User interface language: English | Español

Date November 2019 Marks available 2 Reference code 19N.2.SL.TZ0.S_8
Level Standard Level Paper Paper 2 Time zone Time zone 0
Command term Find Question number S_8 Adapted from N/A

Question

Let  f ( x ) = x 4 54 x 2 + 60 x , for  1 x 6 . The following diagram shows the graph of f .

There are x -intercepts at x = 0 and at x = p . There is a maximum at point A where x = a , and a point of inflexion at point B where x = b .

Find the value of p .

[2]
a.

Write down the coordinates of A .

[2]
b.i.

Find the equation of the tangent to the graph of f at A .

[2]
b.ii.

Find the coordinates of B .

[5]
c.i.

Find the rate of change of f at B .

[2]
c.ii.

Let R be the region enclosed by the graph of f , the x -axis and the lines x = p and x = b . The region R is rotated 360º about the x -axis. Find the volume of the solid formed.

[3]
d.

Markscheme

evidence of valid approach        (M1)

eg        f ( x ) = 0 ,   y = 0

1.13843

p = 1.14        A1  N2

[2 marks]

a.

0.562134 16.7641

( 0.562 16.8 )        A2  N2

[2 marks]

b.i.

valid approach        (M1)

eg      tangent at maximum point is horizontal,  f = 0

y = 16.8  (must be an equation)       A1  N2

[2 marks]

b.ii.

METHOD 1 (using GDC)

valid approach         M1

eg       f = 0 ,  max/min on  f x = 3

sketch of either  f or  f , with max/min or root (respectively)       (A1)

x = 3        A1  N1

substituting their x value into f         (M1)

eg       f ( 3 )

y = 225 (exact)  (accept   ( 3 225 ) )       A1  N1

 

METHOD 2 (analytical)

f = 12 x 2 108        A1

valid approach       (M1)

eg       f = 0 ,   x = ± 3

x = 3        A1  N1

substituting their  x value into f         (M1)

eg       f ( 3 )

y = 225 (exact)  (accept   ( 3 225 ) )       A1  N1

 

[5 marks]

c.i.

recognizing rate of change is  f         (M1)

eg       y ,   f ( 3 )

rate of change is 156 (exact)       A1  N2

[2 marks]

c.ii.

attempt to substitute either their limits or the function into volume formula        (M1)

eg        1.14 3 f 2 ,   π ( x 4 54 x 2 + 60 x ) 2 d x ,   25 752.0

80 902.3

volume  = 80 900        A2   N3

[3 marks]

d.

Examiners report

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

Syllabus sections

Topic 5 —Calculus » SL 5.8—Testing for max and min, optimisation. Points of inflexion
Show 202 related questions
Topic 5 —Calculus » AHL 5.17—Areas under curve onto y-axis, volume of revolution (about x and y axes)
Topic 5 —Calculus

View options