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Date May 2014 Marks available 3 Reference code 14M.1.hl.TZ2.14
Level HL only Paper 1 Time zone TZ2
Command term Find, Hence, and State Question number 14 Adapted from N/A

Question

Consider the following functions:

     h(x)=arctan(x), xR

     g(x)=1x, xR,  x0

Sketch the graph of y=h(x).

[2]
a.

Find an expression for the composite function hg(x) and state its domain.

[2]
b.

Given that f(x)=h(x)+hg(x),

(i)     find f(x) in simplified form;

(ii)     show that f(x)=π2 for x>0.

[7]
c.

Nigel states that f is an odd function and Tom argues that f is an even function.

(i)     State who is correct and justify your answer.

(ii)     Hence find the value of f(x) for x<0.

[3]
d.

Markscheme

    A1A1

 

Note:     A1 for correct shape, A1 for asymptotic behaviour at y=±π2.

 

[2 marks]

a.

hg(x)=arctan(1x)     A1

domain of hg is equal to the domain of g:x, x0     A1

[2 marks]

b.

(i)     f(x)=arctan(x)+arctan(1x)

f(x)=11+x2+11+1x2×1x2     M1A1

f(x)=11+x2+1x2x2+1x2     (A1)

=11+x211+x2

=0     A1

(ii)     METHOD 1

f is a constant     R1

when x>0

f(1)=π4+π4     M1A1

=π2     AG

METHOD 2

from diagram

θ=arctan1x     A1

α=arctanx     A1

θ+α=π2     R1

hence f(x)=π2     AG

METHOD 3

tan(f(x))=tan(arctan(x)+arctan(1x))     M1

=x+1x1x(1x)     A1

denominator = 0, so f(x)=π2 (for x>0)     R1

[7 marks]

c.

(i)     Nigel is correct.     A1

METHOD 1

arctan(x) is an odd function and 1x is an odd function

composition of two odd functions is an odd function and sum of two odd functions is an odd function     R1

METHOD 2

f(x)=arctan(x)+arctan(1x)=arctan(x)arctan(1x)=f(x)

therefore f is an odd function.     R1

(ii)     f(x)=π2     A1

[3 marks]

d.

Examiners report

[N/A]
a.
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b.
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c.
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d.

Syllabus sections

Topic 2 - Core: Functions and equations » 2.1 » Odd and even functions.

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