Date | November 2009 | Marks available | 17 | Reference code | 09N.1.hl.TZ0.12 |
Level | HL only | Paper | 1 | Time zone | TZ0 |
Command term | Determine, Explain, Find, Show that, and Sketch | Question number | 12 | Adapted from | N/A |
Question
A tangent to the graph of y=lnx passes through the origin.
(a) Sketch the graphs of y=lnx and the tangent on the same set of axes, and hence find the equation of the tangent.
(b) Use your sketch to explain why lnx⩽xe for x>0 .
(c) Show that xe⩽ex for x>0 .
(d) Determine which is larger, πe or eπ .
Markscheme
(a)
A3
Note: Award A1 for each graph
A1 for the point of tangency.
point on curve and line is (a, lna) (M1)
y=ln(x)
dydx=1x⇒dydx=1a(when x=a) (M1)A1
EITHER
gradient of line, m, through (0, 0) and (a, lna) is lnaa (M1)A1
⇒lnaa=1a⇒lna=1⇒a=e⇒m=1e M1A1
OR
y−lna=1a(x−a) (M1)A1
passes through 0 if
lna−1=0 M1
a=e⇒m=1e A1
THEN
∴y=1ex A1
[11 marks]
(b) the graph of lnx never goes above the graph of y=1ex , hence lnx⩽xe R1
[1 mark]
(c) lnx⩽xe⇒elnx⩽x⇒lnxe⩽x M1A1
exponentiate both sides of lnxe⩽x⇒xe⩽ex R1AG
[3 marks]
(d) equality holds when x=e R1
letting x=π⇒πe<eπ A1 N0
[2 marks]
Total [17 marks]
Examiners report
This was the least accessible question in the entire paper, with very few candidates achieving high marks. Sketches were generally done poorly, and candidates failed to label the point of intersection. A ‘dummy’ variable was seldom used in part (a), hence in most cases it was not possible to get more than 3 marks. There was a lot of good guesswork as to the coordinates of the point of intersection, but no reasoning showed. Many candidates started with the conclusion in part (c). In part (d) most candidates did not distinguish between the inequality and strict inequality.