22M.1.SL.TZ1.5

pestleMathematicsAASLPaper 122M· sl-5-4-tangents-and-normalsource ↗

Consider the curve with equation $y = (2 x - 1) e^{k x}$ , where $x \in ℝ$ and $k \in ℚ$ .

The tangent to the curve at the point where $x = 1$ is parallel to the line $y = 5 e^{k} x$ .

Find the value of $k$ .

Markscheme / solution

evidence of using product rule (M1)

$\frac{d y}{d x} = (2 x - 1) \times (k e^{k x}) + 2 \times e^{k x} (= e^{k x} (2 k x - k + 2))$ A1

correct working for one of (seen anywhere) A1

$\frac{d y}{d x}$ at $x = 1 \Rightarrow k e^{k} + 2 e^{k}$

slope of tangent is $5 e^{k}$

their $\frac{d y}{d x}$ at $x = 1$ equals the slope of $y = 5 e^{k} x (= 5 e^{k})$ (seen anywhere) (M1)

$k e^{k} + 2 e^{k} = 5 e^{k}$

$k = 3$ A1

[5 marks]

Examiners’ report

The product rule was well recognised and used with 𝑥=1 properly substituted into this expression. Although the majority of the candidates tried equating the derivative to the slope of the tangent line, the slope of the tangent line was not correctly identified; many candidates incorrectly substituted 𝑥=1 into the tangent equation, thus finding the y-coordinate instead of the slope.