linear interpolation once on the interval [2.5, 3.5] to find an approximation to $\alpha$ , giving your answer to 2 decimal places.

[ 2 ]

E.g. $\frac{\alpha-2.5}{|\mathrm{f}(2.5)|}=\frac{3.5-\alpha}{\mathrm{f}(3.5)} \Rightarrow \alpha=\ldots$ or $\frac{\alpha-2.5}{0-\mathrm{f}(2.5)}=\frac{3.5-2.5}{\mathrm{f}(3.5)-\mathrm{f}(2.5)} \Rightarrow \alpha=\ldots$ $\alpha=$ awrt 3.24
(2)

(b)

$\mathrm g(x)=\frac{1}{10}x^2-\frac{1}{2x^2}+x-11,\qquad x>0.$

[ 2 ]

$\mathrm{g}(x)=\frac{1}{10} x^{2}-\frac{1}{2 x^{2}}+x-11 \quad x>0$

(i)

Using $x_{0}=6$ as a first approximation to $\beta$ , apply the Newton-Raphson procedure once to g(x) to find a second approximation to $\beta$ , giving your answer to 3 decimal places.

[ 2 ]

Newton-Raphson:
$x_1=6-\frac{\mathrm g(6)}{\mathrm g'(6)} =6-\frac{-1.41388\ldots}{2.20462\ldots}.$
M1

$x_1=6.641.$
A1cao

[Maximum number: 8]

The rectangular hyperbola H has equation $x y=c^{2}$ where c is a positive constant.

The point $P\left(c t, \frac{c}{t}\right)$ , where t>0, lies on H

(a)

Use calculus to show that an equation of the normal to H at P is

t^{3} x-t y=c\left(t^{4}-1\right)

The parabola C has equation $y^{2}=6 x$
The normal to H at the point with coordinates (8,2) meets C at the point Q where y>0

[ 4 ]

$xy=c^2\Rightarrow y=c^2x^{-1}$ $\frac{dy}{dx}=-c^2x^{-2}$
At P(ct,c/t), $\frac{dy}{dx}=-\frac{1}{t^2}$ , so the normal gradient is $t^2$ .
$y-\frac ct=t^2(x-ct)$ $ty-c=t^3x-ct^4$ $t^3x-ty=c(t^4-1)$

(b)

Determine the exact coordinates of Q

Given that
- the point R is the focus of C
- the line l is the directrix of C
- the line through Q and R meets l at the point S

[ 4 ]

At (8,2), $c^2=16$ , so c=4, and ct=8, so t=2.
Normal: 8x-2y=4(16-1)=60, so y=4x-30.
With $y^2=6x$ : $(4x-30)^2=6x$ .
$16x^2-246x+900=0$ $x=6\text{ or }\frac{75}{8}$
Since y>0, $Q=\left(\frac{75}{8},\frac{15}{2}\right)$ .

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Covered syllabus points

Question 1

Question (a)

Question (b)

Question (c)

Question 2

Question (a)

Question (b)

Question (c)

Question 3

Question (a)

Question (i)

Question (ii)

Question (b)

Question (i)

Question 4

Question (a)

Question (b)