Moderate -0.3 This is a straightforward differentiation and equation-solving question. Students differentiate y = e^(2x) to get dy/dx = 2e^(2x), set equal to 1/2, solve for x using logarithms, then find y. While it requires multiple steps and exact form (logarithms), it's a standard AS-level calculus exercise with no conceptual challenges, making it slightly easier than average.
A curve has equation \(y = e^{2x}\)
Find the coordinates of the point on the curve where the gradient of the curve is \(\frac{1}{2}\)
Give your answer in an exact form.
[5 marks]
Question 9:
9 | States the correct gradient of the
curve | AO1.2 | B1 | Grad of curve = 2e2x
1
= grad of tangent so 2e2x =
2
1 1
e2x = ⇒2x=ln
4 4
1 1 1
⇒ x= ln =ln =−ln2
2 4 2
1
y =e2x =
4
1
−ln2,
4
Forms an equation using ‘their’
gradient of the curve and puts it equal
1
to
2 | AO1.1a | M1
Takes a log of each side of ‘their’
equation and uses law of logs to
obtain equation in x | AO1.1a | M1
Obtains a correct exact value for x | AO1.1b | A1
Substitutes ‘their’ value of x and
obtains y value and hence the
coordinates
(follow through provided values are
exact) | AO1.1b | A1F
Total | 5
Q | Marking Instructions | AO | Marks | Typical Solution
A curve has equation $y = e^{2x}$
Find the coordinates of the point on the curve where the gradient of the curve is $\frac{1}{2}$
Give your answer in an exact form.
[5 marks]
\hfill \mbox{\textit{AQA AS Paper 2 Q9 [5]}}