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Date May 2021 Marks available 5 Reference code 21M.1.SL.TZ1.8
Level Standard Level Paper Paper 1 (without calculator) Time zone Time zone 1
Command term Find Question number 8 Adapted from N/A

Question

Let y=lnxx4 for x>0.

Consider the function defined by fxlnxx4 for x>0 and its graph y=fx.

Show that dydx=1-4lnxx5.

[3]
a.

The graph of f has a horizontal tangent at point P. Find the coordinates of P.

[5]
b.

Given that f''x=20lnx-9x6, show that P is a local maximum point.

[3]
c.

Solve fx>0 for x>0.

[2]
d.

Sketch the graph of f, showing clearly the value of the x-intercept and the approximate position of point P.

[3]
e.

Markscheme

attempt to use quotient or product rule        (M1)

dydx=x41x-lnx4x3x42  OR  lnx-4x-5+x-41x         A1

correct working         A1

=x31-4lnxx8  OR  cancelling x3  OR  -4lnxx5+1x5

=1-4lnxx5         AG

 

[3 marks]

a.

f'x=dydx=0        (M1)

1-4lnxx5=0

lnx=14        (A1)

x=e14         A1

substitution of their x to find y        (M1)

y=lne14e144

=14e=14e-1         A1

Pe14,14e

 

[5 marks]

b.

f''e14=20lne14-9e146        (M1)

=5-9e1.5  =-4e1.5         A1

which is negative         R1

hence P is a local maximum         AG

 

Note: The R1 is dependent on the previous A1 being awarded.

 

[3 marks]

c.

lnx>0        (A1)

x>1        A1

 

[2 marks]

d.

        A1A1A1

 

 

Note: Award A1 for one x-intercept only, located at 1

     A1 for local maximum, P, in approximately correct position
     A1 for curve approaching x-axis as x (including change in concavity).

 

[3 marks]

e.

Examiners report

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b.
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d.
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e.

Syllabus sections

Topic 5 —Calculus » SL 5.6—Differentiating polynomials n E Q. Chain, product and quotient rules
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Topic 5 —Calculus

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