A-LevelMathematicsDifferentiation9709/33 · Oct/Nov 2025 · Q810 Marks

The curve with equation y = e⁻⁵ˣln 5x has a stationary point at x = p.

(a)Show that p satisfies the equation ln 5p = 1/(5p)[3]
(b)By sketching a suitable pair of graphs, show that the equation in part (a) has only one root.[2]
(c)Show by calculation that 0.2 < p < 0.6.[2]
(d)It is given that the equation in part (a) can be written in the form p = (1/5)exp(1/(5p)) where exp(x) denotes e^x. Use an iterative formula based on this rearrangement to calculate p correct to 2 decimal places. Give the result of each iteration to 4 decimal places.[3]

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2025). The mark or mark type is shown in brackets.

Use the correct product or quotient rule, e.g. e⁻⁵ˣ(d/dx ln 5x) + ln 5x (d/dx e⁻⁵ˣ)
Obtain the correct derivative in any form e.g. (1/x)e⁻⁵ˣ – 5e⁻⁵ˣ ln 5x
Obtain the given answer ln 5p = 1/(5p) after full and correct working
Sketch an acceptable graph, e.g. y = ln 5x or y = 1/(5x)
Sketch a second acceptable graph, e.g. y = 1/(5x) or ln 5x, and justify the given statement by dot, cross or statement only one intersection.
Calculate the values of a relevant expression or pair of expressions at p = 0.2 and p = 0.6
Complete the argument correctly with correct calculated values
Use the iterative formula correctly at least twice
Obtain final answer p = 0.35, Answer = 0.35, or just 0.35 stated
Show sufficient iterations to 4 dp to justify 0.35 to 2 dp or show there is a sign change in the interval (0.345, 0.355)

On Oracle Prep you can write or photograph your own answer and have it marked by AI against this mark scheme, with feedback on what you missed.