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+π2−π.
Hence find the equation of the normal to C at the point (1, 1).
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(1−π2)=π2+1 A1
dydx=2+π2−π AG
[5 marks]
attempt to use gradient of normal =−1dydx (M1)
=π−2π+2
so equation of normal is y−1=π−2π+2(x−1) or y=π−2π+2x+4π+2 A1
[2 marks]