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Date May 2018 Marks available 1 Reference code 18M.2.AHL.TZ2.H_10
Level Additional Higher Level Paper Paper 2 Time zone Time zone 2
Command term Explain Question number H_10 Adapted from N/A

Question

Consider the expression  f ( x ) = tan ( x + π 4 ) cot ( π 4 x ) .

The expression  f ( x ) can be written as  g ( t ) where  t = tan x .

Let  α β be the roots of  g ( t ) = k , where 0 < k < 1.

Sketch the graph of  y = f ( x ) for  5 π 8 x π 8 .

[2]
a.i.

With reference to your graph, explain why  f  is a function on the given domain.

[1]
a.ii.

Explain why f has no inverse on the given domain.

[1]
a.iii.

Explain why f is not a function for 3 π 4 x π 4 .

[1]
a.iv.

Show that  g ( t ) = ( 1 + t 1 t ) 2 .

[3]
b.

Sketch the graph of  y = g ( t ) for t ≤ 0. Give the coordinates of any intercepts and the equations of any asymptotes.

[3]
c.

Find  α and β in terms of k .

[5]
d.i.

Show that  α  + β < −2.

[2]
d.ii.

Markscheme

     A1A1

A1 for correct concavity, many to one graph, symmetrical about the midpoint of the domain and with two axes intercepts.

Note: Axes intercepts and scales not required.

A1 for correct domain

[2 marks]

a.i.

for each value of x there is a unique value of f ( x )       A1

Note: Accept “passes the vertical line test” or equivalent.

[1 mark]

a.ii.

no inverse because the function fails the horizontal line test or equivalent      R1

Note: No FT if the graph is in degrees (one-to-one).

[1 mark]

a.iii.

the expression is not valid at either of x = π 4 ( or 3 π 4 )        R1

[1 mark]

a.iv.

METHOD 1

f ( x ) = tan ( x + π 4 ) tan ( π 4 x )      M1

= tan x + tan π 4 1 tan x tan π 4 tan π 4 tan x 1 + tan π 4 tan x       M1A1

= ( 1 + t 1 t ) 2       AG

 

METHOD 2

f ( x ) = tan ( x + π 4 ) tan ( π 2 π 4 + x )       (M1)

= ta n 2 ( x + π 4 )      A1

g ( t ) = ( tan x + tan π 4 1 tan x tan π 4 ) 2      A1

= ( 1 + t 1 t ) 2       AG

[3 marks]

b.

 

for t ≤ 0, correct concavity with two axes intercepts and with asymptote y  = 1      A1

t intercept at (−1, 0)      A1

y intercept at (0, 1)       A1

[3 marks]

c.

METHOD 1

α β satisfy ( 1 + t ) 2 ( 1 t ) 2 = k      M1

1 + t 2 + 2 t = k ( 1 + t 2 2 t )      A1

( k 1 ) t 2 2 ( k + 1 ) t + ( k 1 ) = 0      A1

attempt at using quadratic formula      M1

α β  = k + 1 ± 2 k k 1 or equivalent     A1

 

METHOD 2

α β satisfy  1 + t 1 t = ( ± ) k       M1

t + k t = k 1       M1

t = k 1 k + 1  (or equivalent)      A1

t k t = ( k + 1 )      M1

t = k + 1 k 1  (or equivalent)       A1

so for eg α = k 1 k + 1 β = k + 1 k 1

[5 marks]

d.i.

α  + β  = 2 ( k + 1 ) ( k 1 ) ( = 2 ( 1 + k ) ( 1 k ) )      A1

since  1 + k > 1 k      R1

α  + β < −2     AG

Note: Accept a valid graphical reasoning.

[2 marks]

d.ii.

Examiners report

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

Syllabus sections

Topic 2—Functions » SL 2.2—Functions, notation domain, range and inverse as reflection
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Topic 2—Functions » SL 2.4—Key features of graphs, intersections using technology
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