Date | November 2018 | Marks available | 7 | Reference code | 18N.2.AHL.TZ0.H_2 |
Level | Additional Higher Level | Paper | Paper 2 | Time zone | Time zone 0 |
Command term | Find | Question number | H_2 | Adapted from | N/A |
Question
A function f satisfies the conditions f(0)=−4, f(1)=0 and its second derivative is f″(x)=15√x+1(x+1)2, x ≥ 0.
Find f(x).
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
f′(x)=∫(15√x+1(x+1)2)dx=10x32−1x+1(+c) (M1)A1A1
Note: A1 for first term, A1 for second term. Withhold one A1 if extra terms are seen.
f(x)=∫(10x32−1x+1+c)dx=4x52−ln(x+1)+cx+d A1
Note: Allow FT from incorrect f′(x) if it is of the form f′(x)=Ax32+Bx+1+c.
Accept ln|x+1|.
attempt to use at least one boundary condition in their f(x) (M1)
x=0, y=−4
⇒ d=−4 A1
x=1, y=0
⇒ 0=4−ln2+c−4
⇒ c=ln2(=0.693) A1
f(x)=4x52−ln(x+1)+xln2−4
[7 marks]