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=sin (xy) , y≠0.
Show that dydx=y cos (xy)2y-x cos (xy).
Prove that, when dydx=0 , y=±1.
Hence find the coordinates of all points on C, for 0<x<4π, where dydx=0.
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= cos (xy)⌊xdydx+y⌋ A1M1A1
Note: Award A1 for LHS, M1 for attempt at chain rule, A1 for RHS.
2ydydx= xdydxcos (xy)+y cos (xy)
2ydydx- xdydxcos (xy)=y cos (xy)
dydx(2y-x cos (xy))=y cos (xy) M1
Note: Award M1 for collecting derivatives and factorising.
dydx=y cos (xy)2y-x cos (xy) AG
[5 marks]
setting dydx=0
y cos (xy)=0 (M1)
(y≠0)⇒cos (xy)=0 A1
⇒sin (xy)(=±√1-cos2 (xy)=±√1-0)=±1 OR xy=(2n+1)π2 (n∈ℤ) 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 (=sin (xy))>0 R1
⇒y2=1 A1
⇒y=±1 AG
[5 marks]
y=±1⇒1=sin (±x)⇒sin x=±1 OR y=±1⇒0=cos (±x)⇒cos x=0 (M1)
(sin x=1⇒)(π2, 1), (5π2, 1) A1A1
(sin x=-1⇒)(3π2, -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 sin x.
[5 marks]