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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 and k.

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  =ekx2kx-k+2            A1

correct working for one of (seen anywhere)            A1

dydx at x=1kek+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.

Syllabus sections

Topic 5 —Calculus » SL 5.4—Tangents and normal
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