By solving the differential equation $\frac{dy}{dt}=y$, show that $y=A{\text{e}}^t$ where $A$ is a constant.

Show that $\frac{dx}{dt}-x=-A{\text{e}}^t$.

Solve the differential equation in part (a)(ii) to find $x$ as a function of $t$.

By differentiating $\frac{dy}{dt}=-x+y$ with respect to $t$, show that $\frac{d^{2}y}{d{t}^2}=2\frac{{\displaystyle dy}}{{\displaystyle dt}}$.

By substituting $Y=\frac{dy}{dt}$, show that $Y=B{\text{e}}^{2t}$ where $B$ is a constant.

Hence find $y$ as a function of $t$.

Hence show that $x=-\frac{B}{2}{\text{e}}^{2t}+C$, where $C$ is a constant.

Show that $\frac{d^{2}y}{d{t}^2}-2\frac{dy}{dt}-3y=0$.

Find the two values for $\lambda$ that satisfy $\frac{d^{2}y}{d{t}^2}-2\frac{dy}{dt}-3y=0$.

Let the two values found in part (c)(ii) be ${\lambda }_1$ and ${\lambda }_2$.

Verify that $y=F{\text{e}}^{{\lambda }_{1}t}+G{\text{e}}^{{\lambda }_{2}t}$ is a solution to the differential equation in (c)(i),where $G$ is a constant.

METHOD 1

$\frac{dy}{d\text{t}}=y$

$\int \frac{dy}{y}=\int d\text{t}$               (M1)

$\ln y=t+c$  OR  $\ln \left|y\right|=t+c$             A1A1

Note: Award A1 for $\ln y$ and A1 for $t$ and $c$.

$y=A{\text{e}}^t$             AG

METHOD 2

rearranging to $\frac{dy}{d\text{t}}-y=0$ AND multiplying by integrating factor ${\text{e}}^{-t}$               M1

$y{\text{e}}^{-t}=A$             A1A1

$y=A{\text{e}}^t$             AG

[3 marks]

substituting $y=A{\text{e}}^t$ into differential equation in $x$               M1

$\frac{dx}{d\text{t}}=x-A{\text{e}}^t$

$\frac{dx}{d\text{t}}-x=-A{\text{e}}^t$             AG

[1 mark]

integrating factor (IF) is ${\text{e}}^{\int -1dt}$               (M1)

$={\text{e}}^{-t}$               (A1)

${\text{e}}^{-t}\frac{dx}{d\text{t}}-x{\text{e}}^{-t}=-A$

$x{\text{e}}^{-t}=-At+D$               (A1)

$x=\left(-At+D\right){\text{e}}^t$               A1

Note: The first constant must be $A$, and the second can be any constant for the final A1 to be awarded. Accept a change of constant applied at the end.

[4 marks]

$\frac{d^{2}y}{d{t}^2}=-\frac{{\displaystyle dx}}{{\displaystyle dt}}+\frac{{\displaystyle dy}}{{\displaystyle dt}}$               A1

EITHER

$=-x+y+\frac{dy}{dt}$               (M1)

$=\frac{dy}{dt}+\frac{dy}{dt}$               A1

OR

$=-x+y+\left(-x+y\right)$               (M1)

$=2\left(-x+y\right)$               A1

THEN

$=2\frac{{\displaystyle dy}}{{\displaystyle dt}}$               AG

[3 marks]

$\frac{dY}{dt}=2Y$               A1

$\int \frac{dY}{Y}=\int 2dt$               M1

$\ln \left|Y\right|=2t+c$  OR  $\ln Y=2t+c$               A1

$Y=B{\text{e}}^{2t}$               AG

[3 marks]

$\frac{dy}{dt}=B{\text{e}}^{2t}$

$y=\int B{\text{e}}^{2t}dt$              M1

$y=\frac{B}{2}{\text{e}}^{2t}+C$              A1

Note: The first constant must be $B$, and the second can be any constant for the final A1 to be awarded. Accept a change of constant applied at the end.

[2 marks]

METHOD 1

substituting $\frac{dy}{dt}=B{\text{e}}^{2t}$ and their (iii) into $\frac{dy}{dt}=-x+y$              M1(M1)

$B{\text{e}}^{2t}=-x+\frac{B}{2}{\text{e}}^{2t}+C$              A1

$x=-\frac{B}{2}{\text{e}}^{2t}+C$              AG

Note: Follow through from incorrect part (iii) cannot be awarded if it does not lead to the AG.

