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Date November 2014 Marks available 3 Reference code 14N.1.sl.TZ0.15
Level SL only Paper 1 Time zone TZ0
Command term Find Question number 15 Adapted from N/A

Question

Consider the curve y=x2+ax1, x0.

Find dydx.

[3]
a.

The gradient of the tangent to the curve is 14 when x=1.

Find the value of a.

[3]
b.

Markscheme

2xax2     (A1)(A1)(A1)     (C3)

Notes: Award (A1) for 2x, (A1) for a and (A1) for x2.

Award at most (A1)(A1)(A0) if extra terms are present.

a.

2(1)a12=14     (M1)(M1)

Note: Award (M1) for substituting 1 into their gradient function, (M1) for equating their gradient function to 14.

Award (M0)(M0)(A0) if the original function is used instead of the gradient function.

 

a=16     (A1)(ft)     (C3)

Note: Follow through from their gradient function from part (a).

b.

Examiners report

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

Syllabus sections

Topic 7 - Introduction to differential calculus » 7.2 » The principle that f(x)=axnf(x)=anxn1 .
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