f: $x \mapsto x^2 - 2x$, g: $x \mapsto 2x + 3$

**(i)** $x^2 - 2x - 15 = 0$ End-points $-3$ and $5$
| M1 A1 | Equation set to 0 and solved; Correct end-points, however used |

$\to x < -3$ and $x > 5$
| A1 [3] | Co-inequalities – not $\leq$ or $\geq$ |

**(ii)** Uses $dy/dx = 2x - 2 = 0$ or $(x-1)^2 - 1$ Minimum at $x = 1$ or correct form
| M1 A1 | Any valid complete method for $x$ value; Correct only |

Range of $y$ is $f(x) \geq -1$
| A1 | Correct for his value of "$x$" – must be $\geq$ |

No inverse since not 1 : 1 (or equivalent)
| B1 [4] | Any valid statement. |

**(iii)** $gf(x) = 2(x^2 - 2x) + 3$ $(2x^2 - 4x + 3)$
| M1 | Must be $gf$ not $fg$ – for unsimplified ans. |

$b^2 - 4ac = 16 - 24 = -8 \to -ve$
| M1 | Used on quadratic=0, even if $fg$ used. |

$\to$ No real solutions.
| A1 [3] | Must be using $gf$ and correct assumption and statement needed. |

[or $gf(x)=0 \to f(x)=-3/2$. Imposs from (ii) ]

**(iv)** $y = 2x + 3$ correct line on diagram | Either inverse as mirror image in $y=x$ or $y = g^{-1}(x) = \frac{1}{2}(x-3)$ drawn
| B2,1,0 [2] | 3 things needed – B1 if one missing: • $g$ correct; • $g^{-1}$ correct – not parallel to $g$; • $y=x$ drawn or statement re symmetry |

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## General Marking Notes:

**DM1 for quadratic equation:** Equation must be set to 0. Formula → must be correct and correctly used – allow for numerical errors though in $b^2$ and $-4ac$. Factors → attempt to find 2 brackets. Each bracket then solved to 0.