Date | May 2013 | Marks available | 2 | Reference code | 13M.1.hl.TZ2.12 |
Level | HL only | Paper | 1 | Time zone | TZ2 |
Command term | Hence and Show that | Question number | 12 | Adapted from | N/A |
Question
The function f is defined by \(f(x) = \frac{{2x - 1}}{{x + 2}}\), with domain \(D = \{ x: - 1 \leqslant x \leqslant 8\} \).
Express \(f(x)\) in the form \(A + \frac{B}{{x + 2}}\), where \(A\) and \(B \in \mathbb{Z}\).
Hence show that \(f'(x) > 0\) on D.
State the range of f.
(i) Find an expression for \({f^{ - 1}}(x)\).
(ii) Sketch the graph of \(y = f(x)\), showing the points of intersection with both axes.
(iii) On the same diagram, sketch the graph of \(y = f'(x)\).
(i) On a different diagram, sketch the graph of \(y = f(|x|)\) where \(x \in D\).
(ii) Find all solutions of the equation \(f(|x|) = - \frac{1}{4}\).
Markscheme
by division or otherwise
\(f(x) = 2 - \frac{5}{{x + 2}}\) A1A1
[2 marks]
\(f'(x) = \frac{5}{{{{(x + 2)}^2}}}\) A1
> 0 as \({(x + 2)^2} > 0\) (on D) R1AG
Note: Do not penalise candidates who use the original form of the function to compute its derivative.
[2 marks]
\(S = \left[ { - 3,\frac{3}{2}} \right]\) A2
Note: Award A1A0 for the correct endpoints and an open interval.
[2 marks]
(i) EITHER
rearrange \(y = f(x)\) to make x the subject M1
obtain one-line equation, e.g. \(2x - 1 = xy + 2y\) A1
\(x = \frac{{2y + 1}}{{2 - y}}\) A1
OR
interchange x and y M1
obtain one-line equation, e.g. \(2y - 1 = xy + 2x\) A1
\(y = \frac{{2x + 1}}{{2 - x}}\) A1
THEN
\({f^{ - 1}}(x) = \frac{{2x + 1}}{{2 - x}}\) A1
Note: Accept \(\frac{5}{{2 - x}} - 2\)
(ii), (iii)
A1A1A1A1
[8 marks]
Note: Award A1 for correct shape of \(y = f(x)\).
Award A1 for x intercept \(\frac{1}{2}\) seen. Award A1 for y intercept \( - \frac{1}{2}\) seen.
Award A1 for the graph of \(y = {f^{ - 1}}(x)\) being the reflection of \(y = f(x)\) in the line \(y = x\). Candidates are not required to indicate the full domain, but \(y = f(x)\) should not be shown approaching \(x = - 2\). Candidates, in answering (iii), can FT on their sketch in (ii).
(i)
A1A1A1
Note: A1 for correct sketch \(x > 0\), A1 for symmetry, A1 for correct domain (from –1 to +8).
Note: Candidates can FT on their sketch in (d)(ii).
(ii) attempt to solve \(f(x) = - \frac{1}{4}\) (M1)
obtain \(x = \frac{2}{9}\) A1
use of symmetry or valid algebraic approach (M1)
obtain \(x = - \frac{2}{9}\) A1
[7 marks]
Examiners report
Generally well done.
In their answers to Part (b), most candidates found the derivative, but many assumed it was obviously positive.
Part (d)(i) Generally well done, but some candidates failed to label their final expression as \({f^{ - 1}}(x)\). Part (d)(ii) Marks were lost by candidates who failed to mark the intercepts with values.
Marks were also lost in this part and in part (e)(i) for graphs that went beyond the explicitly stated domain.