Date | November 2020 | Marks available | 3 | Reference code | 20N.3.AHL.TZ0.Hca_3 |
Level | Additional Higher Level | Paper | Paper 3 | Time zone | Time zone 0 |
Command term | Label and Sketch | Question number | Hca_3 | Adapted from | N/A |
Question
The curve y=f(x) has a gradient function given by
dydx=x-y.
The curve passes through the point (1, 1).
On the same set of axes, sketch and label isoclines for dydx=-1, 0 and 1, and clearly indicate the value of each y-intercept.
Hence or otherwise, explain why the point (1, 1) is a local minimum.
Find the solution of the differential equation dydx=x-y, which passes through the point (1, 1). Give your answer in the form y=f(x).
Explain why the graph of y=f(x) does not intersect the isocline dydx=1.
Sketch the graph of y=f(x) on the same set of axes as part (a)(i).
Markscheme
attempt to find equation of isoclines by setting x-y=-1,0,1 M1
3 parallel lines with positive gradient A1
y-intercept =-c for dydx=c A1
Note: To award A1, each y-intercept should be clear, but condone a missing label (eg. (0, 0)).
If candidates represent the lines using slope fields, but omit the lines, award maximum of M1A0A1.
[3 marks]
at point (1, 1), dydx=0 A1
EITHER
to the left of (1, 1), the gradient is negative R1
to the right of (1, 1), the gradient is positive R1
Note: Accept any correct reasoning using gradient, isoclines or slope field.
If a candidate uses left/right or x<1 / x>1 without explicitly referring to the point (1, 1) or a correct region on the diagram, award R0R1.
OR
d2ydx2=1-dydx A1
d2ydx2=1(>0) A1
Note: accept correct reasoning dydx that is increasing as x increases.
THEN
hence (1, 1) is a local minimum AG
[3 marks]
integrating factor =e∫dx (M1)
=ex (A1)
dydxex+yex=xex (M1)
yex=∫xex dx A1
=xex-∫ex dx (M1)
=xex-ex(+c) A1
Note: Award A1 for the correct RHS.
substituting (1, 1) gives
e=e-e+c M1
c=e
y=x-1+e1-x A1
[8 marks]
METHOD 1
EITHER
attempt to solve for the intersection x-1+e1-x=x-1 (M1)
OR
attempt to find the difference x-1+e1-x-(x-1) (M1)
THEN
e1-x>0 for all x R1
Note: Accept e1-x≠0 or equivalent reasoning.
therefore the curve does not intersect the isocline AG
METHOD 2
y=x-1 is an (oblique) asymptote to the curve R1
Note: Do not accept “the curve is parallel to y=x-1"
y=x-1 is the isocline for dydx=1 R1
therefore the curve does not intersect the isocline AG
METHOD 3
The initial point is above y=x-1, so dydx<1 R1
⇒x-y<1
⇒y>x-1 R1
therefore the curve does not intersect the isocline AG
[2 marks]
concave up curve with minimum at approximately (1, 1) A1
asymptote of curve is isocline y=x-1 A1
Note: Only award FT from (b) if the above conditions are satisfied.
[2 marks]