User interface language: English | Español

Date May 2019 Marks available 3 Reference code 19M.2.SL.TZ1.S_4
Level Standard Level Paper Paper 2 Time zone Time zone 1
Command term Sketch Question number S_4 Adapted from N/A

Question

Let  f ( x ) = ( cos 2 x ) ( sin 6 x ) , for 0 ≤ x  ≤ 1.

Sketch the graph of f on the grid below:

[3]
a.

Find the x -coordinates of the points of inflexion of the graph of f .

[3]
b.

Hence find the values of x for which the graph of f is concave-down.

[2]
c.

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

  A1A1A1 N3

Note: Only if the shape is approximately correct with exactly 2 maximums and 1 minimum on the interval 0 ≤ x ≤ 0, award the following:
A1 for correct domain with both endpoints within circle and oval.
A1 for passing through the other x -intercepts within the circles.
A1 for passing through the three turning points within circles (ignore x -intercepts and extrema outside of the domain).

[3 marks]

a.

evidence of reasoning (may be seen on graph)      (M1)

eg  f = 0 ,  (0.524, 0),  (0.785, 0)

0.523598,  0.785398

x = 0.524 ( = π 6 ) ,   x = 0.785 ( = π 4 )      A1A1  N3

Note: Award M1A1A0 if any solution outside domain (eg x = 0 ) is also included.

[3 marks]

b.

0.524 < x < 0.785 ( π 6 < x < π 4 )      A2  N2

Note: Award A1 if any correct interval outside domain also included, unless additional solutions already penalized in (b).
Award A0 if any incorrect intervals are also included.

[2 marks]

c.

Examiners report

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

Syllabus sections

Topic 5—Calculus » SL 5.6—Stationary points, local max and min
Show 140 related questions
Topic 5—Calculus » SL 5.7—Optimisation
Topic 5—Calculus

View options