METHOD 2

$\frac{dx}{dt}=x-\frac{B}{2}{\text{e}}^{2t}-C$

$\frac{dx}{dt}-x=-\frac{B}{2}{\text{e}}^{2t}-C$

$\frac{d\left(x{\text{e}}^{-t}\right)}{dt}=-\frac{B}{2}{\text{e}}^t-C{\text{e}}^{-t}$              M1

$x{\text{e}}^{-t}=\int -\frac{B}{2}{\text{e}}^t-C{\text{e}}^{-t}dt$

$x{\text{e}}^{-t}=-\frac{B}{2}{\text{e}}^t-C{\text{e}}^{-t}+D$              A1

$x=-\frac{B}{2}{\text{e}}^{2t}+C+D{\text{e}}^t$

$\frac{dy}{dt}=-x+y\Rightarrow B{\text{e}}^{2t}=\frac{B}{2}{\text{e}}^{2t}-C-D{\text{e}}^t+\frac{B}{2}{\text{e}}^{2t}+C\Rightarrow D=0$              M1

$x=-\frac{B}{2}{\text{e}}^{2t}+C$              AG

[3 marks]

$\frac{dy}{dt}=-4x+y$

$\frac{d^{2}y}{d{t}^2}=-4\frac{dx}{dt}+\frac{dy}{dt}$ seen anywhere              M1

METHOD 1

$\frac{d^{2}y}{d{t}^2}=-4\left(x-y\right)+\frac{dy}{dt}$

attempt to eliminate $x$              M1

$=-4\left(\frac{1}{4}\left(y-\frac{dy}{dt}\right)-y\right)+\frac{dy}{dt}$

$=2\frac{dy}{dt}+3y$              A1

$\frac{d^{2}y}{d{t}^2}-2\frac{dy}{dt}-3y=0$              AG

METHOD 2

rewriting LHS in terms of $x$ and $y$              M1

$\frac{d^{2}y}{d{t}^2}-2\frac{dy}{dt}-3y=\left(-8x+5y\right)-2\left(-4x+y\right)-3y$              A1

$=0$              AG

[3 marks]

$\frac{dy}{dt}=F\lambda {\text{e}}^{\lambda t}, \frac{d^{2}y}{d{t}^2}=F{\lambda }^2{\text{e}}^{\lambda t}$               (A1)

$F{\lambda }^2{\text{e}}^{\lambda t}-2F\lambda {\text{e}}^{\lambda t}-3F{\text{e}}^{\lambda t}=0$               (M1)

${\lambda }^2-2\lambda -3=0$  (since ${\text{e}}^{\lambda t}\ne 0$)              A1

${\lambda }_1$ and ${\lambda }_2$ are $3$ and $-1$ (either order)              A1

[4 marks]

METHOD 1

$y=F{\text{e}}^{3t}+G{\text{e}}^{-t}$

$\frac{dy}{dt}=3F{\text{e}}^{3t}-G{\text{e}}^{-t}, \frac{d^{2}y}{d{t}^2}=9F{\text{e}}^{3t}-G{\text{e}}^{-t}$                      (A1)(A1)

$\frac{d^{2}y}{d{t}^2}-2\frac{dy}{dt}-3y=9F{\text{e}}^{3t}+G{\text{e}}^{-t}-2\left(3F{\text{e}}^{3t}-G{\text{e}}^{-t}\right)-3\left(F{\text{e}}^{3t}-G{\text{e}}^{-t}\right)$              M1

$=9F{\text{e}}^{3t}+G{\text{e}}^{-t}-6F{\text{e}}^{3t}+2G{\text{e}}^{-t}-3F{\text{e}}^{3t}-3G{\text{e}}^{-t}$              A1

$=0$              AG

METHOD 2

$y=F{\text{e}}^{{\lambda }_{1}t}+G{\text{e}}^{{\lambda }_{2}t}$

$\frac{dy}{dt}=F{\lambda }_1{\text{e}}^{{\lambda }_{1}t}+G{\lambda }_2{\text{e}}^{{\lambda }_{2}t}, \frac{{\displaystyle d^{2}y}}{{\displaystyle d{t}^2}}=F{{\lambda }_1}^2{\text{e}}^{{\lambda }_{1}t}+G{{\lambda }_2}^2{\text{e}}^{{\lambda }_{2}t}$                      (A1)(A1)

$\frac{d^{2}y}{d{t}^2}-2\frac{dy}{dt}-3y=F{{\lambda }_1}^2{\text{e}}^{{\lambda }_{1}t}+G{{\lambda }_2}^2{\text{e}}^{{\lambda }_{2}t}-2\left(F{\lambda }_1{\text{e}}^{{\lambda }_{1}t}+G{\lambda }_2{\text{e}}^{{\lambda }_{2}t}\right)-3\left(F{\text{e}}^{{\lambda }_{1}t}+G{\text{e}}^{{\lambda }_{2}t}\right)$              M1

$=F{\text{e}}^{{\lambda }_{1}t}\left({\lambda }^2-2\lambda -3\right)+G{\text{e}}^{{\lambda }_{2}t}\left({\lambda }^2-2\lambda -3\right)$              A1

$=0$              AG

[4 marks]

Verify that $u=f\left(t\right)$ satisfies the differential equation $\frac{d^{2}u}{d{t}^2}=u$.

