CAIE A-Level Further Math 1. Further Pure Mathematics 1 Question Bank

$y=x^{2}$ so, $x=y^{\frac{1}{2}}$

B1

Correct substitution.

$y^{2}-y+2 y^{\frac{1}{2}}+5=0$ so, $\left(2 y^{\frac{1}{2}}\right)^{2}=\left(-y^{2}+y-5\right)^{2}$

M1

Obtains an equation which eliminates radicals.

$y^{4}-2 y^{3}+11 y^{2}-14 y+25=0$

A1

$\alpha^{2}+\beta^{2}+\gamma^{2}+\delta^{2}=2$

B1FT

4

$\alpha^{2} \beta^{2} \gamma^{2} \delta^{2}=25$

B1FT

$\frac{1}{\alpha^{2}}+\frac{1}{\beta^{2}}+\frac{1}{\gamma^{2}}+\frac{1}{\delta^{2}}=\frac{\alpha^{2} \beta^{2} \delta^{2}+\alpha^{2} \beta^{2} \gamma^{2}+\beta^{2} \gamma^{2} \delta^{2}+\alpha^{2} \gamma^{2} \delta^{2}}{\alpha^{2} \beta^{2} \gamma^{2} \delta^{2}}$

M1

Relates to coefficients.

$\frac{14}{25}$

A1

CAO

3

$$\begin{aligned} \alpha^{4}+\beta^{4}+\gamma^{4}+\delta^{4}= \left(\alpha^{2}+\beta^{2}+\gamma^{2}+\delta^{2}\right)^{2} \quad-2\left(\alpha^{2} \beta^{2}+\alpha^{2} \gamma^{2}+\alpha^{2} \delta^{2}+\beta^{2} \gamma^{2}+\beta^{2} \delta^{2}+\gamma^{2} \delta^{2}\right) = 2^{2}-2(11) \end{aligned}$$

M1

Uses formula for sum of squares or uses original equation.

-18

A1

2

$x=-1, \quad x=2$

B1

Vertical asymptotes.

y=1

B1

Horizontal asymptote.

2

$\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{\left(x^{2}-x-2\right)(2 x)-\left(x^{2}+2\right)(2 x-1)}{\left(x^{2}-x-2\right)^{2}}$

M1*

Finds $\frac{\mathrm{d} y}{\mathrm{~d} x}$.

$x^{2}+8 x-2=0$

DM1

Sets equal to 0 and forms equation.

(-8.2,0.9), (0.2, -0.9).

A1 A1

Condone
$$\left(-4-3 \sqrt{2}, \frac{2}{3} \sqrt{2}\right),\left(-4+3 \sqrt{2},-\frac{2}{3} \sqrt{2}\right) .$$

4

B1

Axes and all three asymptotes.

B1

Correct shape and position, crossing horizontal
asymptote.

B1

States (0,-1) coordinates of intersection with axes,
may be seen on diagram.

3

B1 FT

FT from sketch in (c)

B1

All correct.

2

$$\begin{aligned} \frac{x^{2}+2}{x^{2}-x-2}=1 \text { or } \frac{x^{2}+2}{x^{2}-x-2}=-1 x+4=0 \quad \text { or } \quad 2 x^{2}-x=0 \end{aligned}$$

M2

Finds critical points, award M1 for each case.

$x=-4 \quad$ or $x=0, \quad x=\frac{1}{2}$

A1

$-4<x<-1, \quad 0<x<\frac{1}{2}, x>2$

B1

Must have three distinct regions. Condone $\leqslant-1$
and $\geqslant 2$.

4

$\left|\begin{array}{ll}1 2 5\end{array}\right|-k\left|\begin{array}{ll}2 3 5\end{array}\right|+3\left|\begin{array}{ll}2 3 2\end{array}\right|=0 \Rightarrow-1-k+3=0 \Rightarrow k=2$

M1 A1

Sets determinant of A equal to zero.

$\left(\begin{array}{ccc}-2 -1 1 1 3\end{array}\right)\left(\begin{array}{ccc}1 2 2 1 3 2 5\end{array}\right)\left(\begin{array}{cc}0 -2 -1 0 0\end{array}\right)=\left(\begin{array}{ccc}-2 -1 1 1 3\end{array}\right)\left(\begin{array}{cc}-2 -1 -1 -2 0\end{array}\right)$

M1

Multiplying two matrices correctly, correct
dimensions.

$\left(\begin{array}{cc}3 -7 -9 3\end{array}\right)$

M1 A1

Completing matrix multiplication, AG.

5

$\left(\begin{array}{cc}3 -7 -9 3\end{array}\right)\binom{x}{y}=\binom{3 x-7 y}{-9 x+3 y}$

B1

Transforms $\binom{x}{y}$ to $\binom{X}{Y}$.

-9 x+3 m x=m(3 x-7 m x)

M1 A1

Uses y=m x and Y=m X.

$-9+3 m=3 m-7 m^{2} \Rightarrow 7 m^{2}=9$

A1

$y=\frac{3}{\sqrt{7}} x$ and $y=-\frac{3}{\sqrt{7}} x$

A1

5

$\mathbf{D}=\left(\begin{array}{ll}\alpha 0 \alpha\end{array}\right)$

B1

$\mathbf{E}=\left(\begin{array}{ll}\beta 0 1\end{array}\right)$

B1

$\mathbf{F}=\left(\begin{array}{ll}0 1 0\end{array}\right)$

B1

$\left(\begin{array}{cc}3 -7 -9 3\end{array}\right)=\left(\begin{array}{ll}\alpha 0 \alpha\end{array}\right)-9\left(\begin{array}{ll}0 \beta 1 0\end{array}\right)$

M1

Setting up simultaneous equations using their D and
E.

$\mathbf{D}=\left(\begin{array}{ll}3 0 3\end{array}\right) \quad \mathbf{E}=\left(\begin{array}{ll}\frac{7}{9} 0 1\end{array}\right)$

A1

Condone $\alpha=3, \beta=\frac{7}{9}$ if it is clear that they refer to
the correct matrices.

5