Date | May 2018 | Marks available | 2 | Reference code | 18M.2.sl.TZ1.1 |
Level | SL only | Paper | 2 | Time zone | TZ1 |
Command term | Find | Question number | 1 | Adapted from | N/A |
Question
Let f(x) = ln x − 5x , for x > 0 .
Find f '(x).
Find f "(x).
Solve f '(x) = f "(x).
Markscheme
\(f'\left( x \right) = \frac{1}{x} - 5\) A1A1 N2
[2 marks]
f "(x) = −x−2 A1 N1
[1 mark]
METHOD 1 (using GDC)
valid approach (M1)
eg
0.558257
x = 0.558 A1 N2
Note: Do not award A1 if additional answers given.
METHOD 2 (analytical)
attempt to solve their equation f '(x) = f "(x) (do not accept \(\frac{1}{x} - 5 = - \frac{1}{{{x^2}}}\)) (M1)
eg \(5{x^2} - x - 1 = 0,\,\,\frac{{1 \pm \sqrt {21} }}{{10}},\,\,\frac{1}{x} = \frac{{ - 1 \pm \sqrt {21} }}{2},\,\, - 0.358\)
0.558257
x = 0.558 A1 N2
Note: Do not award A1 if additional answers given.
[2 marks]