Show that ${\left(f\left(t\right)\right)}^2+{\left(g\left(t\right)\right)}^2=f\left(2t\right)$.

$f\left(\text{i}u\right)$.

$g\left(\text{i}u\right)$.

Hence find, and simplify, an expression for ${\left(f\left(\text{i}u\right)\right)}^2+{\left(g\left(\text{i}u\right)\right)}^2$.

Show that ${\left(f\left(t\right)\right)}^2-{\left(g\left(t\right)\right)}^2={\left(f\left(\text{i}u\right)\right)}^2-{\left(g\left(\text{i}u\right)\right)}^2$.

Sketch the graph of $x^2-y^2=1$, stating the coordinates of any axis intercepts and the equation of each asymptote.

The hyperbola with equation $x^2-y^2=1$ can be rotated to coincide with the curve defined by $xy=k, k\in \mathrm{ℝ}$.

Find the possible values of $k$.

$f\text{'}\left(t\right)=\frac{{\text{e}}^t-{\text{e}}^{-t}}{2}$                       A1

$f\text{'}\text{'}\left(t\right)=\frac{{\text{e}}^t+{\text{e}}^{-t}}{2}$                       A1

$=f\left(t\right)$                       AG

[2 marks]

METHOD 1

${\left(f\left(t\right)\right)}^2+{\left(g\left(t\right)\right)}^2$

substituting $f$ and $g$                      M1

$=\frac{{\left({\text{e}}^t+{\text{e}}^{-t}\right)}^2+{\left({\text{e}}^t-{\text{e}}^{-t}\right)}^2}{4}$

$=\frac{{\left({\text{e}}^t\right)}^2+2+{\left({\text{e}}^{-t}\right)}^2+{\left({\text{e}}^t\right)}^2-2+{\left({\text{e}}^{-t}\right)}^2}{4}$                      (M1)

$=\frac{{\left({\text{e}}^t\right)}^2+{\left({\text{e}}^{-t}\right)}^2}{2} \left(=\frac{{\text{e}}^{2t}+{\text{e}}^{-2t}}{2}\right)$                      A1

$=f\left(2t\right)$                      AG

METHOD 2

$f\left(2t\right)=\frac{{\text{e}}^{2t}+{\text{e}}^{-2t}}{2}$

$=\frac{{\left({\text{e}}^t\right)}^2+{\left({\text{e}}^{-t}\right)}^2}{2}$                     M1

$=\frac{{\left({\text{e}}^t+{\text{e}}^{-t}\right)}^2+{\left({\text{e}}^t-{\text{e}}^{-t}\right)}^2}{4}$                     M1A1

$={\left(f\left(t\right)\right)}^2+{\left(g\left(t\right)\right)}^2$                      AG

Note: Accept combinations of METHODS 1 & 2 that meet at equivalent expressions.

[3 marks]

substituting ${\text{e}}^{\text{i}u}=\cos u+\text{i} \sin u$ into the expression for $f$                      (M1)

obtaining ${\text{e}}^{\text{-i}u}=\cos u-\text{i} \sin u$                      (A1)

$f\left(\text{i}u\right)=\frac{\cos u+\text{i} \sin u+\cos u-\text{i} \sin u}{2}$

Note: The M1 can be awarded for the use of sine and cosine being odd and even respectively.

$=\frac{2 \cos u}{2}$

$=\cos u$                      A1

[3 marks]

$g\left(\text{i}u\right)=\frac{\cos u+\text{i} \sin u-\cos u+\text{i} \sin u}{2}$

substituting and attempt to simplify                      (M1)

$=\frac{2\text{i} \sin u}{2}$

$=\text{i} \sin u$                      A1

[2 marks]

METHOD 1

${\left(f\left(\text{i}u\right)\right)}^2+{\left(g\left(\text{i}u\right)\right)}^2$

substituting expressions found in part (c)                     (M1)

$={\cos }^2 u-{\sin }^2 u \left(=\cos 2u\right)$                      A1

METHOD 2

$f\left(2\text{i}u\right)=\frac{{\text{e}}^{2\text{i}u}+{\text{e}}^{-2\text{i}u}}{2}$

$=\frac{\cos 2u+\text{i} \sin 2u+\cos 2u-\text{i} \sin 2u}{2}$                     M1

$=\cos 2u$                      A1

Note: Accept equivalent final answers that have been simplified removing all imaginary parts eg $2 {\cos }^2 u-1$etc

[2 marks]

${\left(f\left(t\right)\right)}^2-{\left(g\left(t\right)\right)}^2=\frac{{\left({\text{e}}^t+{\text{e}}^{-t}\right)}^2-{\left({\text{e}}^t-{\text{e}}^{-t}\right)}^2}{4}$                      M1

$=\frac{\left({\text{e}}^{2t}+{\text{e}}^{-2t}+2\right)-\left({\text{e}}^{2t}+{\text{e}}^{-2t}-2\right)}{4}$                      A1

$=\frac{4}{4}=1$                      A1

Note: Award A1 for a value of 1 obtained from either LHS or RHS of given expression.

