Processing math: 100%

User interface language: English | Español

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 lnxxe for x>0 .

(c)     Show that xeex 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=1xdydx=1a(when x=a)     (M1)A1

EITHER

gradient of line, m, through (0, 0) and (a, lna) is lnaa     (M1)A1

lnaa=1alna=1a=em=1e     M1A1

OR

ylna=1a(xa)     (M1)A1

passes through 0 if

lna1=0     M1

a=em=1e     A1

THEN

y=1ex     A1

[11 marks]

 

(b)     the graph of lnx never goes above the graph of y=1ex , hence lnxxe     R1

[1 mark]

 

(c)     lnxxeelnxxlnxex     M1A1

exponentiate both sides of lnxexxeex     R1AG

[3 marks]

 

(d)     equality holds when x=e     R1

letting x=ππe<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.

Syllabus sections

Topic 6 - Core: Calculus » 6.1 » Finding equations of tangents and normals.
Show 37 related questions

View options