## Question 10(i):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $y' = 3x^2 - 5$ | M1 | |
| their $y' = 0$ | M1 | |
| $(1.3,\ -4.3)$ cao | A1 | or A1 for $x = \pm\sqrt{\frac{5}{3}}$ oe soi |
| $(-1.3,\ 4.3)$ cao | A1 | allow if not written as co-ordinates if pairing is clear; ignore any work relating to second derivative |

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## Question 10(ii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| crosses axes at $(0, 0)$ | B1 | condone $x$ and $y$ intercepts not written as co-ordinates; may be on graph |
| and $(\pm\sqrt{5},\ 0)$ | B1 | $\pm(2.23$ to $2.24)$ implies $\pm\sqrt{5}$ |
| sketch of cubic with turning points in correct quadrants and of correct orientation and passing through origin | B1 | must meet the $x$-axis three times; B0 e.g. if more than 1 point of inflection |
| $x$-intercepts $\pm\sqrt{5}$ marked | B1 | may be in decimal form $(\pm 2.2\ldots)$ |

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## Question 10(iii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| substitution of $x = 1$ in $f'(x) = 3x^2 - 5$ | M1 | sight of $-2$ does not necessarily imply M1; check $f'(x) = 3x^2 - 5$ is correct in part (i) |
| $-2$ | A1 | |
| $y - {-4} = (\text{their } f'(1)) \times (x - 1)$ oe | M1* | or $-4 = -2x(1) + c$ |
| $-2x - 2 = x^3 - 5x$ and completion to given result www | M1dep* | |
| use of Factor theorem in $x^3 - 3x + 2$ with $-1$ or $\pm 2$ | M1 | or any other valid method; must be shown; e.g. long division or comparing coefficients to find $(x-1)(x^2 + x - 2)$ or $(x+2)(x^2 - 2x + 1)$ is enough for M1 with both factors correct |
| $x = -2$ obtained correctly | A1 | NB M0A0 for $x(x^2 - 3) = -2$ so $x = -2$ or $x^2 - 3 = -2$ oe |

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