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Date November 2012 Marks available 4 Reference code 12N.1.sl.TZ0.4
Level SL only Paper 1 Time zone TZ0
Command term Find Question number 4 Adapted from N/A

Question

Part of the graph of f(x)=ax36x2f(x)=ax36x2 is shown below.

 

 

The point P lies on the graph of ff . At P,  x = 1.

Find f(x) .

[2]
a.

The graph of f has a gradient of 3 at the point P. Find the value of a .

[4]
b.

Markscheme

f(x)=3ax212x     A1A1     N2

Note: Award A1 for each correct term.

[2 marks]

a.

setting their derivative equal to 3 (seen anywhere)     A1

e.g. f(x)=3

attempt to substitute x=1 into f(x)     (M1)

e.g. 3a(1)212(1)

correct substitution into f(x)     (A1)

e.g. 3a12 , 3a=15

a=5    A1     N2

[4 marks]

b.

Examiners report

A majority of candidates answered part (a) correctly, and a good number earned full marks on both parts of this question. In part (b), some common errors included setting the derivative equal to zero, or substituting 3 for x in their derivative. There were also a few candidates who incorrectly tried to work with f(x) , rather than f(x) , in part (b).

a.

A majority of candidates answered part (a) correctly, and a good number earned full marks on both parts of this question. In part (b), some common errors included setting the derivative equal to zero, or substituting 3 for x in their derivative. There were also a few candidates who incorrectly tried to work with f(x) , rather than f(x) , in part (b).

b.

Syllabus sections

Topic 6 - Calculus » 6.1 » Derivative interpreted as gradient function and as rate of change.
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