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Date May 2019 Marks available 5 Reference code 19M.1.AHL.TZ2.H_6
Level Additional Higher Level Paper Paper 1 (without calculator) Time zone Time zone 2
Command term Show that Question number H_6 Adapted from N/A

Question

The curve C is given by the equation y = x tan ( π x y 4 ) .

At the point (1, 1) , show that  d y d x = 2 + π 2 π .

[5]
a.

Hence find the equation of the normal to C at the point (1, 1).

[2]
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

d y d x = x se c 2 ( π x y 4 ) [ ( π 4 x d y d x + π 4 y ) ] + tan ( π x y 4 )      A1A1

Note: Award A1 for each term.

attempt to substitute x = 1 , y = 1 into their equation for  d y d x      M1

d y d x = π 2 d y d x + π 2 + 1

d y d x ( 1 π 2 ) = π 2 + 1      A1

d y d x = 2 + π 2 π      AG

[5 marks]

a.

attempt to use gradient of normal  = 1 d y d x        (M1)

= π 2 π + 2

so equation of normal is  y 1 = π 2 π + 2 ( x 1 ) or  y = π 2 π + 2 x + 4 π + 2        A1

[2 marks]

b.

Examiners report

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

Syllabus sections

Topic 5 —Calculus » SL 5.4—Tangents and normal
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Topic 5 —Calculus

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