Date | November 2017 | Marks available | 3 | Reference code | 17N.2.sl.TZ0.5 |
Level | SL only | Paper | 2 | Time zone | TZ0 |
Command term | Find | Question number | 5 | Adapted from | N/A |
Question
Let f(x)=6−ln(x2+2), for x∈R. The graph of f passes through the point (p, 4), where p>0.
Find the value of p.
The following diagram shows part of the graph of f.
The region enclosed by the graph of f, the x-axis and the lines x=−p and x=p is rotated 360° about the x-axis. Find the volume of the solid formed.
Markscheme
valid approach (M1)
egf(p)=4, intersection with y=4, ±2.32
2.32143
p=√e2−2 (exact), 2.32 A1 N2
[2 marks]
attempt to substitute either their limits or the function into volume formula (must involve f2, accept reversed limits and absence of π and/or dx, but do not accept any other errors) (M1)
eg∫2.32−2.32f2, π∫(6−ln(x2+2))2dx, 105.675
331.989
volume=332 A2 N3
[3 marks]