Date | November 2018 | Marks available | 3 | Reference code | 18N.2.AHL.TZ0.H_9 |
Level | Additional Higher Level | Paper | Paper 2 | Time zone | Time zone 0 |
Command term | Hence or otherwise and Solve | Question number | H_9 | Adapted from | N/A |
Question
The function f is defined by f(x)=2lnx+1x−3, 0 < x < 3.
Draw a set of axes showing x and y values between −3 and 3. On these axes
Hence, or otherwise, find the coordinates of the point of inflexion on the graph of y=f(x).
sketch the graph of y=f(x), showing clearly any axis intercepts and giving the equations of any asymptotes.
sketch the graph of y=f−1(x), showing clearly any axis intercepts and giving the equations of any asymptotes.
Hence, or otherwise, solve the inequality f(x)>f−1(x).
Markscheme
finding turning point of y=f′(x) or finding root of y=f″(x) (M1)
x=0.899 A1
y=f(0.899048…)=−0.375 (M1)A1
(0.899, −0.375)
Note: Do not accept x=0.9. Accept y-coordinates rounding to −0.37 or −0.375 but not −0.38.
[4 marks]
smooth curve over the correct domain which does not cross the y-axis
and is concave down for x > 1 A1
x-intercept at 0.607 A1
equations of asymptotes given as x = 0 and x = 3 (the latter must be drawn) A1A1
[4 marks]
attempt to reflect graph of f in y = x (M1)
smooth curve over the correct domain which does not cross the x-axis and is concave down for y > 1 A1
y-intercept at 0.607 A1
equations of asymptotes given as y = 0 and y = 3 (the latter must be drawn) A1
Note: For FT from (i) to (ii) award max M1A0A1A0.
[4 marks]
solve f(x)=f−1(x) or f(x)=x to get x = 0.372 (M1)A1
0 < x < 0.372 A1
Note: Do not award FT marks.
[3 marks]