Problems
Filters
2 problems found
2009 Paper 3 Q3
The function \(\f(t)\) is defined, for \(t\ne0\), by \[ \f(t) = \frac t {\e^t-1}\,. \] \begin{questionparts} \item By expanding \(\e^t\), show that \(\displaystyle \lim _{t\to0} \f(t) = 1\,\). Find \(\f'(t)\) and evaluate \(\displaystyle \lim _{t\to0} \f'(t)\,\). \item Show that \(\f(t) +\frac12 t\) is an even function. [{\bf Note:} A function \(\g(t)\) is said to be {\em even} if \(\g(t) \equiv \g(-t)\).] \item Show with the aid of a sketch that \( \e^t( 1-t)\le 1\,\) and deduce that \(\f'(t)\ne 0\) for \(t\ne0\). \end{questionpart} Sketch the graph of \(\f(t)\).
Solution:
- Claim \(f(t) + \frac12 t\) is an even function. Proof: Consider \(f(-t) - \frac12t\), then \begin{align*} f(-t) - \frac12t &= \frac{-t}{e^{-t}-1} - \frac12t \\ &= \frac{-te^t}{1-e^t} - \frac12 t \\ &= \frac{t(1-e^t) -t}{1-e^t} - \frac12 t \\ &= t - \frac{t}{1-e^t} - \frac12 t \\ &= \frac{t}{e^t-1} + \frac12 t \end{align*} So it is even.
- Drawing the tangent to \(y = e^{-x}\) at \((0,1)\) we find that \(e^{-t} \geq (1-t)\) for all \(t\), in particular, \(e^t(1-t) \leq 1\) \(f'(t) = \frac{(e^t(1-t) -1}{(e^t-1)^2} \leq 0\) and \(f'(t) = -\frac12\) when \(t = 0\)
1994 Paper 2 Q3
The function \(\mathrm{f}\) satisfies \(\mathrm{f}(0)=1\) and \[ \mathrm{f}(x-y)=\mathrm{f}(x)\mathrm{f}(y)-\mathrm{f}(a-x)\mathrm{f}(a+y) \] for some fixed number \(a\) and all \(x\) and \(y\). Without making any further assumptions about the nature of the function show that \(\mathrm{f}(a)=0\). Show that, for all \(t\),
- \(\mathrm{f}(t)=\mathrm{f}(-t)\),
- \(\mathrm{f}(2a)=-1\),
- \(\mathrm{f}(2a-t)=-\mathrm{f}(t)\),
- \(\mathrm{f}(4a+t)=\mathrm{f}(t)\).
Solution: Let \(P(x,y)\) be the statement that the functional equation holds, then: \begin{align*} P(0,0): && f(0) &= f(0)f(0)-f(a)f(a) \\ \Rightarrow && 1 &= 1 - f(a)^2 \\ \Rightarrow && f(a)^2 &= 0 \\ \Rightarrow && f(a) &= 0 \end{align*}
- \begin{align*} P(0,t): && f(-t) &= f(0)f(t) - f(a)f(a-t) \\ \Rightarrow && f(-t) &= f(t) - 0 \\ \Rightarrow && f(t) &= f(-t) \end{align*}
- \begin{align*} P(a,a): && f(0) &= f(a)f(a)-f(0)f(2a) \\ \Rightarrow && 1 &= 0 - f(2a) \\ \Rightarrow && f(2a) &= -1 \end{align*}
- \begin{align*} P(2a,t): && f(2a-t) &= f(2a)f(t) - f(-a)f(a+t) \\ \Rightarrow && f(2a-t) &= -f(t)-f(a)f(a+t) \\ &&&= -f(t)-0 \\ \Rightarrow && f(2a-t) &= -f(t) \end{align*}
- \begin{align*} && f(4a+t) &= f(2a-(-2a-t)) \\ &&&=-f(2a+t) \\ &&&=-f(2a-(-t)) \\ &&&=f(-t) \\ &&&=f(t) \end{align*}