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Date November 2016 Marks available 2 Reference code 16N.1.SL.TZ0.T_14
Level Standard Level Paper Paper 1 (with calculator from previous syllabus) Time zone Time zone 0
Command term Find Question number T_14 Adapted from N/A

Question

The equation of a curve is y = 1 2 x 4 3 2 x 2 + 7 .

The gradient of the tangent to the curve at a point P is 10 .

Find d y d x .

[2]
a.

Find the coordinates of P.

[4]
b.

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

2 x 3 3 x      (A1)(A1)     (C2)

 

Note:     Award (A1) for 2 x 3 , award (A1) for 3 x .

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

 

[2 marks]

a.

2 x 3 3 x = 10    (M1)

 

Note:     Award (M1) for equating their answer to part (a) to 10 .

 

x = 2    (A1)(ft)

 

Note:     Follow through from part (a). Award (M0)(A0) for 2 seen without working.

 

y = 1 2 ( 2 ) 4 3 2 ( 2 ) 2 + 7    (M1)

 

Note:     Award (M1) substituting their 2 into the original function.

 

y = 9    (A1)(ft)     (C4)

 

Note:     Accept ( 2 ,   9 ) .

 

[4 marks]

b.

Examiners report

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

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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