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Date May 2014 Marks available 2 Reference code 14M.1.hl.TZ2.14
Level HL only Paper 1 Time zone TZ2
Command term Sketch Question number 14 Adapted from N/A

Question

Consider the following functions:

     \(h(x) = \arctan (x),{\text{ }}x \in \mathbb{R}\)

     \(g(x) = \frac{1}{x}\), \(x\in \mathbb{R}\), \({\text{ }}x \ne 0\)

Sketch the graph of \(y = h(x)\).

[2]
a.

Find an expression for the composite function \(h \circ g(x)\) and state its domain.

[2]
b.

Given that \(f(x) = h(x) + h \circ g(x)\),

(i)     find \(f'(x)\) in simplified form;

(ii)     show that \(f(x) = \frac{\pi }{2}\) 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 \(y =  \pm \frac{\pi }{2}\).

 

[2 marks]

a.

\(h \circ g(x) = \arctan \left( {\frac{1}{x}} \right)\)     A1

domain of \(h \circ g\) is equal to the domain of \(g:x \in  \circ ,{\text{ }}x \ne 0\)     A1

[2 marks]

b.

(i)     \(f(x) = \arctan (x) + \arctan \left( {\frac{1}{x}} \right)\)

\(f'(x) = \frac{1}{{1 + {x^2}}} + \frac{1}{{1 + \frac{1}{{{x^2}}}}} \times  - \frac{1}{{{x^2}}}\)     M1A1

\(f'(x) = \frac{1}{{1 + {x^2}}} + \frac{{ - \frac{1}{{{x^2}}}}}{{\frac{{{x^2} + 1}}{{{x^2}}}}}\)     (A1)

\( = \frac{1}{{1 + {x^2}}} - \frac{1}{{1 + {x^2}}}\)

\( = 0\)     A1

(ii)     METHOD 1

f is a constant     R1

when \(x > 0\)

\(f(1) = \frac{\pi }{4} + \frac{\pi }{4}\)     M1A1

\( = \frac{\pi }{2}\)     AG

METHOD 2

from diagram

\(\theta  = \arctan \frac{1}{x}\)     A1

\(\alpha  = \arctan x\)     A1

\(\theta  + \alpha  = \frac{\pi }{2}\)     R1

hence \(f(x) = \frac{\pi }{2}\)     AG

METHOD 3

\(\tan \left( {f(x)} \right) = \tan \left( {\arctan (x) + \arctan \left( {\frac{1}{x}} \right)} \right)\)     M1

\( = \frac{{x + \frac{1}{x}}}{{1 - x\left( {\frac{1}{x}} \right)}}\)     A1

denominator = 0, so \(f(x) = \frac{\pi }{2}{\text{ (for }}x > 0)\)     R1

[7 marks]

c.

(i)     Nigel is correct.     A1

METHOD 1

\(\arctan (x)\) is an odd function and \(\frac{1}{x}\) 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

\(f( - x) = \arctan ( - x) + \arctan \left( { - \frac{1}{x}} \right) =  - \arctan (x) - \arctan \left( {\frac{1}{x}} \right) =  - f(x)\)

therefore f is an odd function.     R1

(ii)     \(f(x) =  - \frac{\pi }{2}\)     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 3 - Core: Circular functions and trigonometry » 3.5 » The inverse functions \(x \mapsto \arcsin x\) , \(x \mapsto \arccos x\) , \(x \mapsto \arctan x\) ; their domains and ranges; their graphs.

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