19M.2.AHL.TZ1.H_1

pestleMathematicsAAHLPaper 219M· sl-5-1-introduction-of-differential-calculussource ↗

Let $l$ be the tangent to the curve $y = x{{\text{e}}^{2x}}$ at the point (1, ${{\text{e}}^2}$ ).

Find the coordinates of the point where $l$ meets the $x$ -axis.

Markscheme / solution

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

METHOD 1

equation of tangent is $y = 22.167 \ldots x - 14.778 \ldots$ OR $y = - 7.389 \ldots = 22.167 \ldots \left( {x - 1} \right)$ (M1)(A1)

meets the $x$ -axis when $y = 0$

$x = 0.667$

meets $x$ -axis at (0.667, 0) $\left( { = \left( {\frac{2}{3},\,\,0} \right)} \right)$ A1A1

Note: Award A1 for $x = \frac{2}{3}$ or $x = 0.667$ seen and A1 for coordinates ( $x$ , 0) given.

METHOD 1

Attempt to differentiate (M1)

$\frac{{{\text{d}}y}}{{{\text{d}}x}} = {{\text{e}}^{2x}} + 2x{{\text{e}}^{2x}}$

when $x = 1$ , $\frac{{{\text{d}}y}}{{{\text{d}}x}} = 3{{\text{e}}^2}$ (M1)

equation of the tangent is $y - {{\text{e}}^2} = 3{{\text{e}}^2}\left( {x - 1} \right)$

$y = 3{{\text{e}}^2}x - 2{{\text{e}}^2}$

meets $x$ -axis at $x = \frac{2}{3}$

${\left( {\frac{2}{3},\,\,0} \right)}$ A1A1

Note: Award A1 for $x = \frac{2}{3}$ or $x = 0.667$ seen and A1 for coordinates ( $x$ , 0) given.

[4 marks]

Examiners’ report

[N/A]