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Date November 2016 Marks available 5 Reference code 16N.1.AHL.TZ0.H_9
Level Additional Higher Level Paper Paper 1 Time zone Time zone 0
Command term Find Question number H_9 Adapted from N/A

Question

A curve has equation 3 x 2 y 2 e x 1 = 2 .

Find an expression for d y d x  in terms of x and y .

[5]
a.

Find the equations of the tangents to this curve at the points where the curve intersects the line x = 1 .

[4]
b.

Markscheme

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

attempt to differentiate implicitly     M1

3 ( 4 y d y d x + 2 y 2 ) e x 1 = 0    A1A1A1

 

Note: Award A1 for correctly differentiating each term.

 

d y d x = 3 e 1 x 2 y 2 4 y    A1

 

Note: This final answer may be expressed in a number of different ways.

 

[5 marks]

a.

3 2 y 2 = 2 y 2 = 1 2 y = ± 1 2    A1

d y d x = 3 2 1 2 ± 4 1 2 = ± 2 2    M1

at ( 1 ,   1 2 )  the tangent is y 1 2 = 2 2 ( x 1 )  and     A1

at ( 1 ,   1 2 )  the tangent is y + 1 2 = 2 2 ( x 1 )      A1

 

Note: These equations simplify to y = ± 2 2 x .

 

Note: Award A0M1A1A0 if just the positive value of y is considered and just one tangent is found.

 

[4 marks]

b.

Examiners report

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

Syllabus sections

Topic 5—Calculus » SL 5.1—Introduction of differential calculus
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Topic 5—Calculus » SL 5.3—Differentiating polynomials, n E Z
Topic 5—Calculus » SL 5.4—Tangents and normal
Topic 5—Calculus

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