${\left(f\left(\text{i}u\right)\right)}^2-{\left(g\left(\text{i}u\right)\right)}^2={\cos }^2 u+{\sin }^2 u$                      M1

$=1$  (hence ${\left(f\left(t\right)\right)}^2-{\left(g\left(t\right)\right)}^2={\left(f\left(\text{i}u\right)\right)}^2-{\left(g\left(\text{i}u\right)\right)}^2$)                      AG

Note: Award full marks for showing that ${\left(f\left(z\right)\right)}^2-{\left(g\left(z\right)\right)}^2=1, \forall z\in \mathrm{ℂ}$.


[4 marks]

        A1A1A1A1

Note: Award A1 for correct curves in the upper quadrants, A1 for correct curves in the lower quadrants, A1 for correct $x$-intercepts of $(-1, 0)$ and $(1, 0)$ (condone $x=-1$ and $1$), A1 for $y=x$ and $y=-x$.

[4 marks]

attempt to rotate by $45°$ in either direction               (M1)

Note: Evidence of an attempt to relate to a sketch of $xy=k$ would be sufficient for this (M1).

attempting to rotate a particular point, eg $(1, 0)$               (M1)

$(1, 0)$ rotates to $\left(\frac{1}{\sqrt{2}}, \pm \frac{{\displaystyle 1}}{{\displaystyle \sqrt{2}}}\right)$ (or similar)               (A1)

hence $k=\pm \frac{1}{2}$             A1A1

[5 marks]

Find $\left(f\circ g\right)\left(\left(x\text{,}y\right)\right)$.

Find $\left(g\circ f\right)\left(\left(x\text{,}y\right)\right)$.

State with a reason whether or not $f$ and $g$ commute.

Find the inverse of $f$.

$\left(f\circ g\right)\left(\left(x\text{,}y\right)\right)=f\left(g\left(\left(x\text{,}y\right)\right)\right)$  ($=f\left(\left(xy,x+y\right)\right)$)       (M1)

$=\left(xy+x+y,xy-x-y\right)$       A1A1

[3 marks]

$\left(g\circ f\right)\left(\left(x\text{,}y\right)\right)=g\left(f\left(\left(x\text{,}y\right)\right)\right)$

$=g\left(\left(x+y,x-y\right)\right)$

$=\left(\left(x+y\right)\left(x-y\right),x+y+x-y\right)$

$=\left(x^2-y^2\text{,}2x\right)$       A1A1

[2 marks]

no because $f\circ g\ne g\circ f$        R1

Note: Accept counter example.

[1 mark]

$f\left(\left(x\text{,}y\right)\right)=\left(a\text{,}b\right)\Rightarrow \left(x+y,x-y\right)=\left(a\text{,}b\right)$       (M1)

$\{\begin{array}{c}x=\frac{a+b}{2}\\ y=\frac{a-b}{2}\end{array}$       (M1)

$f^{-1}\left(\left(x\text{,}y\right)\right)=\left(\frac{x+y}{2},\frac{x-y}{2}\right)$        A1

[3 marks]

Given that $1$ and $4+\text{i}$ are roots of the equation, write down the third root.

Verify that the mean of the two complex roots is $4$.

Show that the line $y=x-1$ is tangent to the curve $y=f\left(x\right)$ at the point $\text{A}\left(4, 3\right)$.

Sketch the curve $y=f\left(x\right)$ and the tangent to the curve at point $\text{A}$, clearly showing where the tangent crosses the $x$-axis.

Show that $g\text{'}\left(x\right)=2\left(x-r\right)\left(x-a\right)+x^2-2ax+a^2+b^2$.

Hence, or otherwise, prove that the tangent to the curve $y=g\left(x\right)$ at the point $\text{A}\left(a, g\left(a\right)\right)$ intersects the $x$-axis at the point $\text{R}\left(r, 0\right)$.

Deduce from part (d)(i) that the complex roots of the equation $\left(z-r\right)\left(z^2-2az+a^2+b^2\right)=0$ can be expressed as $a\pm \text{i}\sqrt{g\text{'}\left(a\right)}$.

Use this diagram to determine the roots of the corresponding equation of the form $\left(z-r\right)\left(z^2-2az+a^2+16\right)=0$ for $z\in \mathrm{ℂ}$.

State the coordinates of ${\text{C}}_2$.

Show that the $x$-coordinate of $\text{P}$ is $\frac{1}{3}\left(2a+r\right)$.

You are not required to demonstrate a change in concavity.

Hence describe numerically the horizontal position of point $\text{P}$ relative to the horizontal positions of the points $\text{R}$ and $\text{A}$.

Sketch the curve $y=\left(x-r\right)\left(x^2-2ax+a^2+b^2\right)$ for $a=r=1$ and $b=2$.

