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Date May 2021 Marks available 3 Reference code 21M.3.AHL.TZ2.1
Level Additional Higher Level Paper Paper 3 Time zone Time zone 2
Command term Show that Question number 1 Adapted from N/A

Question

This question asks you to explore the behaviour and some key features of the function fn(x)=xn(a-x)n , where a+ and n+.

In parts (a) and (b), only consider the case where a=2.

Consider f1(x)=x(2-x).

Consider fnx=xn2-xn, where n+, n>1.

Now consider fnx=xna-xn where a+ and n+, n>1.

By using the result from part (f) and considering the sign of fn'-1, show that the point 0,0 on the graph of y=fnx is

Sketch the graph of y=f1(x), stating the values of any axes intercepts and the coordinates of any local maximum or minimum points.

[3]
a.

Use your graphic display calculator to explore the graph of y=fn(x) for

•   the odd values n=3 and n=5;

•   the even values n=2 and n=4.

Hence, copy and complete the following table.

[6]
b.

Show that fn'x=nxn-1a-2xa-xn-1.

[5]
c.

State the three solutions to the equation fn'x=0.

[2]
d.

Show that the point a2, fna2 on the graph of y=fnx is always above the horizontal axis.

[3]
e.

Hence, or otherwise, show that fn'a4>0, for n+.

[2]
f.

a local minimum point for even values of n, where n>1 and a+.

[3]
g.i.

a point of inflexion with zero gradient for odd values of n, where n>1 and a+.

[2]
g.ii.

Consider the graph of y=xna-xn-k, where n+a+ and k.

State the conditions on n and k such that the equation xna-xn=k has four solutions for x.

[5]
h.

Markscheme

inverted parabola extended below the x-axis             A1

x-axis intercept values x=0,2         A1


Note: Accept a graph passing through the origin as an indication of x=0.

local maximum at 1,1                 A1


Note: Coordinates must be stated to gain the final A1.
        Do not accept decimal approximations.


[3 marks]

a.

             A1A1A1A1A1A1


Note:
Award A1 for each correct value.

For a table not sufficiently or clearly labelled, assume that their values are in the same order as the table in the question paper and award marks accordingly.


[6 marks]

b.

METHOD 1

attempts to use the product rule            (M1)

fn'x=-nxna-xn-1+nxn-1a-xn            A1A1


Note: Award A1 for a correct udvdx and A1 for a correct vdudx.


EITHER

attempts to factorise fn'x (involving at least one of nxn-1 or a-xn-1)           (M1)

=nxn-1a-xn-1a-x-x            A1


OR

attempts to express fn'x as the difference of two products with each product containing at least one of nxn-1 or a-xn-1           (M1)

=-xnxn-1a-xn-1+a-xnxn-1a-xn-1            A1


THEN

fn'x=nxn-1a-2xa-xn-1            AG


Note: Award the final (M1)A1 for obtaining any of the following forms: 

fn'x=nxna-xna-x-xxa-x;   fn'x=nxna-xnxa-xa-x-x;

        fn'x=nxn-1a-xn-xa-xn-1;

        fn'x=a-xn-1nxn-1a-xn-nxn

 

 

METHOD 2

fnx=xa-xn           (M1)

=ax-x2n            A1

attempts to use the chain rule           (M1)

fn'x=na-2xax-x2n-1            A1A1


Note:
Award A1 for na-2x and A1 for ax-x2n-1.


fn'x=nxn-1a-2xa-xn-1            AG

 

[5 marks]

c.

x=0, x=a2, x=a            A2

Note: Award A1 for either two correct solutions or for obtaining x=0, x=-a, x=-a2
       
  Award A0 otherwise.

 

[2 marks]

d.

attempts to find an expression for fna2             (M1)

fna2=a2na-a2n

=a2na2n  =a22n,=a2n2            A1


EITHER

since a+, a22n>0  (for n+, n>1 and so fna2>0)                R1


Note: Accept any logically equivalent conditions/statements on a and n.
        Award R0 if any conditions/statements specified involving a, n or both are incorrect.

