Date | May 2014 | Marks available | 2 | Reference code | 14M.1.hl.TZ1.11 |
Level | HL only | Paper | 1 | Time zone | TZ1 |
Command term | Show that | Question number | 11 | Adapted from | N/A |
Question
Consider the function f(x)=lnxx, x>0.
The sketch below shows the graph of y= f(x) and its tangent at a point A.
Show that f′(x)=1−lnxx2.
Find the coordinates of B, at which the curve reaches its maximum value.
Find the coordinates of C, the point of inflexion on the curve.
The graph of y= f(x) crosses the x-axis at the point A.
Find the equation of the tangent to the graph of f at the point A.
The graph of y= f(x) crosses the x-axis at the point A.
Find the area enclosed by the curve y=f(x), the tangent at A, and the line x=e.
Markscheme
f′(x)=x×1x−lnxx2 M1A1
=1−lnxx2 AG
[2 marks]
1−lnxx2=0 has solution x=e M1A1
y=1e A1
hence maximum at the point (e, 1e)
[3 marks]
f″(x)=x2(−1x)−2x(1−lnx)x4 M1A1
=2lnx−3x3
Note: The M1A1 should be awarded if the correct working appears in part (b).
point of inflexion where f″(x)=0 M1
so x=e32, y=32e−32 A1A1
C has coordinates (e32, 32e−32)
[5 marks]
f(1)=0 A1
f′(1)=1 (A1)
y=x+c (M1)
through (1, 0)
equation is y=x−1 A1
[4 marks]
METHOD 1
area =∫e1x−1−lnxxdx M1A1A1
Note: Award M1 for integration of difference between line and curve, A1 for correct limits, A1 for correct expressions in either order.
∫lnxxdx=(lnx)22(+c) (M1)A1
∫(x−1)dx=x22−x(+c) A1
=[12x2−x−12(lnx)2]e1
=(12e2−e−12)−(12−1)
=12e2−e A1
METHOD 2
area = area of triangle −∫e1lnxxdx M1A1
Note: A1 is for correct integral with limits and is dependent on the M1.
∫lnxxdx=(lnx)22(+c) (M1)A1
area of triangle =12(e−1)(e−1) M1A1
12(e−1)(e−1)−(12)=12e2−e A1
[7 marks]