Standard +0.3 This is a straightforward calculus problem requiring differentiation of an exponential, equating the gradient to the given line's slope (-1/8), solving a simple logarithmic equation, and substituting back. All steps are standard AS-level techniques with no novel insight required, making it slightly easier than average.
A curve has the equation \(y = e^{-2x}\)
At point \(P\) on the curve the tangent is parallel to the line \(x + 8y = 5\)
Find the coordinates of \(P\) stating your answer in the form \((\ln p, q)\), where \(p\) and \(q\) are rational.
[7 marks]
Question 6:
6 | Recalls gradient function for
đđđĽđĽ | 1.2 | B1 | Gradient of is â2
â2đĽđĽ â2đĽđĽ
Gradiente of line is e
1
â8
â2 =
1
â2 đĽđĽ
â8
e
= 16
2đĽđĽ
2ex = ln 16
x = ln 16 = ln 4
1
2
y = = =
1
â2ln4 âln16
ln16
e y = e e
1
16
P is (ln 4, )
1
16
e
Finds gradient of line | 1.1b | B1
Equates their gradient of line to
their gradient of tangent | 3.1a | M1
Solves their equation for x | 1.1a | M1
Obtains correct value for x as
ln 4 | 1.1b | A1
Substitutes their x value to
obtain a value for y in a correct
but unsimplified form
Or
uses gradient = â2y =
1 | 1.1a | M1
â8
Obtains correct value for y | 1.1b | R1
Total | 7
Q | Marking instructions | AO | Marks | Typical solution
A curve has the equation $y = e^{-2x}$
At point $P$ on the curve the tangent is parallel to the line $x + 8y = 5$
Find the coordinates of $P$ stating your answer in the form $(\ln p, q)$, where $p$ and $q$ are rational.
[7 marks]
\hfill \mbox{\textit{AQA AS Paper 1 2021 Q6 [7]}}