Year: 2019
Paper: 3
Question Number: 2
Course: LFM Pure and Mechanics
Section: Differentiation from first principles
There was a significant rise in the total entry this year with an increase of nearly 8.5% on 2018. One question was attempted by over 90%, two others were very popular, and three further questions were attempted by 60% or more. No question was generally avoided and even the least popular attracted more than 10% of the candidates. 88% restricted themselves to attempting no more than 7 questions, and only a handful, but not the very best, scored strongly attempting more than 7 questions.
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
The definition of the derivative $f'$ of a (differentiable) function f is
$$f'(x) = \lim_{h\to 0} \frac{f(x + h) - f(x)}{h}. \quad (*)$$
\begin{questionparts}
\item The function f has derivative $f'$ and satisfies
$$f(x + y) = f(x)f(y)$$
for all $x$ and $y$, and $f'(0) = k$ where $k \neq 0$. Show that $f(0) = 1$.
Using $(*)$, show that $f'(x) = kf(x)$ and find $f(x)$ in terms of $x$ and $k$.
\item The function g has derivative $g'$ and satisfies
$$g(x + y) = \frac{g(x) + g(y)}{1 + g(x)g(y)}$$
for all $x$ and $y$, $|g(x)| < 1$ for all $x$, and $g'(0) = k$ where $k \neq 0$.
Find $g'(x)$ in terms of $g(x)$ and $k$, and hence find $g(x)$ in terms of $x$ and $k$.
\end{questionparts}\begin{questionparts}
\item $\,$ \begin{align*}
&& f(0+x) &= f(0)f(x) \\
\Rightarrow && f(0) &= 0, 1\\
&&\text{since }f'(0) \neq 0 & \text{ there is some non-zero } f(x) \\
\Rightarrow && f(0) &= 1
\end{align*}
\begin{align*}
&& f'(x) &= \lim_{h\to 0} \frac{f(x+h)-f(x)}{h} \\
&&&= \lim_{h\to 0} \frac{f(x)f(h)-f(x)}{h} \\
&&&= f(x) \cdot \lim_{h\to 0} \frac{f(h)-1}{h} \\
&&&= f(x) \cdot \lim_{h\to 0} \frac{f(0+h)-f(0)}{h} \\
&&&= f(x) \cdot f'(0) \\
&&&= kf(x)
\end{align*}
Since $f'(x) = kf(x)$ we must have $\frac{f'(x)}{f(x)} = k \Rightarrow \ln f(x) = kx + c \Rightarrow f(x) = Ae^{kx}$ but $f(0) = 1$ so $f(x) = e^{kx}$
\item Consider
\begin{align*}
&& g(0+0) &= \frac{g(0)+g(0)}{1+(g(0))^2} \\
\Rightarrow && g(0)(1+(g(0))^2)&= 2g(0) \\
\Rightarrow && 0 &= g(0)\left (1- (g(0))^2 \right) \\
\Rightarrow && g(0) &= -1, 0, 1 \\
\Rightarrow && g(0) &= 0 \tag{$|g(0)| < 1$}
\end{align*}
\begin{align*}
&& g'(x) &=\lim_{h\to 0} \frac{g(x+h)-g(x)}{h} \\
&&&= \lim_{h\to 0} \frac{\frac{g(x)+g(h)}{1+g(x)g(h)}-g(x)}{h} \\
&&&= \lim_{h\to 0} \frac{g(x)+g(h)-g(x)(1+g(x)g(h))}{h(1+g(x)g(h))} \\
&&&= \lim_{h\to 0} \frac{g(h)-g(x)(g(x)g(h))}{h(1+g(x)g(h))} \\
&&&= (1-(g(x))^2) \cdot \lim_{h \to 0} \frac{1}{1+g(x)g(h)} \cdot \lim_{h \to 0} \frac{g(h)}{h} \\
&&&= (1-(g(x))^2) \cdot \frac{1}{1+g(x)\cdot 0} \cdot \lim_{h \to 0} \frac{g(h) - g(0)}{h} \\
&&&= (1-(g(x))^2) \cdot g'(0)\\
&&&= k (1-(g(x))^2) \\
\end{align*}
Let $y = g(x)$ so
\begin{align*}
&& y' &= k(1-y^2) \\
\Rightarrow && kx &= \int \frac{1}{1-y^2} \d y \\
\Rightarrow &&&= \int \frac12\left ( \frac{1}{1-y} + \frac{1}{1+y} \right) \d y \\
&&&= \frac12\ln \left ( \frac{1+y}{1-y} \right) + C \\
x = 0, y = 0: && 0 &= \ln 1 + C \\
\Rightarrow && C &= 0 \\
\Rightarrow && \frac{1+y}{1-y} &= e^{2kx} \\
\Rightarrow && 1+y &= e^{2kx} - e^{2kx}y \\
\Rightarrow && y &= \frac{e^{2kx}-1}{e^{2kx}+1} \\
&&&= \tanh kx
\end{align*}
\end{questionparts}
The most popular question, it was also the most successful with an average score of over 11/20 and many fully correct solutions. Most candidates that noticed that the equation for f(xy) implied f(x) = 0 ∀x, or f(0) = 1 correctly eliminated the former, but quite a few did not realise that it was a possibility to consider. Nearly every candidate successfully found f′(x) = f(x) lim and most also proceeded correctly from there to find the required differential equation. Finding f(x) was generally successful although some did not check the boundary conditions. In part (ii), there were fewer issues demonstrating that g(0) = 0 than there had been with f(0) = 1 in part (i). The simplification in order to find the limit to obtain g′(x) was usually successful. Solution of that differential equation was often well done, either using partial fractions or as a hyperbolic function, although some mistakenly identified the solution as a tan function.