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Date May Specimen paper Marks available 3 Reference code SPM.1.SL.TZ0.9
Level Standard Level Paper Paper 1 (without calculator) Time zone Time zone 0
Command term Find Question number 9 Adapted from N/A

Question

Let f ( x ) = ln 5 x k x where  x > 0 k R + .

The graph of f has exactly one maximum point P.

The second derivative of  f is given by  f ( x ) = 2 ln 5 x 3 k x 3 . The graph of f has exactly one point of inflexion Q.

Show that f ( x ) = 1 ln 5 x k x 2 .

[3]
a.

Find the x-coordinate of P.

[3]
b.

Show that the x-coordinate of Q is  1 5 e 3 2 .

[3]
c.

The region R is enclosed by the graph of f , the x-axis, and the vertical lines through the maximum point P and the point of inflexion Q.

Given that the area of R is 3, find the value of k .

[7]
d.

Markscheme

attempt to use quotient rule       (M1)

correct substitution into quotient rule

f ( x ) = 5 k x ( 1 5 x ) k ln 5 x ( k x ) 2  (or equivalent)        A1

= k k ln 5 x k 2 x 2 , ( k R + )        A1

= 1 ln 5 x k x 2         AG

[3 marks]

a.

f ( x ) = 0      M1

1 ln 5 x k x 2 = 0

ln 5 x = 1       (A1)

x = e 5       A1

[3 marks]

b.

f ( x ) = 0     M1

2 ln 5 x 3 k x 3 = 0

ln 5 x = 3 2      A1

5 x = e 3 2      A1

so the point of inflexion occurs at  x = 1 5 e 3 2      AG

[3 marks]

c.

attempt to integrate   (M1)

u = ln 5 x d u d x = 1 x

ln 5 x k x d x = 1 k u d u      (A1)

EITHER

= u 2 2 k      A1

so 1 k 1 3 2 u d u = [ u 2 2 k ] 1 3 2      A1

OR

= ( ln 5 x ) 2 2 k      A1

so e 5 1 5 e 3 2 ln 5 x k x d x = [ ( ln 5 x ) 2 2 k ] e 5 1 5 e 3 2      A1

THEN

= 1 2 k ( 9 4 1 )

= 5 8 k      A1

setting their expression for area equal to 3       M1 

5 8 k = 3

k = 5 24      A1

[7 marks]

d.

Examiners report

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

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 » SL 5.8—Testing for max and min, optimisation. Points of inflexion
Topic 5 —Calculus » SL 5.10—Indefinite integration, reverse chain, by substitution
Topic 5 —Calculus » SL 5.11—Definite integrals, areas under curve onto x-axis and areas between curves
Topic 5 —Calculus

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