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>0, k∈R+.
The graph of f has exactly one maximum point P.
The second derivative of f is given by f″(x)=2ln5x−3kx3. The graph of f has exactly one point of inflexion Q.
Show that f′(x)=1−ln5xkx2.
Find the x-coordinate of P.
Show that the x-coordinate of Q is 15e32.
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.
Markscheme
attempt to use quotient rule (M1)
correct substitution into quotient rule
f′(x)=5kx(15x)−kln5x(kx)2 (or equivalent) A1
=k−kln5xk2x2, (k∈R+) A1
=1−ln5xkx2 AG
[3 marks]
f′(x)=0 M1
1−ln5xkx2=0
ln5x=1 (A1)
x=e5 A1
[3 marks]
f″(x)=0 M1
2ln5x−3kx3=0
ln5x=32 A1
5x=e32 A1
so the point of inflexion occurs at x=15e32 AG
[3 marks]
attempt to integrate (M1)
u=ln5x⇒dudx=1x
∫ln5xkxdx=1k∫udu (A1)
EITHER
=u22k A1
so 1k32∫1udu=[u22k]321 A1
OR
=(ln5x)22k A1
so 15e32∫e5ln5xkxdx=[(ln5x)22k]15e32e5 A1
THEN
=12k(94−1)
=58k A1
setting their expression for area equal to 3 M1
58k=3
k=524 A1
[7 marks]