For $a=r$ and $b>0$, state in terms of $r$, the coordinates of points $\text{P}$ and $\text{A}$.

$4-\text{i}$        A1

[1 mark]

mean$=\frac{1}{2}\left(4+\text{i}+4-\text{i}\right)$          A1

$=4$          AG

[1 mark]

METHOD 1

attempts product rule differentiation        (M1)

Note: Award (M1) for attempting to express $f\left(x\right)$ as $f\left(x\right)=x^3-9x^2+25x-17$

$f\text{'}\left(x\right)=\left(x-1\right)\left(2x-8\right)+x^2-8x+17 \left(f\text{'}\left(x\right)=3x^2-18x+25\right)$        A1

$f\text{'}\left(4\right)=1$        A1

Note: Where $f\text{'}\left(x\right)$ is correct, award A1 for solving $f\text{'}\left(x\right)=1$ and obtaining $x=4$.

EITHER

$y-3=1\left(x-4\right)$        A1

OR

$y=x+c$

$3=4+c\Rightarrow c=-1$        A1

OR

states the gradient of $y=x-1$ is also $1$ and verifies that $\left(4, 3\right)$ lies on the line $y=x-1$        A1

THEN

so $y=x-1$ is the tangent to the curve at $\text{A}\left(4, 3\right)$        AG

Note: Award a maximum of (M0)A0A1A1 to a candidate who does not attempt to find $f\text{'}\left(x\right)$.

METHOD 2

sets $f\left(x\right)=x-1$ to form $x-1=\left(x-1\right)\left(x^2-8x+17\right)$        (M1)

EITHER

$\left(x-1\right)\left(x^2-8x+16\right)=0 \left(x^3-9x^2+24x-16=0\right)$        A1

attempts to solve a correct cubic equation        (M1)

$\left(x-1\right){\left(x-4\right)}^2=0\Rightarrow x=1, 4$

OR

recognises that $x\ne 1$ and forms $x^2-8x+17=1 \left(x^2-8x+16=0\right)$        A1

attempts to solve a correct quadratic equation        (M1)

${\left(x-4\right)}^2=0\Rightarrow x=4$

THEN

$x=4$ is a double root        R1

so $y=x-1$ is the tangent to the curve at $\text{A}\left(4, 3\right)$        AG

Note: Candidates using this method are not required to verify that $y=3$.

[4 marks]

a positive cubic with an  $x$-intercept $\left(x=1\right)$, and a local maximum and local minimum in the first quadrant both positioned to the left of $\text{A}$        A1

Note: As the local minimum and point A are very close to each other, condone graphs that seem to show these points coinciding.
For the point of tangency, accept labels such as $\text{A}, \left(4,3\right)$ or the point labelled from both axes. Coordinates are not required.

a correct sketch of the tangent passing through $\text{A}$ and crossing the $x$-axis at the same point $\left(x=1\right)$ as the curve        A1

Note: Award A1A0 if both graphs cross the $x$-axis at distinctly different points.

[2 marks]

EITHER

$g\text{'}\left(x\right)=\left(x-r\right)\left(2x-2a\right)+x^2-2ax+a^2+b^2$         (M1)A1

OR

$g\left(x\right)=x^3-\left(2a+r\right)x^2+\left(a^2+b^2+2ar\right)x-\left(a^2+b^2\right)r$

attempts to find $g\text{'}\left(x\right)$        M1

$g\text{'}\left(x\right)=3x^2-2\left(2a+r\right)x+a^2+b^2+2ar$

$=2x^2-2\left(a+r\right)x+2ar+x^2-2ax+a^2+b^2$        A1

$\left(=2\left(x^2-ax-rx+ar\right)+x^2-2ax+a^2+b^2\right)$

THEN

$g\text{'}\left(x\right)=2\left(x-r\right)\left(x-a\right)+x^2-2ax+a^2+b^2$        AG

[2 marks]

METHOD 1

$g\left(a\right)=b^2\left(a-r\right)$         (A1)

$g\text{'}\left(a\right)=b^2$         (A1)

attempts to substitute their $g\left(a\right)$ and $g\text{'}\left(a\right)$ into $y-g\left(a\right)=g\text{'}\left(a\right)\left(x-a\right)$        M1

$y-b^2\left(a-r\right)=b^2\left(x-a\right)$

EITHER

$y=b^2\left(x-r\right) \left(y=b^{2}x-b^{2}r\right)$        A1

sets $y=0$ so $b^2\left(x-r\right)=0$        M1

$b>0\Rightarrow x=r$ OR $b\ne 0\Rightarrow x=r$        R1

OR 

sets $y=0$ so $-b^2\left(a-r\right)=b^2\left(x-a\right)$        M1

$b>0$ OR $b\ne 0\Rightarrow -\left(a-r\right)=x-a$        R1

$x=r$        A1

THEN

so the tangent intersects the $x$-axis at the point $\text{R}\left(r, 0\right)$        AG

