Consider the curve $y = x{{\text{e}}^x}$ and the line $y = kx,{\text{ }}k \in \mathbb{R}$.

(a)     Let k = 0.

(i)     Show that the curve and the line intersect once.

(ii)     Find the angle between the tangent to the curve and the line at the point of intersection.

(b)     Let k =1. Show that the line is a tangent to the curve.

(c)     (i)     Find the values of k for which the curve $y = x{{\text{e}}^x}$ and the line $y = kx$ meet in two distinct points.

(ii)     Write down the coordinates of the points of intersection.

(iii)     Write down an integral representing the area of the region A enclosed by the curve and the line.

(iv)     Hence, given that $0 < k < 1$, show that $A < 1$.

(a)     (i)     $x{{\text{e}}^x} = 0 \Rightarrow x = 0$     A1

so, they intersect only once at (0, 0)

(ii)     $y' = {{\text{e}}^x} + x{{\text{e}}^x} = (1 + x){{\text{e}}^x}$     M1A1

$y'(0) = 1$     A1

$\theta = \arctan 1 = \frac{\pi }{4}{\text{ }}(\theta = 45^\circ )$     A1

[5 marks]

(b)     when $k = 1,{\text{ }}y = x$

$x{{\text{e}}^x} = x \Rightarrow x({{\text{e}}^x} - 1) = 0$     M1

$\Rightarrow x = 0$     A1

$y'(0) = 1$ which equals the gradient of the line $y = x$     R1

so, the line is tangent to the curve at origin     AG

Note: Award full credit to candidates who note that the equation $x({{\text{e}}^x} - 1) = 0$ has a double root x = 0 so y = x is a tangent.

[3 marks]

(c)     (i)     $x{{\text{e}}^x} = kx \Rightarrow x({{\text{e}}^x} - k) = 0$     M1

$\Rightarrow x = 0{\text{ or }}x = \ln k$     A1

$k > 0{\text{ and }}k \ne 1$     A1

(ii)     (0, 0) and $(\ln k,{\text{ }}k\ln k)$     A1A1

(iii)     $A = \left| {\int_0^{\ln k} {kx - x{{\text{e}}^x}{\text{d}}x} } \right|$     M1A1

Note: Do not penalize the omission of absolute value.

(iv)     attempt at integration by parts to find $\int {x{{\text{e}}^x}{\text{d}}x}$     M1

$\int {x{{\text{e}}^x}{\text{d}}x}  = x{{\text{e}}^x} - \int {{{\text{e}}^x}{\text{d}}x = {{\text{e}}^x}(x - 1)}$     A1

as $0 < k < 1 \Rightarrow \ln k < 0$     R1

$A = \int_{\ln k}^0 {kx - x{{\text{e}}^x}{\text{d}}x = \left[ {\frac{k}{2}{x^2} - (x - 1){{\text{e}}^x}} \right]_{\ln k}^0}$     A1

$= 1 - \left( {\frac{k}{2}{{(\ln k)}^2} - (\ln k - 1)k} \right)$     A1

$= 1 - \frac{k}{2}\left( {{{(\ln k)}^2} - 2\ln k + 2} \right)$

$= 1 - \frac{k}{2}\left( {{{(\ln k - 1)}^2} + 1} \right)$     M1A1

since $\frac{k}{2}\left( {{{(\ln k - 1)}^2} + 1} \right) > 0$     R1

$A < 1$     AG

[15 marks]

Total [23 marks]

Many candidates solved (a) and (b) correctly but in (c), many failed to realise that the equation $x{{\text{e}}^x} = kx$ has two roots under certain conditions and that the point of the question was to identify those conditions. Most candidates made a reasonable attempt to write down the appropriate integral in (c)(iii) with the modulus signs and limits often omitted but no correct solution has yet been seen to (c)(iv).