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Date May 2022 Marks available 2 Reference code 22M.1.AHL.TZ1.8
Level Additional Higher Level Paper Paper 1 Time zone Time zone 1
Command term Find Question number 8 Adapted from N/A

Question

Consider the curve y=2x4-ex.

Find dydx.

[2]
a.i.

Find d2ydx2.

[2]
a.ii.

The curve has a point of inflexion at a, b.

Find the value of a.

[2]
b.

Markscheme

use of product rule         (M1)

dydx=24-ex+2x-ex         A1

=8-2ex-2xex


[2 marks]

a.i.

use of product rule         (M1)

d2ydx2=-2ex-2ex-2xex         A1

=-4ex-2xex

=-22+xex


[2 marks]

a.ii.

-22+aea=0  OR  sketch of d2ydx2 with x-intercept indicated  OR  finding the local maximum of dydx at -2, 8.27         (M1)

a=  -2         A1


[2 marks]

b.

Examiners report

Some candidates attempted to apply the product rule in parts (a)(i) and (ii) but often incorrectly, particularly in part (ii) when finding d2ydx2. In part (b) there was little understanding shown of the point of inflexion. There were some attempts, some of which were correct, but many where either the function or the first derivative were set to zero rather than the second derivative.

a.i.
[N/A]
a.ii.
[N/A]
b.

Syllabus sections

Topic 5—Calculus » AHL 5.10—Second derivatives, testing for max and min
Topic 5—Calculus

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