| Date | May 2018 | Marks available | 2 | Reference code | 18M.2.sl.TZ1.1 |
| Level | SL only | Paper | 2 | Time zone | TZ1 |
| Command term | Solve | Question number | 1 | Adapted from | N/A |
Question
Let f(x) = ln x − 5x , for x > 0 .
Find f '(x).
[2]
a.
Find f "(x).
[1]
b.
Solve f '(x) = f "(x).
[2]
c.
Markscheme
\(f'\left( x \right) = \frac{1}{x} - 5\) A1A1 N2
[2 marks]
a.
f "(x) = −x−2 A1 N1
[1 mark]
b.
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]
c.
Examiners report
[N/A]
a.
[N/A]
b.
[N/A]
c.
Syllabus sections
Topic 6 - Calculus » 6.2 » Derivative of \({x^n}\left( {n \in \mathbb{Q}} \right)\) , \(\sin x\) , \(\cos x\) , \(\tan x\) , \({{\text{e}}^x}\) and \(\ln x\) .
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