Date | May 2017 | Marks available | 3 | Reference code | 17M.2.sl.TZ2.8 |
Level | SL only | Paper | 2 | Time zone | TZ2 |
Command term | Find | Question number | 8 | Adapted from | N/A |
Question
Let f(x)=−0.5x4+3x2+2xf(x)=−0.5x4+3x2+2x. The following diagram shows part of the graph of ff.
There are xx-intercepts at x=0x=0 and at x=px=p. There is a maximum at A where x=ax=a, and a point of inflexion at B where x=bx=b.
Find the value of pp.
Write down the coordinates of A.
Write down the rate of change of ff at A.
Find the coordinates of B.
Find the the rate of change of ff at B.
Let RR be the region enclosed by the graph of ff , the xx-axis, the line x=bx=b and the line x=ax=a. The region RR is rotated 360° about the xx-axis. Find the volume of the solid formed.
Markscheme
evidence of valid approach (M1)
egf(x)=0, y=0f(x)=0, y=0
2.73205
p=2.73p=2.73 A1 N2
[2 marks]
1.87938, 8.11721
(1.88, 8.12)(1.88, 8.12) A2 N2
[2 marks]
rate of change is 0 (do not accept decimals) A1 N1
[1 marks]
METHOD 1 (using GDC)
valid approach M1
egf″=0, max/min on f′, x=−1
sketch of either f′ or f″, with max/min or root (respectively) (A1)
x=1 A1 N1
Substituting their x value into f (M1)
egf(1)
y=4.5 A1 N1
METHOD 2 (analytical)
f″=−6x2+6 A1
setting f″=0 (M1)
x=1 A1 N1
substituting their x value into f (M1)
egf(1)
y=4.5 A1 N1
[4 marks]
recognizing rate of change is f′ (M1)
egy′, f′(1)
rate of change is 6 A1 N2
[3 marks]
attempt to substitute either limits or the function into formula (M1)
involving f2 (accept absence of π and/or dx)
egπ∫(−0.5x4+3x2+2x)2dx, ∫1.881f2
128.890
volume=129 A2 N3
[3 marks]