Show that \(f'(x) = {{\rm{e}}^{2x}}(2\cos x - \sin x)\) .

Let the line L be the normal to the curve of f at \(x = 0\) .

Find the equation of L .

The graph of f and the line L intersect at the point (0, 1) and at a second point P.

(i)     Find the x-coordinate of P.

(ii)    Find the area of the region enclosed by the graph of f and the line L .

correctly finding the derivative of  \({{\rm{e}}^{2x}}\) , i.e. \(2{{\rm{e}}^{2x}}\)     A1

correctly finding the derivative of  \(\cos x\) , i.e. \( - \sin x\)     A1

evidence of using the product rule, seen anywhere     M1

e.g. \(f'(x) = 2{{\rm{e}}^{2x}}\cos x - {{\rm{e}}^{2x}}\sin x\)

\(f'(x) = 2{{\rm{e}}^{2x}}(2\cos x - \sin x)\)     AG     N0

[3 marks]

evidence of finding \(f(0) = 1\) , seen anywhere     A1

attempt to find the gradient of f     (M1)

e.g. substituting \(x = 0\) into \(f'(x)\)

value of the gradient of f     A1

e.g. \(f'(0) = 2\) , equation of tangent is \(y = 2x + 1\)

gradient of normal \( = - \frac{1}{2}\)     (A1)

\(y - 1 = - \frac{1}{2}x\left( {y = - \frac{1}{2}x + 1} \right)\)     A1     N3

[5 marks]

(i) evidence of equating correct functions     M1

e.g. \({{\rm{e}}^{2x}}\cos x = - \frac{1}{2}x + 1\) , sketch showing intersection of graphs    

\(x = 1.56\)     A1     N1

(ii) evidence of approach involving subtraction of integrals/areas     (M1)

e.g. \(\int {\left[ {f(x) - g(x)} \right]} {\rm{d}}x\) , \(\int {f(x)} {\rm{d}}x - {\text{area under trapezium}}\)

fully correct integral expression     A2

e.g. \(\int_0^{1.56} {\left[ {{{\rm{e}}^{2x}}\cos x - \left( { - \frac{1}{2}x + 1} \right)} \right]} {\rm{d}}x\) , \(\int_0^{1.56} {{{\rm{e}}^{2x}}\cos x} {\rm{d}}x - 0.951 \ldots \)

\({\rm{area}} = 3.28\)     A1     N2

[6 marks]

A good number of candidates demonstrated the ability to apply the product and chain rules to obtain the given derivative.

Where candidates recognized that the gradient of the tangent is the derivative, many went on to correctly find the equation of the normal.

Few candidates showed the setup of the equation in part (c) before writing their answer from the GDC. Although a good number of candidates correctly expressed the integral to find the area between the curves, surprisingly few found a correct answer. Although this is a GDC paper, some candidates attempted to integrate this function analytically.