Date | November Example questions | Marks available | 4 | Reference code | EXN.2.AHL.TZ0.12 |
Level | Additional Higher Level | Paper | Paper 2 | Time zone | Time zone 0 |
Command term | Show that | Question number | 12 | Adapted from | N/A |
Question
Consider the differential equation
dydx=f(yx), x>0
The curve y=f(x) for x>0 has a gradient function given by
dydx=y2+3xy+2x2x2.
The curve passes through the point (1, -1).
Use the substitution y=vx to show that ∫dvf(v)-v=ln x+C where C is an arbitrary constant.
By using the result from part (a) or otherwise, solve the differential equation and hence show that the curve has equation y=x(tan (ln x)-1).
The curve has a point of inflexion at (x1, y1) where e-π2<x1<eπ2. Determine the coordinates of this point of inflexion.
Use the differential equation dydx=y2+3xy+2x2x2 to show that the points of zero gradient on the curve lie on two straight lines of the form y=mx where the values of m are to be determined.
Markscheme
* This sample question was produced by experienced DP mathematics senior examiners to aid teachers in preparing for external assessment in the new MAA course. There may be minor differences in formatting compared to formal exam papers.
y=vx⇒dydx=v+xdvdx M1
v+xdvdx=f(v) A1
∫dvf(v)-v=∫dxx A1
integrating the RHS, ∫dvf(v)-v=ln x+C AG
[3 marks]
EITHER
attempts to find f(v) M1
f(v)=v2+3v+2 (A1)
substitutes their f(v) into ∫dvf(v)-v M1
∫dvf(v)-v=∫dvv2+2v+2
attempts to complete the square (M1)
∫dv(v+1)2+1 A1
arctan (v+1) (=ln x+C) A1
OR
attempts to find f(v) M1
v+xdvdx=v2+3v+2 A1
∫dvv2+2v+2=dxx M1
attempts to complete the square (M1)
∫dv(v+1)2+1(=∫dxx) A1
arctan (v+1) (=ln x+C) A1
THEN
when x=1, v=-1 (or y=-1) and so C=0 M1
substitutes for v into their expression M1
arctan (yx+1)=ln x
yx+1=tan (ln x) A1
so y=x(tan (ln x)-1) AG
[9 marks]
METHOD 1
EITHER
a correct graph of y=f'(x) (for approximately e-π2<x<eπ2) with a local minimum point below the x-axis A2
Note: Award M1A1 for dydx=tan (ln x)+sec2 (ln x)-1.
attempts to find the x-coordinate of the local minimum point on the graph of y=f'(x) (M1)
OR
a correct graph of y=f''(x) (for approximately e-π2<x<eπ2) showing the location of the x-intercept A2
Note: Award M1A1 for d2ydx2=sec2 (ln x)x+2 sec2 (ln x) tan (ln x)x.
attempts to find the -intercept (M1)
THEN
A1
attempts to find (M1)
the coordinates are A1
METHOD 2
attempts implicit differentiation on to find M1
(or equivalent)
() A1
attempts to solve for where M1
A1
attempts to find (M1)
the coordinates are A1
[6 marks]
M1
attempts to solve for M1
or A1
and A1
Note: Award M1 for stating , M1 for substituting into , A1 for and A1 for and .
[4 marks]