$y = x^2 \ln x$ | M1 | Product rule
$\frac{dy}{dx} = x^2 \cdot \frac{1}{x} + 2x \ln x$ | B1 | $\frac{d}{dx}(\ln x) = 1/x$ soi oe
 | A1 |
$= x + 2x \ln x$ | M1 | their deriv = 0 or attempt to verify
$\frac{dy}{dx} = 0$ when $x + 2 \ln x = 0$ | M1 | their deriv = 0 or attempt to verify
$\Rightarrow x(1 + 2\ln x) = 0$ | M1 |
$\Rightarrow \ln x = -\frac{1}{2} \Rightarrow x = e^{-1/2} = 1/\sqrt{e}$ | E1 |
 | [6] |

# Question 4(i)

$y = \frac{x^3}{2x + 1}$ | M1 | Use of quotient rule (or product rule)
$\frac{dy}{dx} = \frac{(2x+1) \cdot 3x^2 - x^3 \cdot 2}{(2x+1)^2}$ | A1 | Correct numerator – condone missing bracket provided it is treated as present
$= \frac{2x^2 + 2x}{(2x+1)^2} = \frac{2x(x+1)}{(2x+1)^2}$ | A1 | Correct denominator
 | E1 | www – do not condone missing brackets
 | [4] |

# Question 4(ii)

$\frac{dy}{dx} = 0$ when $2x(x+1) = 0$ | B1 B1 | Must be from correct working:
$\Rightarrow x = 0$ or $x = -1$ | B1 B1 | SC –1 if denominator = 0
$y = 0$ or $y = -1$ | [4] |

# Question 5(i)

$\frac{1}{2}(1 + 2x)^{-1/2} \times 2$ | M1 | Chain rule
$= \frac{1}{\sqrt{1+2x}}$ | B1 | $\frac{1}{2}u^{-1/2}$ or $\frac{1}{2}(1+2x)^{-1/2}$
 | A1 | oe, but must resolve $\frac{1}{2} \times 2 = 1$
 | [3] |

# Question 5(ii)

$y = \ln(1 - e^{-x})$ | M1 | Chain rule
$\frac{dy}{dx} = \frac{1}{1-e^{-x}} \cdot (-e^{-x})(-1)$ | B1 | $\frac{1}{1-e^{-x}}$ or $\frac{1}{u}$ if substituting $u = 1 - e^{-x}$
$= \frac{e^{-x}}{1-e^{-x}}$ | A1 | $\times(-e^{-x})(-1)$ or $e^{-x}$
$= \frac{1}{e^x - 1}$ | E1 | www (may imply $xe^s$ top and bottom)
 | [4] |

# Question 6(i)

$f(-x) = \frac{\sin(-x)}{2-\cos(-x)}$ | M1 | substituting $-x$ for $x$ in $f(x)$
$= \frac{-\sin x}{2-\cos x}$ | A1 |
$= -f(x)$ | [3] | Graph completed with rotational symmetry about O.
 | B1 |

# Question 6(ii)

$f'(x) = \frac{(2-\cos x)\cos x - \sin x \cdot \sin x}{(2-\cos x)^2}$ | M1 | Quotient or product rule consistent with their derivatives
$= \frac{2\cos x - \cos^2 x - \sin^2 x}{(2-\cos x)^2}$ | A1 | Correct expression
$= \frac{2\cos x - 1}{(2-\cos x)^2}$ | E1 |
$f'(x) = 0$ when $2\cos x - 1 = 0$ | M1 | numerator = 0
$\Rightarrow \cos x = \frac{1}{2} \Rightarrow x = \pi/3$ | A1 |
When $x = \pi/3$, $y = \frac{\sin(\pi/3)}{2-\cos(\pi/3)} = \frac{\sqrt{3}/2}{2-1/2}$ | M1 | Substituting their $\pi/3$ into $y$
$= \frac{\sqrt{3}}{3}$ | A1 | o.e. but exact
So range is $-\frac{\sqrt{3}}{3} \leq y \leq \frac{\sqrt{3}}{3}$ | B1ft | fit their $\frac{\sqrt{3}}{3}$
 | [8] |

# Question 6(iii)

$\int_0^\pi \frac{\sin x}{2-\cos x}dx$ let $u = 2 - \cos x$ | M1 | $\int \frac{1}{u}du$
$\Rightarrow \frac{du}{dx} = \sin x$ | B1 | $u = 1$ to $3$
When $x = 0$, $u = 1$; when $x = \pi$, $u = 3$ | A1ft | $[\ln u]$
$= \int_1^3 \frac{du}{u}$ | [4] |
$= [\ln u]_1^3$ | A1cao |
$= \ln 3 - \ln 1 = \ln 3$ | [4] |
or $= [\ln(2-\cos x)]_0^\pi$ | M2 | $[k\ln(2-\cos x)]$
$= \ln 3 - \ln 1 = \ln 3$ | A1 | $k = 1$
 | A1 cao |

# Question 6(iv)

Graph showing evidence of stretch s.f. $\frac{1}{2}$ in $x$-direction | B1ft | [1]

# Question 6(v)

Area is stretched with scale factor $\frac{1}{2}$ | M1 | soi
So area is $\frac{1}{2}\ln 3$ | A1ft | $\frac{1}{2}$ their $\ln 3$
 | [2] |

# Question 7(i)

$x - 1 = \sin y$ | M1 |
$\Rightarrow x = 1 + \sin y$ | A1 |
$\Rightarrow \frac{dx}{dy} = \cos y$ | E1 | www

# Question 7(ii)

When $x = 1.5$, $y = \arcsin(0.5) = \pi/6$ | M1 A1 | condone $30°$ or $0.52$ or better
$\frac{dy}{dx} = \frac{1}{\cos y}$ | M1 | or $\frac{dy}{dx} = \frac{1}{\sqrt{1-(x-1)^2}}$
$= \frac{1}{\cos \pi/6}$ | A1 | or equivalent, but must be exact
$= \frac{2}{\sqrt{3}}$ | [7] |