Loading [MathJax]/jax/element/mml/optable/GeneralPunctuation.js

User interface language: English | Español

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=xtan(πxy4).

At the point (1, 1) , show that dydx=2+π2π.

[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

dydx=xsec2(πxy4)[(π4xdydx+π4y)]+tan(πxy4)     A1A1

Note: Award A1 for each term.

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

dydx=π2dydx+π2+1

dydx(1π2)=π2+1     A1

dydx=2+π2π     AG

[5 marks]

a.

attempt to use gradient of normal =1dydx       (M1)

=π2π+2

so equation of normal is y1=π2π+2(x1) or y=π2π+2x+4π+2       A1

[2 marks]

b.

Examiners report

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

Syllabus sections

Topic 5 —Calculus » SL 5.4—Tangents and normal
Show 95 related questions
Topic 5 —Calculus

View options