Date | May 2008 | Marks available | 7 | Reference code | 08M.2.hl.TZ2.6 |
Level | HL only | Paper | 2 | Time zone | TZ2 |
Command term | Find and Justify | Question number | 6 | Adapted from | N/A |
Question
Consider the curve with equation f(x)=e−2x2 for x<0f(x)=e−2x2 for x<0 .
Find the coordinates of the point of inflexion and justify that it is a point of inflexion.
Markscheme
METHOD 1
EITHER
Using the graph of y=f′(x) (M1)
A1
The maximum of f′(x) occurs at x = −0.5 . A1
OR
Using the graph of y=f″(x). (M1)
A1
The zero of f″(x) occurs at x = −0.5 . A1
THEN
Note: Do not award this A1 for stating x = ±0.5 as the final answer for x .
f(−0.5)=0.607(=e−0.5) A2
Note: Do not award this A1 for also stating (0.5, 0.607) as a coordinate.
EITHER
Correctly labelled graph of f′(x) for x<0 denoting the maximum f′(x) R1
(e.g. f′(−0.6)=1.17 and f′(−0.4)=1.16 stated) A1 N2
OR
Correctly labelled graph of f″(x) for x<0 denoting the maximum f′(x) R1
(e.g. f″(−0.6)=0.857 and f″(−0.4)=−1.05 stated) A1 N2
OR
f′(0.5)≈1.21. f′(x)<1.21 just to the left of x=−12
and f′(x)<1.21 just to the right of x=−12 R1
(e.g. f′(−0.6)=1.17 and f′(−0.4)=1.16 stated) A1 N2
OR
f″(x)>0 just to the left of x=−12 and f″(x)<0 just to the right of x=−12 R1
(e.g. f″(−0.6)=0.857 and f″(−0.4)=−1.05 stated) A1 N2
[7 marks]
METHOD 2
f′(x)=−4xe−2x2 A1
f″(x)=−4e−2x2+16x2e−2x2(=(16x2−4)e−2x2) A1
Attempting to solve f″(x)=0 (M1)
x=−12 A1
Note: Do not award this A1 for stating x=±12 as the final answer for x .
f(−12)=1√e (=0.607) A1
Note: Do not award this A1 for also stating (12,1√e) as a coordinate.
EITHER
Correctly labelled graph of f′(x) for x<0 denoting the maximum f′(x) R1
(e.g. f′(−0.6)=1.17 and f′(−0.4)=1.16 stated) A1 N2
OR
Correctly labelled graph of f″(x) for x<0 denoting the maximum f′(x) R1
(e.g. f″(−0.6)=0.857 and f″(−0.4)=−1.05 stated) A1 N2
OR
f′(0.5)≈1.21. f′(x)<1.21 just to the left of x=−12
and f′(x)<1.21 just to the right of x=−12 R1
(e.g. f′(−0.6)=1.17 and f′(−0.4)=1.16 stated) A1 N2
OR
f″(x)>0 just to the left of x=−12 and f″(x)<0 just to the right of x=−12 R1
(e.g. f″(−0.6)=0.857 and f″(−0.4)=−1.05 stated) A1 N2
[7 marks]
Examiners report
Most candidates adopted an algebraic approach rather than a graphical approach. Most candidates found f′(x) correctly, however when attempting to find f″(x), a surprisingly large number either made algebraic errors using the product rule or seemingly used an incorrect form of the product rule. A large number ignored the domain restriction and either expressed x=±12 as the x-coordinates of the point of inflection or identified x=12 rather than x=−12. Most candidates were unsuccessful in their attempts to justify the existence of the point of inflection.