(i)     Sketch the graph of f.

(ii)     By referring to your graph, show that f is a bijection.

Find \({f^{ - 1}}(x)\).

(i)

     A1A1

Note: Award A1 for each part of the piecewise function. Award A1A0 if the two parts of the graph are of the correct shape but f is not continuous at x = 2. Do not penalise a discontinuity in the derivative at x = 2.

(ii)     demonstrating the need to show that f is both an injection and a surjection (seen anywhere)     (R1)

f is an injection by any valid reason eg horizontal line test, strictly increasing function     R1

the range of f is \(\mathbb{R}\) so that f is a surjection     R1

f is therefore a bijection     AG

[5 marks]

considering the linear section, put

\(y = 2x + 1\) or \(x = 2y + 1\)     (M1)

\(x = \frac{{y - 1}}{2}\) or \(y = \frac{{x - 1}}{2}\)     A1

so \({f^{ - 1}}(x) = \frac{{x - 1}}{2},{\text{ }}x \leqslant 5\)     A1

EITHER

\(y = {(x - 1)^2} + 4\)     M1A1

\({(x - 1)^2} = y - 4\)

\(x = 1 \pm \sqrt {y - 4} \)     A1

\(x = 1 + \sqrt {y - 4} \)

taking the + sign to give the right hand half of the parabola     R1

so \({f^{ - 1}}(x) = 1 + \sqrt {x - 4} ,{\text{ }}x > 5\)     A1

OR

considering the quadratic section, put

\(y = {x^2} - 2x + 5\)

\({x^2} - 2x + 5 - y = 0\)     M1

\(x = \frac{{2 \pm \sqrt {4 - 4(5 - y)} }}{2}{\text{ }}( = 1 \pm \sqrt {y - 4} )\)     M1A1

taking the + sign to give the right hand half of the parabola     R1

so \({f^{ - 1}}(x) = \frac{{2 + \sqrt {4 - 4(5 - x)} }}{2},{\text{ }}x > 5{\text{ }}({f^{ - 1}}(x) = 1 + \sqrt {x - 4} ,{\text{ }}x > 5)\)     A1

Note: Award A0 for omission of \({f^{ - 1}}(x)\) or omission of the domain. Penalise the omission of the notation \({f^{ - 1}}(x)\) only once. The domain must be seen in both cases.

[8 marks]

For the most part the piecewise function was correctly graphed. Even though the majority of candidates knew that it is required to establish that the function is an injection and a surjection in order to prove it is a bijection, many just quoted the definition of injection or surjection and did not relate their reason to the graph.

The majority of candidates found the inverse of the first part of the piecewise function but some struggled with the algebra of the second part. In finding the inverse of the quadratic part of the function some candidates omitted the plus or minus sign in front of the square root. Others who had it often forgot to eliminate the negative sign and so did not gain the reasoning mark. Most did not state the correct domain for either part of the inverse function.