1998 Paper 3 Q7

Year: 1998
Paper: 3
Question Number: 7

Course: UFM Pure
Section: Taylor series

Difficulty: 1700.0 Banger: 1500.0

Problem

Sketch the graph of \({\rm f}(s)={ \e}^s(s-3)+3\) for \(0\le s < \infty\). Taking \({\e\approx 2.7}\), find the smallest positive integer, \(m\), such that \({\rm f}(m) > 0\). Now let $$ {\rm b}(x) = {x^3 \over \e^{x/T} -1} \, $$ where \(T\) is a positive constant. Show that \({\rm b}(x)\) has a single turning point in \(0 < x < \infty\). By considering the behaviour for small \(x\) and for large \(x\), sketch \({\rm b}(x)\) for \(0\le x < \infty\). Let $$ \int_0^\infty {\rm b}(x)\,\d x =B, $$ which may be assumed to be finite. Show that \(B = K T^n\) where \(K\) is a constant, and \(n\) is an integer which you should determine. Given that \(\displaystyle{B \approx 2 \int_0^{Tm} {\rm b}(x) {\,\rm d }x}\), use your graph of \({\rm b}(x)\) to find a rough estimate for \(K\).

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Difficulty Rating: 1700.0

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Banger Rating: 1500.0

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Problem source
Sketch the graph of ${\rm f}(s)={ \e}^s(s-3)+3$ for $0\le s < \infty$. Taking
${\e\approx 2.7}$, find the smallest positive integer, $m$, such that
${\rm f}(m) > 0$.
Now let
$$
{\rm b}(x) = {x^3 \over \e^{x/T} -1} \,
$$
where $T$ is a positive constant. 
Show that ${\rm b}(x)$ has a single turning
point in $0 <  x < \infty$. By considering the behaviour for small $x$
and for large $x$, sketch ${\rm b}(x)$ for $0\le x < \infty$.
Let 
$$
 \int_0^\infty {\rm b}(x)\,\d x =B,
$$
which may be assumed to be finite.
Show that $B = K  T^n$ where $K$ is a constant,  and $n$ is an
integer which you should determine.
Given that $\displaystyle{B \approx 2 \int_0^{Tm} {\rm b}(x) {\,\rm d }x}$,   
use your graph of ${\rm b}(x)$ to  find a rough estimate for $K$.