| Date | May 2013 | Marks available | 3 | Reference code | 13M.1.hl.TZ2.9 |
| Level | HL only | Paper | 1 | Time zone | TZ2 |
| Command term | Show that | Question number | 9 | Adapted from | N/A |
Question
The function f is given by \(f(x) = \frac{{{3^x} + 1}}{{{3^x} - {3^{ - x}}}}\), for x > 0.
Show that \(f(x) > 1\) for all x > 0.
Solve the equation \(f(x) = 4\).
Markscheme
EITHER
\(f(x) - 1 = \frac{{1 + {3^{ - x}}}}{{{3^x} - {3^{ - x}}}}\) M1A1
> 0 as both numerator and denominator are positive R1
OR
\({3^x} + 1 > {3^x} > {3^x} - {3^{ - x}}\) M1A1
Note: Accept a convincing valid argument the numerator is greater than the denominator.
numerator and denominator are positive R1
hence \(f(x) > 1\) AG
[3 marks]
one line equation to solve, for example, \(4({3^x} - {3^{ - x}}) = {3^x} + 1\), or equivalent A1
\(\left( {3{y^2} - y - 4 = 0} \right)\)
attempt to solve a three-term equation M1
obtain \(y = \frac{4}{3}\) A1
\(x = {\log _3}\left( {\frac{4}{3}} \right)\) or equivalent A1
Note: Award A0 if an extra solution for x is given.
[4 marks]
Examiners report
(a) This is a question where carefully organised reasoning is crucial. It is important to state that both the numerator and the denominator are positive for \(x > 0\). Candidates were more successful with part (b) than with part (a).
(a) This is a question where carefully organised reasoning is crucial. It is important to state that both the numerator and the denominator are positive for \(x > 0\). Candidates were more successful with part (b) than with part (a).
