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Date May 2019 Marks available 3 Reference code 19M.2.SL.TZ2.T_5
Level Standard Level Paper Paper 2 Time zone Time zone 2
Command term Find Question number T_5 Adapted from N/A

Question

Consider the function  f ( x ) = 1 3 x 3 + 3 4 x 2 x 1 .

Find  f ( x ) .

[3]
d.

Find the gradient of the graph of  y = f ( x ) at  x = 2 .

[2]
e.

Find the equation of the tangent line to the graph of y = f ( x ) at  x = 2 . Give the equation in the form  a x + b y + d = 0 where,  a b , and d Z .

[2]
f.

Markscheme

x 2 + 3 2 x 1       (A1)(A1)(A1)

Note: Award (A1) for each correct term. Award at most (A1)(A1)(A0) if there are extra terms.

[3 marks]

d.

2 2 + 3 2 × 2 1      (M1)

Note: Award (M1) for correct substitution of 2 in their derivative of the function.

6     (A1)(ft)(G2)

Note: Follow through from part (d).

[2 marks]

e.

8 3 = 6 ( 2 ) + c      (M1)

Note: Award (M1) for 2, their part (a) and their part (e) substituted into equation of a straight line.

c = 28 3

OR

( y 8 3 ) = 6 ( x 2 )      (M1)

Note: Award (M1) for 2, their part (a) and their part (e) substituted into equation of a straight line.

OR

y = 6 x 28 3 ( y = 6 x 9.33333 )      (M1)

Note: Award (M1) for their answer to (e) and intercept  28 3  substituted in the gradient-intercept line equation.

18 x + 3 y + 28 = 0   (accept integer multiples)     (A1)(ft)(G2)

Note: Follow through from parts (a) and (e).

[2 marks]

f.

Examiners report

[N/A]
d.
[N/A]
e.
[N/A]
f.

Syllabus sections

Topic 5—Calculus » SL 5.1—Introduction of differential calculus
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