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Date November 2018 Marks available 4 Reference code 18N.2.AHL.TZ0.H_9
Level Additional Higher Level Paper Paper 2 Time zone Time zone 0
Command term Find Question number H_9 Adapted from N/A

Question

The function f is defined by f(x)=2lnx+1x3, 0 < x < 3.

Draw a set of axes showing x and y values between −3 and 3. On these axes

Find f(x).

[4]
a.

Hence, or otherwise, find the coordinates of the point of inflexion on the graph of y=f(x).

[4]
b.

sketch the graph of y=f(x), showing clearly any axis intercepts and giving the equations of any asymptotes.

[4]
c.i.

sketch the graph of y=f1(x), showing clearly any axis intercepts and giving the equations of any asymptotes.

[4]
c.ii.

Hence, or otherwise, solve the inequality f(x)>f1(x).

[3]
d.

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

METHOD 1

f(x)=2(x3)x(2lnx+1)(x3)2(=2(x3)x(2lnx+1)x(x3)2)      (M1)A1A1A1

Note: Award M1 for attempt at quotient rule, A1A1 for numerator and A1 for denominator.

 

METHOD 2

f(x)=(2lnx+1)(x3)1      (A1)

f(x)=(2x)(x3)1(2lnx+1)(x3)2(=2(x3)x(2lnx+1)x(x3)2)      (M1)A1A1

Note: Award M1 for attempt at product rule, A1 for first term, A1 for second term.

 

[4 marks]

a.

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]

b.

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]

c.i.

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]

c.ii.

solve f(x)=f1(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]

d.

Examiners report

[N/A]
a.
[N/A]
b.
[N/A]
c.i.
[N/A]
c.ii.
[N/A]
d.

Syllabus sections

Topic 5—Calculus » AHL 5.9—Differentiating standard functions and derivative rules
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