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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 Show that Question number 9 Adapted from N/A

Question

Let f(x)=ln5xkx where x>0kR+.

The graph of f has exactly one maximum point P.

The second derivative of f is given by f(x)=2ln5x3kx3. The graph of f has exactly one point of inflexion Q.

Show that f(x)=1ln5xkx2.

[3]
a.

Find the x-coordinate of P.

[3]
b.

Show that the x-coordinate of Q is 15e32.

[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)=5kx(15x)kln5x(kx)2 (or equivalent)        A1

=kkln5xk2x2, (kR+)        A1

=1ln5xkx2        AG

[3 marks]

a.

f(x)=0     M1

1ln5xkx2=0

ln5x=1      (A1)

x=e5      A1

[3 marks]

b.

f(x)=0    M1

2ln5x3kx3=0

ln5x=32     A1

5x=e32     A1

so the point of inflexion occurs at x=15e32     AG

[3 marks]

c.

attempt to integrate   (M1)

u=ln5xdudx=1x

ln5xkxdx=1kudu     (A1)

EITHER

=u22k     A1

so 1k321udu=[u22k]321     A1

OR

=(ln5x)22k     A1

so 15e32e5ln5xkxdx=[(ln5x)22k]15e32e5     A1

THEN

=12k(941)

=58k     A1

setting their expression for area equal to 3       M1 

58k=3

k=524     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
Show 127 related questions
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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