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Date November 2019 Marks available 2 Reference code 19N.1.AHL.TZ0.H_10
Level Additional Higher Level Paper Paper 1 (without calculator) Time zone Time zone 0
Command term Find Question number H_10 Adapted from N/A

Question

Consider  f ( x ) = 2 x 4 x 2 1 1 < x < 1 .

For the graph of  y = f ( x ) ,

Find  f ( x ) .

[2]
a.i.

Show that, if  f ( x ) = 0 , then  x = 2 3 .

[3]
a.ii.

find the coordinates of the y -intercept.

[1]
b.i.

show that there are no x -intercepts.

[2]
b.ii.

sketch the graph, showing clearly any asymptotic behaviour.

[2]
b.iii.

Show that 3 x + 1 1 x 1 = 2 x 4 x 2 1 .

[2]
c.

The area enclosed by the graph of y = f ( x ) and the line y = 4 can be expressed as ln v . Find the value of v .

[7]
d.

Markscheme

attempt to use quotient rule (or equivalent)       (M1)

f ( x ) = ( x 2 1 ) ( 2 ) ( 2 x 4 ) ( 2 x ) ( x 2 1 ) 2        A1

= 2 x 2 + 8 x 2 ( x 2 1 ) 2

[2 marks]

a.i.

f ( x ) = 0

simplifying numerator (may be seen in part (i))       (M1)

x 2 4 x + 1 = 0  or equivalent quadratic equation       A1

 

EITHER

use of quadratic formula

x = 4 ± 12 2        A1

 

OR

use of completing the square

( x 2 ) 2 = 3        A1

 

THEN

x = 2 3   (since  2 + 3  is outside the domain)       AG

 

Note: Do not condone verification that x = 2 3 f ( x ) = 0 .

Do not award the final A1 as follow through from part (i).

 

[3 marks]

a.ii.

(0, 4)       A1

[1 mark]

b.i.

2 x 4 = 0 x = 2       A1

outside the domain       R1

[2 marks]

b.ii.

      A1A1

award A1 for concave up curve over correct domain with one minimum point in the first quadrant
award A1 for approaching x = ± 1 asymptotically

[2 marks]

b.iii.

valid attempt to combine fractions (using common denominator)      M1

3 ( x 1 ) ( x + 1 ) ( x + 1 ) ( x 1 )       A1

= 3 x 3 x 1 x 2 1

= 2 x 4 x 2 1       AG

[2 marks]

c.

f ( x ) = 4 2 x 4 = 4 x 2 4       M1

       ( x = 0   or)   x = 1 2       A1

 

area under the curve is  0 1 2 f ( x ) d x       M1

= 0 1 2 3 x + 1 1 x 1 d x

Note: Ignore absence of, or incorrect limits up to this point.

 

= [ 3 ln | x + 1 | ln | x 1 | ] 0 1 2       A1

= 3 ln 3 2 ln 1 2 ( 0 )

= ln 27 4       A1

area is  2 0 1 2 f ( x ) d x   or  0 1 2 4 d x 0 1 2 f ( x ) d x       M1

= 2 ln 27 4

= ln 4 e 2 27       A1

( v = 4 e 2 27 )

 

[7 marks]

d.

Examiners report

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

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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