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]

METHOD 1

${f_n}^{\text{'}}\left(\frac{a}{4}\right)=n{\left(\frac{a}{4}\right)}^{n-1}\left(a-\frac{a}{2}\right){\left(a-\frac{a}{4}\right)}^{n-1} \left(=n{\left(\frac{a}{4}\right)}^{n-1}\left(\frac{a}{2}\right){\left(\frac{3a}{4}\right)}^{n-1}\right)$            A1

EITHER

$n{\left(\frac{a}{4}\right)}^{n-1}\left(\frac{a}{2}\right){\left(\frac{3a}{4}\right)}^{n-1}>0$ as $a\in {\mathrm{ℝ}}^{+}$ and $n\in {\mathrm{\mathbb{Z}}}^{+}$                R1

OR

$n{\left(\frac{a}{4}\right)}^{n-1}, \left(a-\frac{a}{2}\right)$ and ${\left(a-\frac{a}{4}\right)}^{n-1}$ are all $>0$                R1

Note: Do not award A0R1.
        Accept equivalent reasoning on correct alternative expressions for ${f_n}^{\text{'}}\left(\frac{a}{4}\right)$ and accept any logically equivalent conditions/statements on $a$ and $n$.

        Exceptions to the above are condone $n>1$ and condone $n>0$.

        An alternative form for ${f_n}^{\text{'}}\left(\frac{a}{4}\right)$ is $\left(2n\right){\left(3\right)}^{n-1}{\left(\frac{a}{4}\right)}^{2n-1}$.

THEN

hence ${f_n}^{\text{'}}\left(\frac{a}{4}\right)>0$                 AG

METHOD 2

$f_n\left(0\right)=0$ and $f_n\left(\frac{a}{2}\right)>0$            A1

(since $f_n$ is continuous and there are no stationary points between $x=0$ and $x=\frac{a}{2}$)

the gradient (of the curve) must be positive between $x=0$ and $x=\frac{a}{2}$                 R1

Note: Do not award A0R1.

hence ${f_n}^{\text{'}}\left(\frac{a}{4}\right)>0$                 AG

[2 marks]

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

for $n$ even:

$n{\left(-1\right)}^{n-1}\left(=-n\right)<0$  (and $\left(a+2\right), {\left(a+1\right)}^{n-1}$ are both $>0$)                R1

${f_n}^{\text{'}}\left(-1\right)<0$            A1

${f_n}^{\text{'}}\left(0\right)=0$ and ${f_n}^{\text{'}}\left(\frac{a}{4}\right)>0$  (seen anywhere)            A1

Note: Candidates can give arguments based on the sign of ${\left(-1\right)}^{n-1}$ to obtain the R mark.
        For example, award R1 for the following:
        If $n$ is even, then $n-1$ is odd and hence ${\left(-1\right)}^{n-1}<0 \left(=-1\right)$.
        Do not award R0A1.
        The second A1 is independent of the other two marks.
        The A marks can be awarded for correct descriptions expressed in words.
        Candidates can state $(0, 0)$ as a point of zero gradient from part (d) or show, state or explain (words or diagram) that ${f_n}^{\text{'}}\left(0\right)=0$. The last A mark can be awarded for a clearly labelled diagram showing changes in the sign of the gradient.
        The last A1 can be awarded for use of a specific case (e.g. $n=2$).

hence $(0, 0)$ is a local minimum point            AG

[3 marks]

for $n$ odd:

$n{\left(-1\right)}^{n-1}\left(=n\right)<0$, (and $\left(a+2\right), {\left(a+1\right)}^{n-1}$ are both $>0$)  so ${f_n}^{\text{'}}\left(-1\right)>0$               R1

Note: Candidates can give arguments based on the sign of ${\left(-1\right)}^{n-1}$ to obtain the R mark.
        For example, award R1 for the following:
        If $n$ is odd, then $n-1$ is even and hence ${\left(-1\right)}^{n-1}>0 \left(=1\right)$.

${f_n}^{\text{'}}\left(0\right)=0$ and ${f_n}^{\text{'}}\left(\frac{a}{4}\right)>0$  (seen anywhere)            A1

Note: The A1 is independent of the R1.
         Candidates can state $\left(0, 0\right)$ as a point of zero gradient from part (d) or show, state or explain (words or diagram) that ${f_n}^{\text{'}}\left(0\right)=0$. The last A mark can be awarded for a clearly labelled diagram showing changes in the sign of the gradient.
        The last A1 can be awarded for use of a specific case (e.g. $n=3$).

hence $(0, 0)$ is a point of inflexion with zero gradient           AG

[2 marks]

considers the parity of $n$            (M1)

Note: Award M1 for stating at least one specific even value of $n$.

$n$ must be even (for four solutions)           A1

Note: The above 2 marks are independent of the 3 marks below.

$0<k<{\left(\frac{a}{2}\right)}^{2n}$           A1A1A1

Note: Award A1 for the correct lower endpoint, A1 for the correct upper endpoint and A1 for strict inequality signs.

         The third A1 (strict inequality signs) can only be awarded if A1A1 has been awarded.
         For example, award A1A1A0 for $0\le k\le {\left(\frac{a}{2}\right)}^{2n}$. Award A1A0A0 for $k>0$.

         Award A1A0A0 for $0<k<f_n\left(\frac{a}{2}\right)$.

[5 marks]