| Date | May 2018 | Marks available | 1 | Reference code | 18M.2.hl.TZ2.10 |
| Level | HL only | Paper | 2 | Time zone | TZ2 |
| Command term | Explain | Question number | 10 | Adapted from | N/A |
Question
Consider the expression \(f\left( x \right) = {\text{tan}}\left( {x + \frac{\pi }{4}} \right){\text{cot}}\left( {\frac{\pi }{4} - x} \right)\).
The expression \(f\left( x \right)\) can be written as \(g\left( t \right)\) where \(t = {\text{tan}}\,x\).
Let \(\alpha \), β be the roots of \(g\left( t \right) = k\), where 0 < \(k\) < 1.
Sketch the graph of \(y = f\left( x \right)\) for \( - \frac{{5\pi }}{8} \leqslant x \leqslant \frac{\pi }{8}\).
With reference to your graph, explain why \(f\) is a function on the given domain.
Explain why \(f\) has no inverse on the given domain.
Explain why \(f\) is not a function for \( - \frac{{3\pi }}{4} \leqslant x \leqslant \frac{\pi }{4}\).
Show that \(g\left( t \right) = {\left( {\frac{{1 + t}}{{1 - t}}} \right)^2}\).
Sketch the graph of \(y = g\left( t \right)\) for t ≤ 0. Give the coordinates of any intercepts and the equations of any asymptotes.
Find \(\alpha \) and β in terms of \(k\).
Show that \(\alpha \) + β < −2.
Markscheme
A1A1
A1 for correct concavity, many to one graph, symmetrical about the midpoint of the domain and with two axes intercepts.
Note: Axes intercepts and scales not required.
A1 for correct domain
[2 marks]
for each value of \(x\) there is a unique value of \(f\left( x \right)\) A1
Note: Accept “passes the vertical line test” or equivalent.
[1 mark]
no inverse because the function fails the horizontal line test or equivalent R1
Note: No FT if the graph is in degrees (one-to-one).
[1 mark]
the expression is not valid at either of \(x = \frac{\pi }{4}\,\,\left( {{\text{or}} - \frac{{3\pi }}{4}} \right)\) R1
[1 mark]
METHOD 1
\(f\left( x \right) = \frac{{{\text{tan}}\left( {x + \frac{\pi }{4}} \right)}}{{{\text{tan}}\left( {\frac{\pi }{4} - x} \right)}}\) M1
\( = \frac{{\frac{{{\text{tan}}\,x + {\text{tan}}\,\frac{\pi }{4}}}{{1 - {\text{tan}}\,x\,{\text{tan}}\,\frac{\pi }{4}}}}}{{\frac{{{\text{tan}}\,\frac{\pi }{4} - {\text{tan}}\,x}}{{1 + {\text{tan}}\,\frac{\pi }{4}{\text{tan}}\,x}}}}\) M1A1
\( = {\left( {\frac{{1 + t}}{{1 - t}}} \right)^2}\) AG
METHOD 2
\(f\left( x \right) = {\text{tan}}\left( {x + \frac{\pi }{4}} \right){\text{tan}}\left( {\frac{\pi }{2} - \frac{\pi }{4} + x} \right)\) (M1)
\( = {\text{ta}}{{\text{n}}^2}\left( {x + \frac{\pi }{4}} \right)\) A1
\(g\left( t \right) = {\left( {\frac{{{\text{tan}}\,x + {\text{tan}}\,\frac{\pi }{4}}}{{1 - {\text{tan}}\,x\,{\text{tan}}\,\frac{\pi }{4}}}} \right)^2}\) A1
\( = {\left( {\frac{{1 + t}}{{1 - t}}} \right)^2}\) AG
[3 marks]
for t ≤ 0, correct concavity with two axes intercepts and with asymptote \(y\) = 1 A1
t intercept at (−1, 0) A1
\(y\) intercept at (0, 1) A1
[3 marks]
METHOD 1
\(\alpha \), β satisfy \(\frac{{{{\left( {1 + t} \right)}^2}}}{{{{\left( {1 - t} \right)}^2}}} = k\) M1
\(1 + {t^2} + 2t = k\left( {1 + {t^2} - 2t} \right)\) A1
\(\left( {k - 1} \right){t^2} - 2\left( {k + 1} \right)t + \left( {k - 1} \right) = 0\) A1
attempt at using quadratic formula M1
\(\alpha \), β \( = \frac{{k + 1 \pm 2\sqrt k }}{{k - 1}}\) or equivalent A1
METHOD 2
\(\alpha \), β satisfy \(\frac{{1 + t}}{{1 - t}} = \left( \pm \right)\sqrt k \) M1
\(t + \sqrt k t = \sqrt k - 1\) M1
\(t = \frac{{\sqrt k - 1}}{{\sqrt k + 1}}\) (or equivalent) A1
\(t - \sqrt k t = - \left( {\sqrt k + 1} \right)\) M1
\(t = \frac{{\sqrt k + 1}}{{\sqrt k - 1}}\) (or equivalent) A1
so for eg, \(\alpha = \frac{{\sqrt k - 1}}{{\sqrt k + 1}}\), β \( = \frac{{\sqrt k + 1}}{{\sqrt k - 1}}\)
[5 marks]
\(\alpha \) + β \( = 2\frac{{\left( {k + 1} \right)}}{{\left( {k - 1} \right)}}\,\left( { = - 2\frac{{\left( {1 + k} \right)}}{{\left( {1 - k} \right)}}} \right)\) A1
since \(1 + k > 1 - k\) R1
\(\alpha \) + β < −2 AG
Note: Accept a valid graphical reasoning.
[2 marks]
