(i)   Explain why the inverse function \({f^{ - 1}}\) does not exist.

(ii)     Show that the equation of the normal to the curve at the point P where \(x = \ln 3\) is given by \(9x + 12y - 9\ln 3 - 20 = 0\).

(iii)     Find the x-coordinates of the points Q and R on the curve such that the tangents at Q and R pass through \({\text{(0, 0)}}\).

The domain of \(f\) is now restricted to \(x \geqslant 0\).

(i)     Find an expression for \({f^{ - 1}}(x)\).

(ii)     Find the volume generated when the region bounded by the curve \(y = f(x)\) and the lines \(x = 0\) and \(y = 5\) is rotated through an angle of \(2\pi \) radians about the y-axis.

(i)     either counterexample or sketch or

recognising that \(y = k{\text{ }}(k > 1)\) intersects the graph of \(y = f(x)\) twice     M1

function is not \(1 - 1\) (does not obey horizontal line test)     R1

so \({f^{ - 1}}\) does not exist     AG

(ii)   \(f'(x) = \frac{1}{2}\left( {{{\text{e}}^x} - {{\text{e}}^{ - x}}} \right)\)     (A1)

\(f'(\ln 3) = \frac{4}{3}{\text{ }}( = 1.33)\)     (A1)

\(m =  - \frac{3}{4}\)     M1

\(f(\ln 3) = \frac{5}{3}{\text{ }}( = 1.67)\)     A1

EITHER

\(\frac{{y - \frac{5}{3}}}{{x - \ln 3}} =  - \frac{3}{4}\)     M1

\(4y - \frac{{20}}{3} =  - 3x + 3\ln 3\)     A1

OR

\(\frac{5}{3} =  - \frac{3}{4}\ln 3 + c\)     M1

\(c = \frac{5}{3} + \frac{3}{4}\ln 3\)

\(y =  - \frac{3}{4}x + \frac{5}{3} + \frac{3}{4}\ln 3\)     A1

\(12y =  - 9x + 20 + 9\ln 3\)

THEN

\(9x + 12y - 9\ln 3 - 20 = 0\)     AG

(iii)     The tangent at \(\left( {a,{\text{ }}f(a)} \right)\) has equation \(y - f(a) = f'(a)(x - a)\).     (M1)

\(f'(a) = \frac{{f(a)}}{a}\) (or equivalent)     (A1)

\({{\text{e}}^a} - {{\text{e}}^{ - a}} = \frac{{{{\text{e}}^a} + {{\text{e}}^{ - a}}}}{a}\) (or equivalent)     A1

attempting to solve for a     (M1)

\(a =  \pm 1.20\)     A1A1

[14 marks]

(i)     \(2y = {{\text{e}}^x} + {{\text{e}}^{ - x}}\)

\({{\text{e}}^{2x}} - 2y{{\text{e}}^x} + 1 = 0\)     M1A1

Note:     Award M1 for either attempting to rearrange or interchanging x and y.

\({{\text{e}}^x} = \frac{{2y \pm \sqrt {4{y^2} - 4} }}{2}\)     A1

\({{\text{e}}^x} = y \pm \sqrt {{y^2} - 1} \)

\(x = \ln \left( {y \pm \sqrt {{y^2} - 1} } \right)\)     A1

\({f^{ - 1}}(x) = \ln \left( {x + \sqrt {{x^2} - 1} } \right)\)     A1

Note:     Award A1 for correct notation and for stating the positive “branch”.

(ii)     \(V = \pi \int_1^5 {{{\left( {\ln \left( {y + \sqrt {{y^2} - 1} } \right)} \right)}^2}{\text{d}}y} \)     (M1)(A1)

Note:     Award M1 for attempting to use \(V = \pi \int_c^d {{x^2}{\text{d}}y} \).

\( = 37.1{\text{ }}\left( {{\text{unit}}{{\text{s}}^3}} \right)\)     A1

[8 marks]

In part (a) (i), successful candidates typically sketched the graph of \(y = f(x)\), applied the horizontal line test to the graph and concluded that the function was not \(1 - 1\) (it did not obey the horizontal line test).

In part (a) (ii), a large number of candidates were able to show that the equation of the normal at point P was \(9x + 12y - 9\ln 3 - 20 = 0\). A few candidates used the gradient of the tangent rather than using it to find the gradient of the normal.

Part (a) (iii) challenged most candidates. Most successful candidates graphed \(y = f(x)\) and \(y = xf'(x)\) on the same set of axes and found the x-coordinates of the intersection points.

Part (b) (i) challenged most candidates. While a large number of candidates seemed to understand how to find an inverse function, poor algebra skills (e.g. erroneously taking the natural logarithm of both sides) meant that very few candidates were able to form a quadratic in either \({{\text{e}}^x}\) or \({{\text{e}}^y}\).