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Date November 2020 Marks available 5 Reference code 20N.1.AHL.TZ0.H_11
Level Additional Higher Level Paper Paper 1 (without calculator) Time zone Time zone 0
Command term Find and Hence Question number H_11 Adapted from N/A

Question

Consider the curve C defined by y2=sinxy , y0.

Show that dydx=ycosxy2y-xcosxy.

[5]
a.

Prove that, when dydx=0 , y=±1.

[5]
b.

Hence find the coordinates of all points on C, for 0<x<4π, where dydx=0.

[5]
c.

Markscheme

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

attempt at implicit differentiation       M1

2ydydx=cosxyxdydx+y       A1M1A1


Note: Award A1 for LHS, M1 for attempt at chain rule, A1 for RHS.


2ydydx=xdydxcosxy+ycosxy

2ydydx-xdydxcosxy=ycosxy

dydx2y-xcosxy=ycosxy       M1


Note: Award M1 for collecting derivatives and factorising.


dydx=ycosxy2y-xcosxy       AG


[5 marks]

a.

setting dydx=0

ycosxy=0       (M1)

y0cosxy=0       A1

sinxy=±1-cos2xy=±1-0=±1  OR  xy=2n+1π2n  OR  xy=π2, 3π2,       A1


Note: If they offer values for xy, award A1 for at least two correct values in two different ‘quadrants’ and no incorrect values.


y2=sinxy>0       R1

y2=1       A1

y=±1       AG


[5 marks]

b.

y=±11=sin±xsinx=±1  OR  y=±10=cos±xcosx=0       (M1)

sinx=1π2,1,5π2,1       A1A1

sinx=-13π2,-1,7π2,-1       A1A1


Note:
Allow ‘coordinates’ expressed as x=π2, y=1 for example.
Note: Each of the A marks may be awarded independently and are not dependent on (M1) being awarded.

Note: Mark only the candidate’s first two attempts for each case of sinx.

[5 marks]

c.

Examiners report

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

Syllabus sections

Topic 5 —Calculus » SL 5.4—Tangents and normal
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Topic 5 —Calculus » AHL 5.14—Implicit functions, related rates, optimisation
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