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Date	November 2021	Marks available	4	Reference code	21N.3.AHL.TZ0.1
Level	Additional Higher Level	Paper	Paper 3	Time zone	Time zone 0
Command term	Sketch	Question number	1	Adapted from	N/A

Question

In this question you will explore some of the properties of special functions $f$ and $g$ and their relationship with the trigonometric functions, sine and cosine.

Functions $f$ and $g$ are defined as $f (z) = \frac{e^{z} + e^{- z}}{2}$ and $g (z) = \frac{e^{z} - e^{- z}}{2}$ , where $z \in ℂ$ .

Consider $t$ and $u$ , such that $t, u \in ℝ$ .

Using $e^{i u} = \cos u + i \sin u$ , find expressions, in terms of $\sin u$ and $\cos u$ , for

The functions $\cos x$ and $\sin x$ are known as circular functions as the general point ( $\cos θ, \sin θ$ ) defines points on the unit circle with equation $x^{2} + y^{2} = 1$ .

The functions $f (x)$ and $g (x)$ are known as hyperbolic functions, as the general point ( $f (θ), g (θ)$ ) defines points on a curve known as a hyperbola with equation $x^{2} - y^{2} = 1$ . This hyperbola has two asymptotes.

Verify that $u = f (t)$ satisfies the differential equation $\frac{d^{2} u}{d t^{2}} = u$ .

[2]

Show that ${(f (t))}^{2} + {(g (t))}^{2} = f (2 t)$ .

[3]

$f (i u)$ .

[3]

c.i.

$g (i u)$ .

[2]

c.ii.

Hence find, and simplify, an expression for ${(f (i u))}^{2} + {(g (i u))}^{2}$ .

[2]

Show that ${(f (t))}^{2} - {(g (t))}^{2} = {(f (i u))}^{2} - {(g (i u))}^{2}$ .

[4]

Sketch the graph of $x^{2} - y^{2} = 1$ , stating the coordinates of any axis intercepts and the equation of each asymptote.

[4]

The hyperbola with equation $x^{2} - y^{2} = 1$ can be rotated to coincide with the curve defined by $x y = k, k \in ℝ$ .

Find the possible values of $k$ .

[5]

Markscheme

$f' (t) = \frac{e^{t} - e^{- t}}{2}$ A1

$f'' (t) = \frac{e^{t} + e^{- t}}{2}$ A1

$= f (t)$ AG

[2 marks]

METHOD 1

${(f (t))}^{2} + {(g (t))}^{2}$

substituting $f$ and $g$ M1

$= \frac{{(e^{t} + e^{- t})}^{2} + {(e^{t} - e^{- t})}^{2}}{4}$

$= \frac{{(e^{t})}^{2} + 2 + {(e^{- t})}^{2} + {(e^{t})}^{2} - 2 + {(e^{- t})}^{2}}{4}$ (M1)

$= \frac{{(e^{t})}^{2} + {(e^{- t})}^{2}}{2} (= \frac{e^{2 t} + e^{- 2 t}}{2})$ A1

$= f (2 t)$ AG

METHOD 2

$f (2 t) = \frac{e^{2 t} + e^{- 2 t}}{2}$

$= \frac{{(e^{t})}^{2} + {(e^{- t})}^{2}}{2}$ M1

$= \frac{{(e^{t} + e^{- t})}^{2} + {(e^{t} - e^{- t})}^{2}}{4}$ M1A1

$= {(f (t))}^{2} + {(g (t))}^{2}$ AG

Note: Accept combinations of METHODS 1 & 2 that meet at equivalent expressions.

[3 marks]

substituting $e^{i u} = \cos u + i \sin u$ into the expression for $f$ (M1)

obtaining $e^{-i u} = \cos u - i \sin u$ (A1)

$f (i u) = \frac{\cos u + i \sin u + \cos u - i \sin u}{2}$

Note: The M1 can be awarded for the use of sine and cosine being odd and even respectively.

$= \frac{2 \cos u}{2}$

$= \cos u$ A1

[3 marks]

c.i.

$g (i u) = \frac{\cos u + i \sin u - \cos u + i \sin u}{2}$

substituting and attempt to simplify (M1)

$= \frac{2 i \sin u}{2}$

$= i \sin u$ A1

[2 marks]

c.ii.

METHOD 1

${(f (i u))}^{2} + {(g (i u))}^{2}$

substituting expressions found in part (c) (M1)

$= \cos^{2} u - \sin^{2} u (= \cos 2 u)$ A1

METHOD 2

$f (2 i u) = \frac{e^{2 i u} + e^{- 2 i u}}{2}$

$= \frac{\cos 2 u + i \sin 2 u + \cos 2 u - i \sin 2 u}{2}$ M1

$= \cos 2 u$ A1

Note: Accept equivalent final answers that have been simplified removing all imaginary parts eg $2 \cos^{2} u - 1$ etc

[2 marks]

${(f (t))}^{2} - {(g (t))}^{2} = \frac{{(e^{t} + e^{- t})}^{2} - {(e^{t} - e^{- t})}^{2}}{4}$ M1

$= \frac{(e^{2 t} + e^{- 2 t} + 2) - (e^{2 t} + e^{- 2 t} - 2)}{4}$ A1

$= \frac{4}{4} = 1$ A1

Note: Award A1 for a value of 1 obtained from either LHS or RHS of given expression.

${(f (i u))}^{2} - {(g (i u))}^{2} = \cos^{2} u + \sin^{2} u$ M1

$= 1$ (hence ${(f (t))}^{2} - {(g (t))}^{2} = {(f (i u))}^{2} - {(g (i u))}^{2}$ ) AG

Note: Award full marks for showing that ${(f (z))}^{2} - {(g (z))}^{2} = 1, \forall z \in ℂ$ .

[4 marks]

A1A1A1A1

Note: Award A1 for correct curves in the upper quadrants, A1 for correct curves in the lower quadrants, A1 for correct $x$ -intercepts of $(- 1, 0)$ and $(1, 0)$ (condone $x = - 1$ and $1$ ), A1 for $y = x$ and $y = - x$ .