Solved: **In this question you will explore some of the properties of special functions... [IB Mathematics Analysis and Approaches (AA)]

Question

HLPaper 3

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

Functions $h$ and $j$ are defined as $h(z) = \frac{e^z + e^{-z}}{2}$ and $j(z) = \frac{e^z - e^{-z}}{2}$ , where $z \in \mathbb{C}$ . Consider $p$ and $q$ , such that $p, q \in \mathbb{R}$ .

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

The functions $\cos x$ and $\sin x$ are known as circular functions as the general point $(\cos \theta , \sin \theta )$ defines points on the unit circle with equation $x^2 + y^2 = 1$ . The functions $h(x)$ and $j(x)$ are known as hyperbolic functions, as the general point $(h(\theta), j(\theta))$ defines points on a curve known as a hyperbola with equation $x^2 - y^2 = 1$ . This hyperbola has two asymptotes.

Verify that $u=h(p)$ satisfies the differential equation $\frac{d^2u}{dp^2}=u$ .

[2]

Verified

Solution

$h'(p) = \frac{e^p - e^{-p}}{2}$ A1

$h''(p) = \frac{e^p + e^{-p}}{2}$ A1

$h(p)$ AG

[2 marks]

Show that $h(p)^2 + j(p)^2 = h^2(p)$ .

[3]

Verified

Solution

METHOD 1

${h^2(p) + j^2(p)}$

substituting ${h}$ and ${j}$ M1

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

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

${=h^2(p)}$ AG

METHOD 2

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

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

${=h^2(p) + j^2(p)}$ M1A1

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

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

[3 marks]

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

[4]

Verified

Solution

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 $y=-x$ .

[4 marks]

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

Find the possible values of $m$ .

[5]

Verified

Solution

attempt to rotate by $45^\circ$ in either direction (M1)

Note: Evidence of an attempt to relate to a sketch of $xy=m$ would be sufficient for this (M1).

attempting to rotate a particular point, eg $(1,0)$ (M1)

$(1,0)$ rotates to $\left(\frac{1}{\sqrt{2}}, \pm \frac{1}{\sqrt{2}}\right)$ (or similar) (A1)

hence $m=\pm \frac{1}{2}$ A1A1

[5 marks]

find $j(iq)$ .

[2]

Verified

Solution

$j(iq) = \frac{{\cos q + i \sin q - \cos q + i \sin q}}{{2}}$

substituting and attempt to simplify (M1)

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

$= i \sin q$ A1

[2 marks]

find $h(iq)$ .

[3]

Verified

Solution

substituting $e^{iq} = \cos q + i \sin q$ into the expression for $h$ (M1)

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

$h(iq) = \frac{{\cos q + i \sin q + \cos q - i \sin q}}{{2}}$

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

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

$= \cos q$ (A1)

[3 marks]

Hence find, and simplify, an expression for $h(iq)^2 + j(iq)^2$ .

[2]

Verified

Solution

METHOD 1

$h^{2}(iq) + j^{2}(iq)$

substituting expressions found in part (c) (M1)

$=\cos^{2} q - \sin^{2} q$ A1

METHOD 2

$h^{2}(iq) = \frac{e^{2iq} + e^{-2iq}}{2}$

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

$=\cos^{2}q$ A1

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

[2 marks]

Show that $h(p)^2 - j(p)^2 = h(iq)^2 - j(iq)^2$ .

[4]

Verified

Solution

$h(p)^2 - j(p)^2 = \frac{{(e^p + e^{-p})^2 - (e^p - e^{-p})^2}}{{4}}$ M1

$= \frac{{4e^{2p} + 4e^{-2p} + 8 - 4e^{2p} - 4e^{-2p} + 8}}{{4}}$ A1

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

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

$h(iq)^2 - j(iq)^2 = \cos^2 q + \sin^2 q$ M1

$= 1$ (hence $h(p)^2 - j(p)^2 = h(iq)^2 - j(iq)^2$ ) AG

Note: Award full marks for showing that $h(z)^2 - j(z)^2 = 1, \forall z \in ℂ$ .

[4 marks]

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