EXN.1.SL.TZ0.7

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Consider the curve $y = x^{2} - 4 x + 2$ .

Find an expression for $\frac{d y}{d x}$ .

[1]

Show that the normal to the curve at the point where $x = 1$ is $2 y - x + 3 = 0$ .

[6]

Markscheme / solution

* 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.

$\frac{d y}{d x} = 2 x - 4$ A1

[1 mark]

Gradient at $x = 1$ is $- 2$ M1

Gradient of normal is $\frac{1}{2}$ A1

When $x = 1$ $y = 1 - 4 + 2 = - 1$ (M1)A1

EITHER

$y + 1 = \frac{1}{2} (x - 1)$ M1

$2 y + 2 = x - 1$ or $y + 1 = \frac{1}{2} x - \frac{1}{2}$ A1

$- 1 = \frac{1}{2} \times 1 + c$ M1

$y = \frac{1}{2} x - \frac{3}{2}$ A1

THEN

$2 y - x + 3 = 0$ AG

[6 marks]

Examiners’ report

[N/A]