Pearson Edexcel IAL Mathematics Unit FP1: Further Pure Mathematics 1 Question Bank

$4z^3-8z^2-24z+108=(z+3)\left(4z^2+\ldots z+36\right)$

M1

$=(z+3)\left(4z^2-20z+36\right)$

A1

$4z^2-20z+36=0 \Rightarrow z=\frac{20\pm\sqrt{400-4\times4\times36}}{8}=\ldots$

dM1

Roots are $-3,\;\frac{5\pm \mathrm{i}\sqrt{11}}{2}$

A1

(4)

Product of complex roots is $\frac{36}{4}=9$, so modulus is $\sqrt{9}$.
or
$$\text{Modulus}=\sqrt{\left(\frac{5}{2}\right)^2+\left(\frac{\sqrt{11}}{2}\right)^2}$$
M1
Hence modulus is 3
A1

Complex conjugate pair in correct quadrant for their roots.
M1
All three roots correctly positioned.
A1
M1 plots the complex roots as a conjugate pair in the correct quadrants for their roots.
A1 fully correct diagram with one root on the negative real axis, and the other as a complex pair of roughly the same length in quadrants 1 and 4.

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

B1

(1)

$\alpha^{2}+\beta^{2}=(\alpha+\beta)^{2}-2 \alpha \beta$

M1

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

A1

(2)

For roots $\alpha-\frac1{\beta^2}$ and $\beta-\frac1{\alpha^2}$:
$$\text{Sum}=\alpha+\beta-\frac{\alpha^2+\beta^2}{\alpha^2\beta^2} =\frac32-\frac{-19/4}{(7/2)^2}=\frac{185}{98}.$$
M1 A1

$$\text{Product}=\alpha\beta-\frac{\alpha+\beta}{\alpha\beta}+\frac1{\alpha^2\beta^2} =\frac72-\frac37+\frac4{49}=\frac{309}{98}.$$
M1 A1

$$x^2-\frac{185}{98}x+\frac{309}{98}=0.$$
M1

$$98x^2-185x+309=0.$$
A1

$x_{1}=0.5-\frac{\mathrm{f}(0.5)}{\mathrm{f}^{\prime}(0.5)}=0.5-\frac{0.29289321 \ldots}{46.70710678 \ldots}$

Correct application of Newton-Raphson

M1

=0.494

Correct value 3dp. A correct derivative
must have been used

A1

(2)

$$\frac{4.9-\beta}{|f(4.9)|}=\frac{\beta-4.8}{f(4.8)}$$$$\beta=4.883$$

M1 A1

ALT:
$$\beta=\frac{4.8|f(4.9)|+4.9|f(4.8)|}{|f(4.8)|+|f(4.9)|}$$$$\beta=\frac{4.8\times0.0344+4.9\times0.1627}{0.0344+0.1627}=4.883$$

M1 A1

$$xy=c^2\Rightarrow y=c^2x^{-1}$$$$\frac{dy}{dx}=-c^2x^{-2}$$
At P(ct,c/t), $\frac{dy}{dx}=-\frac{1}{t^2}$, so the normal gradient is $t^2$.
$$y-\frac ct=t^2(x-ct)$$$$ty-c=t^3x-ct^4$$$$t^3x-ty=c(t^4-1)$$
B1 M1 M1 A1

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, $Q=\left(\frac{75}{8},\frac{15}{2}\right)$.
B1 M1 dM1 A1

For $y^2=6x$, 4a=6, so focus $R=\left(\frac32,0\right)$ and directrix $x=-\frac32$.
Line through $Q\left(\frac{75}{8},\frac{15}{2}\right)$ and $R\left(\frac32,0\right)$ has gradient $\frac{20}{21}$.
$$y=\frac{20}{21}\left(x-\frac32\right)$$
At $x=-\frac32$, $y=-\frac{20}{7}$, so $S=\left(-\frac32,-\frac{20}{7}\right)$.
$$QS=\sqrt{\left(\frac{75}{8}+\frac32\right)^2+\left(\frac{15}{2}+\frac{20}{7}\right)^2}=\frac{2925}{56}$$
B1 M1 M1 M1 A1