Show that $\frac{dk}{dT}$ is always positive.

Given that $\underset{T\to \infty }{\lim }k=A$ and $\underset{T\to 0}{\lim }k=0$, sketch the graph of $k$ against $T$.

(i)   the gradient of this line in terms of $c$;

(ii)  the $y$-intercept of this line in terms of $A$.

Find the equation of the regression line for $\ln k$ on $\frac{1}{T}$.

$c$.

It is not required to state units for this value.

$A$.

It is not required to state units for this value.

attempt to use chain rule, including the differentiation of $\frac{1}{T}$          (M1)

$\frac{dk}{dT}=A\times \frac{c}{T^2}\times {\text{e}}^{-\frac{c}{T}}$          A1

this is the product of positive quantities so must be positive          R1

Note: The R1 may be awarded for correct argument from their derivative. R1 is not possible if their derivative is not always positive.

[3 marks]

         A1A1A1

Note: Award A1 for an increasing graph, entirely in first quadrant, becoming concave down for larger values of $T$, A1 for tending towards the origin and A1 for asymptote labelled at $k=A$.

[3 marks]

taking $\ln$ of both sides   OR   substituting $y=\ln x$  and  $x=\frac{1}{T}$           (M1)

$\ln k=\ln A-\frac{c}{T}$  OR  $y=-cx+\ln A$           (A1)

(i)   so gradient is $-c$         A1

(ii)  $y$-intercept is $\ln A$         A1

Note: The implied (M1) and (A1) can only be awarded if both correct answers are seen. Award zero if only one value is correct and no working is seen.

[4 marks]

an attempt to convert data to $\frac{1}{T}$ and $\ln k$           (M1)

e.g. at least one correct row in the following table

line is $\ln k=-13400\times \frac{1}{T}+15.0 \left(=-13383.1\dots \times \frac{1}{T}+15.0107\dots \right)$         A1

[2 marks]

$c=13400 \left(13383.1\dots \right)$         A1

[1 mark]

attempt to rearrange or solve graphically $\ln A=15.0107\dots$          (M1)

$A=3 300 000 \left(3 304 258\dots \right)$         A1

 Note: Accept an $A$ value of $3269017$… from use of $3\mathrm{sf}$ value.

[2 marks]

This question caused significant difficulties for many candidates and many did not even attempt the question. Very few candidates were able to differentiate the expression in part (a) resulting in difficulties for part (b). Responses to parts (c) to (e) illustrated a lack of understanding of linearizing a set of data. Those candidates that were able to do part (d) frequently lost a mark as their answer was given in x and y.