METHOD 2

$g\text{'}\left(a\right)=b^2$         (A1)

$g\left(a\right)=b^2\left(a-r\right)$         (A1)

attempts to substitute their $g\left(a\right)$ and $g\text{'}\left(a\right)$ into $y=g\text{'}\left(a\right)x+c$ and attempts to find $c$        M1

$c=-b^{2}r$

EITHER

$y=b^2\left(x-r\right) \left(y=b^{2}x-b^{2}r\right)$        A1

sets $y=0$ so $b^2\left(x-r\right)=0$        M1

$b>0\Rightarrow x=r$ OR $b\ne 0\Rightarrow x=r$        R1

OR

sets $y=0$ so $b^2\left(x-r\right)=0$        M1

$b>0$ OR $b\ne 0\Rightarrow x-r=0$        R1

$x=r$        A1

METHOD 3

$g\text{'}\left(a\right)=b^2$         (A1)

the line through $R\left(r, 0\right)$ parallel to the tangent at $\text{A}$ has equation
$y=b^2\left(x-r\right)$        A1

sets $g\left(x\right)=b^2\left(x-r\right)$ to form $b^2\left(x-r\right)=\left(x-r\right)\left(x^2-2ax+a^2+b^2\right)$        M1

$b^2=x^2-2ax+a^2+b^2, \left(x\ne r\right)$        A1

${\left(x-a\right)}^2=0$        A1

since there is a double root $\left(x=a\right)$, this parallel line through $R\left(r, 0\right)$ is the required tangent at $\text{A}$        R1

[6 marks]

EITHER

$g\text{'}\left(a\right)=b^2\Rightarrow b=\sqrt{g\text{'}\left(a\right)}$ (since $b>0$)        R1


Note: Accept $b=\pm \sqrt{g\text{'}\left(a\right)}$.

OR

$\left(a\pm b\text{i=}\right)a\pm \text{i}\sqrt{b^2}$ and $g\text{'}\left(a\right)=b^2$        R1

THEN

hence the complex roots can be expressed as $a\pm \text{i}\sqrt{g\text{'}\left(a\right)}$        AG

[1 mark]

$b=4$ (seen anywhere)        A1

EITHER

attempts to find the gradient of the tangent in terms of $a$ and equates to $16$       (M1)


OR

substitutes $r=-2, x=a$  and  $y=80$ to form $80=\left(a-\left(-2\right)\right)\left(a^2-2a^2+a^2+16\right)$       (M1)

OR

substitutes $r=-2, x=a$  and  $y=80$ into $y=16\left(x-r\right)$       (M1)

THEN

$\frac{80}{a+2}=16\Rightarrow a=3$

roots are $-2$ (seen anywhere) and $3\pm 4\text{i}$        A1A1

Note: Award A1 for $-2$ and A1 for $3\pm 4\text{i}$. Do not accept coordinates.

[4 marks]

$\left(3, -4\right)$        A1

Note: Accept “$x=3$ and $y=-4$”.
Do not award A1FT for $(a, -4)$. 

[1 mark]

$g\text{'}\left(x\right)=2\left(x-r\right)\left(x-a\right)+x^2-2ax+a^2+b^2$

attempts to find $g\text{'}\text{'}\left(x\right)$        M1

$g\text{'}\text{'}\left(x\right)=2\left(x-a\right)+2\left(x-r\right)+2x-2a \left(=6x-2r-4a\right)$

sets $g\text{'}\text{'}\left(x\right)=0$ and correctly solves for $x$        A1

for example, obtaining $x-r+2\left(x-a\right)=0$ leading to $3x=2a+r$

so $x=\frac{1}{3}\left(2a+r\right)$        AG

Note: Do not award A1 if the answer does not lead to the AG.

[2 marks]

point $\text{P}$ is $\frac{2}{3}$ of the horizontal distance (way) from point $\text{R}$ to point $\text{A}$       A1

Note: Accept equivalent numerical statements or a clearly labelled diagram displaying the numerical relationship.
Award A0 for non-numerical statements such as “$\text{P}$ is between $\text{R}$ and $\text{A}$, closer to $\text{A}$”.

[1 mark]

$y=\left(x-1\right)\left(x^2-2x+5\right)$       (A1)

a positive cubic with no stationary points and a non-stationary point of inflexion at $x=1$       A1

Note: Graphs may appear approximately linear. Award this A1 if a change of concavity either side of $x=1$ is apparent.
Coordinates are not required and the $y$-intercept need not be indicated.

[2 marks]

$\left(r, 0\right)$         A1

[1 mark]

Part (a) (i) was generally well done with a significant majority of candidates using the conjugate root theorem to state $4-\text{i}$ as the third root. A number of candidates, however, wasted considerable time attempting an algebraic method to determine the third root. Part (a) (ii) was reasonably well done. A few candidates however attempted to calculate the product of $4+\text{i}$ and $4-\text{i}$.

