Date | May 2022 | Marks available | 5 | Reference code | 22M.1.SL.TZ1.5 |
Level | Standard Level | Paper | Paper 1 (without calculator) | Time zone | Time zone 1 |
Command term | Find | Question number | 5 | Adapted from | N/A |
Question
Consider the curve with equation y=(2x-1)ekx, where x∈ℝ and k∈ℚ.
The tangent to the curve at the point where x=1 is parallel to the line y=5ekx.
Find the value of k.
Markscheme
evidence of using product rule (M1)
dydx=(2x-1)×(kekx)+2×ekx (=ekx(2kx-k+2)) A1
correct working for one of (seen anywhere) A1
dydx at x=1⇒kek+2ek
OR
slope of tangent is 5ek
their dydx at x=1 equals the slope of y=5ekx (=5ek) (seen anywhere) (M1)
kek+2ek=5ek
k=3 A1
[5 marks]
Examiners report
The product rule was well recognised and used with 𝑥=1 properly substituted into this expression. Although the majority of the candidates tried equating the derivative to the slope of the tangent line, the slope of the tangent line was not correctly identified; many candidates incorrectly substituted 𝑥=1 into the tangent equation, thus finding the y-coordinate instead of the slope.