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).