Part (b) was reasonably well done by a significant number of candidates. Most were able to find a correct expression for $f\text{'}\left(x\right)$ and a good number of those candidates were able to determine that $f\text{'}\left(4\right)=1$. Candidates that did not determine the equation of the tangent had to state that the gradient of $y=x-1$ is also 1 and verify that the point (4,3) lies on the line. A few candidates only met one of those requirements. Weaker candidates tended to only verify that the point (4,3) lies on the curve and the tangent line without attempting to find $f\text{'}\left(x\right)$.

Part (c) was not answered as well as anticipated. A number of sketches were inaccurate and carelessly drawn with many showing both graphs crossing the x-axis at distinctly different points.

Part (d) (i) was reasonably well done by a good number of candidates. Most successful responses involved use of the product rule. A few candidates obtained full marks by firstly expanding $g\left(x\right)$, then differentiating to find $g\text{'}\left(x\right)$and finally simplifying to obtain the desired result. A number of candidates made elementary mistakes when differentiating. In general, the better candidates offered reasonable attempts at showing the general result in part (d) (ii). A good number gained partial credit by determining that $g\text{'}\left(a\right)=b^2$ and/or $g\left(a\right)=b^2(a-r)$. Only the very best candidates obtained full marks by concluding that as $b>0$ or $b\ne 0$, then $x=r$ when $y=0$.

In general, only the best candidates were able to use the result $g\text{'}\left(a\right)=b^2$ to deduce that the complex roots of the equation can be expressed as $a\pm \text{i}\sqrt{g\text{'}\left(a\right)}$. Although given the complex roots $a\pm b\text{i}$, a significant number of candidates attempted, with mixed success, to use the quadratic formula to solve the equation $z^2-2az+a^2+b^2=0$.

In part (f) (i), only a small number of candidates were able to determine all the roots of the equation. Disappointingly, a large number did not state $-2$ as a root. Some candidates determined that $b=4$ but were unable to use the diagram to determine that $a=3$. Of the candidates who determined all the roots in part (f) (i), very few gave the correct coordinates for C₂ . The most frequent error was to give the y-coordinate as $3-4\text{i}$.

Of the candidates who attempted part (g) (i), most were able to find an expression for $g\text{'}\text{'}\left(x\right)$ and a reasonable number of these were then able to convincingly show that $x=\frac{1}{3}(2a+r)$. It was very rare to see a correct response to part (g) (ii). A few candidates stated that P is between R and A with some stating that P was closer to A. A small number restated $x=\frac{1}{3}(2a+r)$ in words.

Of the candidates who attempted part (h) (i), most were able to determine that $y=(x-1)\left(x^2-2x+5\right)$. However, most graphs were poorly drawn with many showing a change in concavity at $x=0$ rather than at $x=1$. In part (h) (ii), only a very small number of candidates determined that A and P coincide at (r,0).

Sketch the graph of $y=f_1\left(x\right)$, stating the values of any axes intercepts and the coordinates of any local maximum or minimum points.

Use your graphic display calculator to explore the graph of $y=f_n\left(x\right)$ for

•   the odd values $n=3$ and $n=5$;

•   the even values $n=2$ and $n=4$.

Hence, copy and complete the following table.

Show that ${f_n}^{\text{'}}\left(x\right)=n{x}^{n-1}\left(a-2x\right){\left(a-x\right)}^{n-1}$.

State the three solutions to the equation ${f_n}^{\text{'}}\left(x\right)=0$.

Show that the point $\left(\frac{a}{2}, f_n\left(\frac{a}{2}\right)\right)$ on the graph of $y=f_n\left(x\right)$ is always above the horizontal axis.

Hence, or otherwise, show that ${f_n}^{\text{'}}\left(\frac{a}{4}\right)>0$, for $n\in {\mathrm{\mathbb{Z}}}^{+}$.

a local minimum point for even values of $n$, where $n>1$ and $a\in {\mathrm{ℝ}}^{+}$.

a point of inflexion with zero gradient for odd values of $n$, where $n>1$ and $a\in {\mathrm{ℝ}}^{+}$.

Consider the graph of $y=x^n{\left(a-x\right)}^n-k$, where $n\in {\mathrm{\mathbb{Z}}}^{+}$, $a\in {\mathrm{ℝ}}^{+}$ and $k\in \mathrm{ℝ}$.

State the conditions on $n$ and $k$ such that the equation $x^n{\left(a-x\right)}^n=k$ has four solutions for $x$.

inverted parabola extended below the $x$-axis             A1

$x$-axis intercept values $x=0, 2$         A1


Note: Accept a graph passing through the origin as an indication of $x=0$.

local maximum at $\left(1, 1\right)$                 A1

Note: Coordinates must be stated to gain the final A1.
        Do not accept decimal approximations.

[3 marks]

             A1A1A1A1A1A1

Note: Award A1 for each correct value.

