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Date November 2019 Marks available 2 Reference code 19N.1.SL.TZ0.T_14
Level Standard Level Paper Paper 1 Time zone Time zone 0
Command term Find Question number T_14 Adapted from N/A

Question

The diagram shows the curve y=x22+2ax, x0.

The equation of the vertical asymptote of the curve is x=k.

Write down the value of k.

[1]
a.

Find dydx.

[3]
b.

At the point where x=2, the gradient of the tangent to the curve is 0.5.

Find the value of a.

[2]
c.

Markscheme

k= 0         (A1)   (C1) 

Note: Award (A1) for an answer of "x=0".

[1 mark]

a.

x-2ax2        (A1)(A1)(A1)   (C3) 

Note: Award (A1) for x, (A1) for -2a, (A1) for x-2 or 1x2. Award at most (A1)(A1)(A0) if extra terms are seen.

[3 marks]

b.

0.5=2-2a22       (M1)

Note: Award (M1) for their correctly substituted derivative equated to 0.5.

a= 3       (A1)(ft)     (C2)

Note: Follow through from part (b) providing their answer is not a=0 as this value contradicts the graph.

[2 marks]

c.

Examiners report

[N/A]
a.
[N/A]
b.
[N/A]
c.

Syllabus sections

Topic 5—Calculus » SL 5.3—Differentiating polynomials, n E Z
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Topic 5—Calculus » SL 5.4—Tangents and normal
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