Sketch the curve $y=f\left(x\right)$, clearly indicating any asymptotes with their equations and stating the coordinates of any points of intersection with the axes.

Show that $A=\frac{\sqrt{2}\mathrm{\pi }}{2}$.

Find the value of $k$.

Show that $m=-\frac{6x}{{\left(x^2+2\right)}^2}$.

Show that the maximum value of $m$ is $\frac{27}{32}\sqrt{\frac{2}{3}}$.

* This sample question was produced by experienced DP mathematics senior examiners to aid teachers in preparing for external assessment in the new MAA course. There may be minor differences in formatting compared to formal exam papers.

a curve symmetrical about the $y$-axis with correct concavity that has a local maximum point on the positive $y$-axis        A1

a curve clearly showing that $y\to 0$ as $x\to \pm \infty$        A1

$\left(0, \frac{3}{2}\right)$        A1

horizontal asymptote $y=0$ ($x$-axis)        A1

[4 marks]

attempts to find $\int \frac{3}{x^2+2}\mathrm{d}x$        (M1)

$=\left[\frac{3}{\sqrt{2}}\arctan \frac{x}{\sqrt{2}}\right]$        A1

Note: Award M1A0 for obtaining $\left[k \arctan \frac{x}{\sqrt{2}}\right]$ where $k\ne \frac{3}{\sqrt{2}}$.

Note: Condone the absence of or use of incorrect limits to this stage.

$=\frac{3}{\sqrt{2}}\left(\arctan \sqrt{3}-\arctan 0\right)$        (M1)

$=\frac{3}{\sqrt{2}}\times \frac{\mathrm{\pi }}{3}\left(=\frac{\mathrm{\pi }}{\sqrt{2}}\right)$        A1

$A=\frac{\sqrt{2}\mathrm{\pi }}{2}$        AG

[4 marks]

METHOD 1

EITHER

$\underset{0}{\overset{k}{\int }}\frac{3}{x^2+2}\mathrm{d}x=\frac{\sqrt{2}{\displaystyle \mathrm{\pi }}}{4}$

$\frac{3}{\sqrt{2}}\arctan \frac{k}{\sqrt{2}}=\frac{\sqrt{2}\mathrm{\pi }}{4}$        (M1)

OR

$\underset{k}{\overset{\sqrt{6}}{\int }}\frac{3}{x^2+2}\mathrm{d}x=\frac{\sqrt{2}{\displaystyle \mathrm{\pi }}}{4}$

$\frac{3}{\sqrt{2}}\left(\arctan \sqrt{3}-\arctan \frac{k}{\sqrt{2}}\right)=\frac{\sqrt{2}\mathrm{\pi }}{4}$        (M1)

$\arctan \sqrt{3}-\arctan \frac{k}{\sqrt{2}}=\frac{\mathrm{\pi }}{6}$

THEN

$\arctan \frac{k}{\sqrt{2}}=\frac{\mathrm{\pi }}{6}$        A1

$\frac{k}{\sqrt{2}}=\tan \frac{\mathrm{\pi }}{6}\left(=\frac{1}{\sqrt{3}}\right)$        A1

$k=\frac{\sqrt{6}}{3}\left(=\sqrt{\frac{2}{3}}\right)$        A1

METHOD 2

$\underset{0}{\overset{k}{\int }}\frac{3}{x^2+2}\mathrm{d}x=\underset{k}{\overset{\sqrt{6}}{\int }}\frac{3}{x^2+2}\mathrm{d}x$

$\frac{3}{\sqrt{2}}\arctan \frac{k}{\sqrt{2}}=\frac{3}{\sqrt{2}}\left(\arctan \sqrt{3}-\arctan \frac{k}{\sqrt{2}}\right)$        (M1)

$\arctan \frac{k}{\sqrt{2}}=\frac{\mathrm{\pi }}{6}$        A1

$\frac{k}{\sqrt{2}}=\tan \frac{\mathrm{\pi }}{6}\left(=\frac{1}{\sqrt{3}}\right)$        A1

$k=\frac{\sqrt{6}}{3}\left(=\sqrt{\frac{2}{3}}\right)$        A1

[4 marks]

attempts to find $\frac{d}{dx}\left(\frac{3}{x^2+2}\right)$        (M1)

$=\left(3\right)\left(-1\right)\left(2x\right){\left(x^2+2\right)}^{-2}$        A1

so $m=-\frac{6x}{{\left(x^2+2\right)}^2}$        AG

[2 marks]

attempts product rule or quotient rule differentiation        M1

EITHER

$\frac{dm}{dx}=\left(-6x\right)\left(-2\right)\left(2x\right){\left(x^2+2\right)}^{-3}+{\left(x^2+2\right)}^{-2}\left(-6\right)$        A1

OR

$\frac{dm}{dx}=\frac{{\left(x^2+2\right)}^2\left(-6\right)-\left(-6x\right)\left(2\right)\left(2x\right)\left(x^2+2\right)}{{\left(x^2+2\right)}^4}$        A1

Note: Award A0 if the denominator is incorrect. Subsequent marks can be awarded.

THEN

attempts to express their $\frac{dm}{dx}$ as a rational fraction with a factorized numerator        M1

$\frac{dm}{dx}=\frac{6\left(x^2+2\right)\left(3x^2-2\right)}{{\left(x^2+2\right)}^4}\left(=\frac{6\left(3x^2-2\right)}{{\left(x^2+2\right)}^3}\right)$

attempts to solve their $\frac{dm}{dx}=0$ for $x$        M1

$x=\pm \sqrt{\frac{2}{3}}$        A1

from the curve, the maximum value of $m$ occurs at $x=-\sqrt{\frac{2}{3}}$        R1

(the minimum value of $m$ occurs at $x=\sqrt{\frac{2}{3}}$)

Note: Award R1 for any equivalent valid reasoning.

maximum value of $m$ is $-\frac{6\left(-\sqrt{\frac{2}{3}}\right)}{{\displaystyle {\left({\displaystyle {\left(-\sqrt{\frac{2}{3}}\right)}^2}{\displaystyle +}{\displaystyle 2}\right)}^2}}$        A1

leading to a maximum value of $\frac{27}{32}\sqrt{\frac{2}{3}}$        AG

[7 marks]