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Date May 2017 Marks available 2 Reference code 17M.1.AHL.TZ2.H_9
Level Additional Higher Level Paper Paper 1 (without calculator) Time zone Time zone 2
Command term Sketch Question number H_9 Adapted from N/A

Question

Consider the function f defined by f ( x ) = x 2 a 2 ,   x R where a is a positive constant.

The function g is defined by g ( x ) = x f ( x ) for | x | > a .

Showing any x and y intercepts, any maximum or minimum points and any asymptotes, sketch the following curves on separate axes.

y = f ( x ) ;

[2]
a.i.

Showing any x and y intercepts, any maximum or minimum points and any asymptotes, sketch the following curves on separate axes.

y = 1 f ( x ) ;

[4]
a.ii.

Showing any x and y intercepts, any maximum or minimum points and any asymptotes, sketch the following curves on separate axes.

y = | 1 f ( x ) | .

[2]
a.iii.

Find f ( x ) cos x d x .

[5]
b.

By finding g ( x ) explain why g is an increasing function.

[4]
c.

Markscheme

M17/5/MATHL/HP1/ENG/TZ2/09.a.i/M

A1 for correct shape

A1 for correct x and y intercepts and minimum point

[2 marks]

a.i.

M17/5/MATHL/HP1/ENG/TZ2/09.a.ii/M

A1 for correct shape

A1 for correct vertical asymptotes

A1 for correct implied horizontal asymptote

A1 for correct maximum point

[??? marks]

a.ii.

M17/5/MATHL/HP1/ENG/TZ2/09.a.iii/M

A1 for reflecting negative branch from (ii) in the x -axis

A1 for correctly labelled minimum point

[2 marks]

a.iii.

EITHER

attempt at integration by parts     (M1)

( x 2 a 2 ) cos x d x = ( x 2 a 2 ) sin x 2 x sin x d x      A1A1

= ( x 2 a 2 ) sin x 2 [ x cos x + cos x d x ]      A1

= ( x 2 a 2 ) sin x + 2 x cos 2 sin x + c      A1

OR

( x 2 a 2 ) cos x d x = x 2 cos x d x a 2 cos x d x

attempt at integration by parts     (M1)

x 2 cos x d x = x 2 sin x 2 x sin x d x      A1A1

= x 2 sin x 2 [ x cos x + cos x d x ]      A1

= x 2 sin x + 2 x cos x 2 sin x

a 2 cos x d x = a 2 sin x

( x 2 a 2 ) cos x d x = ( x 2 a 2 ) sin x + 2 x cos x 2 sin x + c      A1

[5 marks]

b.

g ( x ) = x ( x 2 a 2 ) 1 2

g ( x ) = ( x 2 a 2 ) 1 2 + 1 2 x ( x 2 a 2 ) 1 2 ( 2 x )     M1A1A1

 

Note:     Method mark is for differentiating the product. Award A1 for each correct term.

 

g ( x ) = ( x 2 a 2 ) 1 2 + x 2 ( x 2 a 2 ) 1 2

both parts of the expression are positive hence g ( x ) is positive     R1

and therefore g is an increasing function (for | x | > a )     AG

[4 marks]

c.

Examiners report

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

Syllabus sections

Topic 5 —Calculus » SL 5.8—Testing for max and min, optimisation. Points of inflexion
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