 

OR

(since a+), a2 raised to an even power (2n) (or equivalent reasoning) is always positive (and so  fna2>0)                R1


Note: The condition a+ is given in the question. Hence some candidates will assume a+ and not state it. In these instances, award R1 for a convincing argument.
        Accept any logically equivalent conditions/statements on on a and n.
        Award R0 if any conditions/statements specified involving an or both are incorrect.


THEN

so a2, fna2 is always above the horizontal axis            AG


Note: Do not award (M1)A0R1.

 

[3 marks]

e.

METHOD 1

fn'a4=na4n-1a-a2a-a4n-1  =na4n-1a23a4n-1            A1


EITHER

na4n-1a23a4n-1>0 as a+ and n+                R1


OR

na4n-1,a-a2 and a-a4n-1 are all >0                R1

 

Note: Do not award A0R1.
        Accept equivalent reasoning on correct alternative expressions for fn'a4 and accept any logically equivalent conditions/statements on a and n.

        Exceptions to the above are condone n>1 and condone n>0.

        An alternative form for fn'a4 is 2n3n-1a42n-1.


THEN

hence fn'a4>0                 AG

 

METHOD 2

fn0=0 and fna2>0            A1

(since fn is continuous and there are no stationary points between x=0 and x=a2)

the gradient (of the curve) must be positive between x=0 and x=a2                 R1


Note: Do not award A0R1.


hence fn'a4>0                 AG

 

[2 marks]

f.

fn'-1=n-1n-1a+2a+1n-1

for n even:

n-1n-1=-n<0  (and a+2,a+1n-1 are both >0)                R1

fn'-1<0            A1

fn'0=0 and fn'a4>0  (seen anywhere)            A1

 

Note: Candidates can give arguments based on the sign of -1n-1 to obtain the R mark.
        For example, award R1 for the following:
        If n is even, then n-1 is odd and hence -1n-1<0 =-1.
        Do not award R0A1.
        The second A1 is independent of the other two marks.
        The A marks can be awarded for correct descriptions expressed in words.
        Candidates can state (0,0) as a point of zero gradient from part (d) or show, state or explain (words or diagram) that fn'0=0. The last mark can be awarded for a clearly labelled diagram showing changes in the sign of the gradient.
        The last A1 can be awarded for use of a specific case (e.g. n=2).


hence (0,0) is a local minimum point            AG

 

[3 marks]

g.i.

for n odd:

n-1n-1=n<0, (and a+2,a+1n-1 are both >0)  so fn'-1>0               R1


Note: Candidates can give arguments based on the sign of -1n-1 to obtain the R mark.
        For example, award R1 for the following:
        If n is odd, then n-1 is even and hence -1n-1>0 =1.


fn'0=0 and fn'a4>0  (seen anywhere)            A1


Note: The A1 is independent of the R1.
         Candidates can state 0,0 as a point of zero gradient from part (d) or show, state or explain (words or diagram) that fn'0=0. The last A mark can be awarded for a clearly labelled diagram showing changes in the sign of the gradient.
        The last A1 can be awarded for use of a specific case (e.g. n=3).

 

hence (0,0) is a point of inflexion with zero gradient           AG

 

[2 marks]

g.ii.

considers the parity of n            (M1)


Note: Award M1 for stating at least one specific even value of n.


n must be even (for four solutions)           A1


Note: The above 2 marks are independent of the 3 marks below.

 

0<k<a22n           A1A1A1

 

Note: Award A1 for the correct lower endpoint, A1 for the correct upper endpoint and A1 for strict inequality signs.

         The third A1 (strict inequality signs) can only be awarded if A1A1 has been awarded.
         For example, award A1A1A0 for 0ka22n. Award A1A0A0 for k>0.

         Award A1A0A0 for 0<k<fna2.

 

[5 marks]

h.

Examiners report

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a.
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b.
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c.
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d.
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e.
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f.
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g.i.
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g.ii.
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h.

Syllabus sections

Topic 5 —Calculus » SL 5.6—Differentiating polynomials n E Q. Chain, product and quotient rules
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Topic 5 —Calculus

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