For a table not sufficiently or clearly labelled, assume that their values are in the same order as the table in the question paper and award marks accordingly.

[6 marks]

METHOD 1

attempts to use the product rule            (M1)

${f_n}^{\text{'}}\left(x\right)=-n{x}^n{\left(a-x\right)}^{n-1}+n{x}^{n-1}{\left(a-x\right)}^n$            A1A1

Note: Award A1 for a correct $u\frac{dv}{dx}$ and A1 for a correct $v\frac{du}{dx}$.

EITHER

attempts to factorise ${f_n}^{\text{'}}\left(x\right)$ (involving at least one of $n{x}^{n-1}$ or ${\left(a-x\right)}^{n-1}$)           (M1)

$=n{x}^{n-1}{\left(a-x\right)}^{n-1}\left(\left(a-x\right)-x\right)$            A1

OR

attempts to express ${f_n}^{\text{'}}\left(x\right)$ as the difference of two products with each product containing at least one of $n{x}^{n-1}$ or ${\left(a-x\right)}^{n-1}$           (M1)

$=\left(-x\right)\left(n{x}^{n-1}\right){\left(a-x\right)}^{n-1}+\left(a-x\right)\left(n{x}^{n-1}\right){\left(a-x\right)}^{n-1}$            A1

THEN

${f_n}^{\text{'}}\left(x\right)=n{x}^{n-1}\left(a-2x\right){\left(a-x\right)}^{n-1}$            AG

Note: Award the final (M1)A1 for obtaining any of the following forms: 

${f_n}^{\text{'}}\left(x\right)=n{x}^n{\left(a-x\right)}^n\left(\frac{a-x-x}{x\left(a-x\right)}\right); {f_n}^{\text{'}}\left(x\right)=\frac{n{x}^n{\left(a-x\right)}^n}{x\left(a-x\right)}\left(a-x-x\right);$

        ${f_n}^{\text{'}}\left(x\right)=n{x}^{n-1}\left({\left(a-x\right)}^n-x{\left(a-x\right)}^{n-1}\right);$

        ${f_n}^{\text{'}}\left(x\right)={\left(a-x\right)}^{n-1}\left(n{x}^{n-1}{\left(a-x\right)}^n-n{x}^n\right)$

METHOD 2

$f_n\left(x\right)={\left(x\left(a-x\right)\right)}^n$           (M1)

$={\left(ax-x^2\right)}^n$            A1

attempts to use the chain rule           (M1)

${f_n}^{\text{'}}\left(x\right)=n\left(a-2x\right){\left(ax-x^2\right)}^{n-1}$            A1A1

Note: Award A1 for $n\left(a-2x\right)$ and A1 for ${\left(ax-x^2\right)}^{n-1}$.

${f_n}^{\text{'}}\left(x\right)=n{x}^{n-1}\left(a-2x\right){\left(a-x\right)}^{n-1}$            AG

[5 marks]

$x=0, x=\frac{a}{2}, x=a$            A2

Note: Award A1 for either two correct solutions or for obtaining $x=0, x=-a, x=-\frac{a}{2}$
          Award A0 otherwise.

[2 marks]

attempts to find an expression for $f_n\left(\frac{a}{2}\right)$             (M1)

$f_n\left(\frac{a}{2}\right)={\left(\frac{a}{2}\right)}^n{\left(a-\frac{a}{2}\right)}^n$

$={\left(\frac{a}{2}\right)}^n{\left(\frac{a}{2}\right)}^n \left(={\left(\frac{a}{2}\right)}^{2n}\right), \left(={\left({\left(\frac{a}{2}\right)}^n\right)}^2\right)$            A1

EITHER

since $a\in {\mathrm{ℝ}}^{+}, {\left(\frac{a}{2}\right)}^{2n}>0$  (for $n\in {\mathrm{\mathbb{Z}}}^{+}, n>1$ and so $f_n\left(\frac{a}{2}\right)>0$)                R1

Note: Accept any logically equivalent conditions/statements on $a$ and $n$.
        Award R0 if any conditions/statements specified involving $a$, $n$ or both are incorrect.

OR

(since $a\in {\mathrm{ℝ}}^{+}$), $\frac{a}{2}$ raised to an even power ($2n$) (or equivalent reasoning) is always positive (and so  $f_n\left(\frac{a}{2}\right)>0$)                R1

Note: The condition $a\in {\mathrm{ℝ}}^{+}$ is given in the question. Hence some candidates will assume $a\in {\mathrm{ℝ}}^{+}$ and not state it. In these instances, award R1 for a convincing argument.
        Accept any logically equivalent conditions/statements on on $a$ and $n$.
        Award R0 if any conditions/statements specified involving $a$, $n$ or both are incorrect.

THEN

so $\left(\frac{a}{2}, f_n\left(\frac{a}{2}\right)\right)$ is always above the horizontal axis            AG

Note: Do not award (M1)A0R1.

[3 marks]