Challenging +1.8 This question requires knowledge that cosh(ix) = cos(x), recognition that cos^n(x) for odd n is an odd function (hence symmetric about the origin), and understanding that the mean value of an odd function over a symmetric interval is zero. While the hyperbolic-complex connection and the symmetry argument require conceptual insight beyond routine calculation, the question is only 3 marks and the key steps are relatively standard for Further Maths students familiar with complex exponentials and function properties.
The function \(f(x) = \cosh(ix)\) is defined over the domain \(\{x \in \mathbb{R} : -a\pi \leq x \leq a\pi\}\), where \(a\) is a positive integer.
By considering the graph of \(y = [f(x)]^n\), find the mean value of \([f(x)]^n\), when \(n\) is an odd positive integer.
Fully justify your answer.
[3 marks]
Question 12:
12 | Correctly manipulates
[ ]n
y= f(x) into a form that
can be sketched. | AO3.1a | M1 | eix +e−ix
cosh(ix)=
2
(x)+isin (x)+cos (−x)+isin (−x)
cos
=
2
(x)+isin (x)+cos (x)−isin (x)
cos
=
2
=cos (x)
∴[ f(x) ]n =cosn(x)
Since the graph has the same shape/area
above and below the x-axis over the given
domain,the mean value of y= [ f(x) ]n must
be 0.
Sketches an appropriate
section of the graph.
Sketch does not need to
be accurate but should
look symmetrical about
y-axis and the area
above and below the
x-axis should look similar | AO1.1b | M1
Deduces, with fully
correct reasoning, the
correct mean value | AO2.2a | R1
Total | 3
Q | Marking Instructions | AO | Marks | Typical Solution
The function $f(x) = \cosh(ix)$ is defined over the domain $\{x \in \mathbb{R} : -a\pi \leq x \leq a\pi\}$, where $a$ is a positive integer.
By considering the graph of $y = [f(x)]^n$, find the mean value of $[f(x)]^n$, when $n$ is an odd positive integer.
Fully justify your answer.
[3 marks]
\hfill \mbox{\textit{AQA Further Paper 1 Q12 [3]}}