Sketch on the same axes the curve \(y = \left| {\frac{7}{{x - 4}}} \right|\) and the line \(y = x + 2\), clearly indicating any axes intercepts and any asymptotes.

Find the exact solutions to the equation \(x + 2 = \left| {\frac{7}{{x - 4}}} \right|\).

A1 for vertical asymptote and for the \(y\)-intercept \(\frac{7}{4}\)

A1 for general shape of \(y = \left| {\frac{7}{{x - 4}}} \right|\) including the \(x\)-axis as asymptote

A1 for straight line with \(y\)-intercept 2 and \(x\)-intercept of \( - 2\)     A1A1A1

[3 marks]

METHOD 1

for \(x > 4\)

\((x + 2)(x - 4) = 7\)     (M1)

\({x^2} - 2x - 8 = 7 \Rightarrow {x^2} - 2x - 15 = 0\)

\((x - 5)(x + 3) = 0\)

\(({\text{as }}x > 4{\text{ then}}){\text{ }}x = 5\)     A1

Note:     Award A0 if \(x =  - 3\) is also given as a solution.

for \(x < 4\)

\((x + 2)(x - 4) =  - 7\)     M1

\( \Rightarrow {x^2} - 2x - 1 = 0\)

\(x = \frac{{2 \pm \sqrt {4 + 4} }}{2} = 1 \pm \sqrt 2 \)     (M1)A1

Note:     Second M1 is dependent on first M1.

[5 marks]

METHOD 2

\({(x + 2)^2} = \frac{{49}}{{{{(x - 4)}^2}}}\)     M1

\({x^4} - 4{x^3} - 12{x^2} + 32x + 15 = 0\)     A1

\((x + 3)(x - 5)({x^2} - 2x - 1) = 0\)

\(x = 5\)     A1

Note:     Award A0 if \(x =  - 3\) is also given as a solution.

\(x = \frac{{2 \pm \sqrt {4 + 4} }}{2} = 1 \pm \sqrt 2 \)     (M1)A1

[5 marks]

Though generally well done, some candidates lost marks unnecessarily by not heeding the instruction to clearly indicate the axes intercepts and asymptotes.

Though this was generally well done, quite a few of the candidates failed to use the graph drawn in part (a) to discount one of the solutions obtained in part (b).