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Date May Specimen paper Marks available 4 Reference code SPM.3.AHL.TZ0.2
Level Additional Higher Level Paper Paper 3 Time zone Time zone 0
Command term Solve, Show that, and Hence Question number 2 Adapted from N/A

Question

This question asks you to investigate some properties of the sequence of functions of the form fn(x)=cos(narccosx), −1 ≤ x ≤ 1 and nZ+.

Important: When sketching graphs in this question, you are not required to find the coordinates of any axes intercepts or the coordinates of any stationary points unless requested.

For odd values of n > 2, use your graphic display calculator to systematically vary the value of n. Hence suggest an expression for odd values of n describing, in terms of n, the number of

For even values of n > 2, use your graphic display calculator to systematically vary the value of n. Hence suggest an expression for even values of ndescribing, in terms of n, the number of

The sequence of functions, fn(x), defined above can be expressed as a sequence of polynomials of degree n.

Consider fn+1(x)=cos((n+1)arccosx).

On the same set of axes, sketch the graphs of y=f1(x) and y=f3(x) for −1 ≤ x ≤ 1.

[2]
a.

local maximum points;

[3]
b.i.

local minimum points;

[1]
b.ii.

On a new set of axes, sketch the graphs of y=f2(x) and y=f4(x) for −1 ≤ x ≤ 1.

[2]
c.

local maximum points;

[3]
d.i.

local minimum points.

[1]
d.ii.

Solve the equation fn(x)=0 and hence show that the stationary points on the graph of y=fn(x) occur at x=coskπn where kZ+ and 0 < k < n.

[4]
e.

Use an appropriate trigonometric identity to show that f2(x)=2x21.

[2]
f.

Use an appropriate trigonometric identity to show that fn+1(x)=cos(narccosx)cos(arccosx)sin(narccosx)sin(arccosx).

[2]
g.

Hence show that fn+1(x)+fn1(x)=2xfn(x)nZ+.

[3]
h.i.

Hence express f3(x) as a cubic polynomial.

[2]
h.ii.

Markscheme

correct graph of y=f1(x)      A1

correct graph of y=f3(x)      A1

[2 marks]

a.

graphical or tabular evidence that n has been systematically varied        M1

eg n = 3, 1 local maximum point and 1 local minimum point

n = 5, 2 local maximum points and 2 local minimum points

n = 7, 3 local maximum points and 3 local minimum points        (A1)

n12 local maximum points      A1

[3 marks]

b.i.

n12 local minimum points      A1

Note: Allow follow through from an incorrect local maximum formula expression.

[1 mark]

b.ii.

correct graph of y=f2(x)       A1

correct graph of y=f4(x)       A1

[2 marks]

c.

graphical or tabular evidence that n has been systematically varied       M1

eg n = 2, 0 local maximum point and 1 local minimum point

n = 4, 1 local maximum points and 2 local minimum points

n = 6, 2 local maximum points and 3 local minimum points       (A1)

n22 local maximum points     A1

[3 marks]

d.i.

n2 local minimum points     A1

[1 mark]

d.ii.

fn(x)=cos(narccos(x))

fn(x)=nsin(narccos(x))1x2      M1A1

Note: Award M1 for attempting to use the chain rule.

fn(x)=0nsin(narccos(x))=0     M1

narccos(x)=kπ(kZ+)     A1

leading to

x=coskπn  (kZ+ and 0 < kn)     AG

[4 marks]

e.

f2(x)=cos(2arccosx)

=2(cos(arccosx))21     M1

stating that (cos(arccosx))=x     A1

so f2(x)=2x21     AG

[2 marks]

f.

 

fn+1(x)=cos((n+1)arccosx)

=cos(narccosx+arccosx)     A1

use of cos(A + B) = cos A cos B − sin A sin B leading to      M1

=cos(narccosx)cos(arccosx)sin(narccosx)sin(arccosx)      AG

[2 marks]

g.

fn1(x)=cos((n1)arccosx)     A1

=cos(narccosx)cos(arccosx)+sin(narccosx)sin(arccosx)    M1

fn+1(x)+fn1(x)=2cos(narccosx)cos(arccosx)     A1

=2xfn(x)     AG

[3 marks]

h.i.

f3(x)=2xf2(x)f1(x)      (M1)

=2x(2x21)x

=4x33x   A1

[2 marks]

h.ii.

Examiners report

[N/A]
a.
[N/A]
b.i.
[N/A]
b.ii.
[N/A]
c.
[N/A]
d.i.
[N/A]
d.ii.
[N/A]
e.
[N/A]
f.
[N/A]
g.
[N/A]
h.i.
[N/A]
h.ii.

Syllabus sections

Topic 5 —Calculus » SL 5.6—Differentiating polynomials n E Q. Chain, product and quotient rules
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Topic 3— Geometry and trigonometry » AHL 3.9—Reciprocal trig ratios and their pythagorean identities. Inverse circular functions
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Topic 5 —Calculus

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