Year: 2019
Paper: 2
Question Number: 1
Course: LFM Pure and Mechanics
Section: Differentiation from first principles
The Pure questions were again the most popular of the paper, with only one of the questions attempted by fewer than half of the candidates (of the remaining four questions, only question 9 was attempted by more than a quarter of the candidates). In many of the questions candidates were often unable to make good use of the results shown in the earlier parts of the question in order to solve the more complex later parts. Nevertheless, some good solutions were seen to all of the questions. For many of the questions, solutions were seen in which the results were reached, but without sufficient justification of some of the steps.
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
Let $f(x) = (x-p)g(x)$, where g is a polynomial. Show that the tangent to the curve $y = f(x)$ at the point with $x = a$, where $a \neq p$, passes through the point $(p, 0)$ if and only if $g'(a) = 0$.
The curve $C$ has equation
$$y = A(x - p)(x - q)(x - r),$$
where $p$, $q$ and $r$ are constants with $p < q < r$, and $A$ is a non-zero constant.
\begin{questionparts}
\item The tangent to $C$ at the point with $x = a$, where $a \neq p$, passes through the point $(p, 0)$. Show that $2a = q + r$ and find an expression for the gradient of this tangent in terms of $A$, $q$ and $r$.
\item The tangent to $C$ at the point with $x = c$, where $c \neq r$, passes through the point $(r, 0)$. Show that this tangent is parallel to the tangent in part (i) if and only if the tangent to $C$ at the point with $x = q$ does not meet the curve again.
\end{questionparts}The tangent to the curve $y = f(x)$ at $x = a$ has the equation $\frac{y-f(a)}{x-a} = f'(a) = g(a)+(a-p)g'(a)$. This passes through $(p,0)$ iff
\begin{align*}
&& \frac{-f(a)}{p-a} &= g(a)+(a-p)g'(a) \\
\Leftrightarrow && -f(a) &= (p-a)g(a) -(a-p)^2g'(a) \\
\Leftrightarrow && -f(a) &= -f(a) -(a-p)^2g'(a) \\
\Leftrightarrow && 0 &= g'(a) \\
\end{align*}
\begin{questionparts}
\item In this case $g(x) = A(x-q)(x-r) = A(x^2-(q+r)x+qr)$ and so we must have that $g'(a) = 0$, ie $A(2a-(q+r)) = 0 \Rightarrow 2a = q+r$
The gradient is $g(a) +(a-p)g'(a) = g(a) = A(a-q)(a-r)$
\item By the same reasoning, but with $g(x) = A(x-p)(x-q)$ we have the gradient is $A(c-p)(c-r)$. This is parallel iff
\begin{align*}
&& (c-p)(c-r) &= (a-q)(a-r)
\end{align*}
The tangent at $x = q$ is $\frac{y-0}{x-q} = A(q-p)(q-r)$ or $ y = A(q-p)(q-r)(x-q)$
\end{questionparts}
This was the question answered by the largest proportion of candidates and many good solutions were seen. However, many candidates did not appreciate the importance of the phrase if and only if in parts of this question. As a result a large number of attempts failed to achieve full marks as it was not made clear that the reasoning presented also worked in the opposite direction. Having shown the first result, many candidates were able to identify the appropriate choice of g(x) when attempting part (i) and successfully showed that 2a = q + r. Many were also able to find a correct expression for the gradient, although some did not find this expression in terms of the variables requested. In part (ii) a pleasing number of candidates were able to recognise that the results from part (i) would be relevant here as well. Again, some of the solutions to this part failed to recognise that the question required the result to be shown in both directions.