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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 Explain and Hence or otherwise 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.

[3]
a.i.

Hence or otherwise, explain why the point (1, 1) is a local minimum.

[3]
a.ii.

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).

[8]
b.

Explain why the graph of y=fx does not intersect the isocline dydx=1.

[2]
c.i.

Sketch the graph of y=fx on the same set of axes as part (a)(i).

[2]
c.ii.

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]

a.i.

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]

a.ii.

integrating factor =edx       (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]

b.

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-x0 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]

c.i.

 

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]

c.ii.

Examiners report

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

Syllabus sections

Topic 5 —Calculus » AHL 5.18—1st order DE’s – Euler method, variables separable, integrating factor, homogeneous DE using sub y=vx
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Topic 5 —Calculus

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