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Date November 2021 Marks available 1 Reference code 21N.1.AHL.TZ0.13
Level Additional Higher Level Paper Paper 1 Time zone Time zone 0
Command term Calculate Question number 13 Adapted from N/A

Question

The slope field for the differential equation dydx=e-x2-y is shown in the following two graphs.

On the second graph,

Calculate the value of dydx at the point (0, 1).

[1]
a.

Sketch, on the first graph, a curve that represents the points where dydx=0.

[2]
b.

(i)   sketch the solution curve that passes through the point (0, 0).

(ii)  sketch the solution curve that passes through the point (0, 0.75).

[4]
c.

Markscheme

dydx=e0-1=0                 A1

 

[1 mark]

a.

gradient =0 at 0, 1                 A1

correct shape                 A1

 
Note: Award second A1 for horizontal asymptote of y=0, and general symmetry about the y-axis.

 

[2 marks]

b.

(i)  positive gradient at origin              A1

      correct shape                 A1

 
Note: Award second A1 for a single maximum in 1st quadrant and tending toward an asymptote.

 

(ii)  positive gradient at (0, 0.75)                 A1

      correct shape                 A1


Note: Award second A1 for a single minimum in 2nd quadrant, single maximum in 1st quadrant and tending toward an asymptote.

 

[4 marks]

c.

Examiners report

There were many good attempts at this question. Care needs to be taken over graph sketching, and the existence of asymptotes or the position of intersections needs to be shown clearly. Many candidates correctly found dydx=0 at 0,1 in part (a). However, they were then misled into finding a solution curve through this point rather than graphing the points where dydx=0 as required in part (b). Part (c) was answered well with a number of correct answers. Often the curve through 0,0.75 had a flat central section and did not show a clear maximum and minimum. The asymptotes were generally poorly drawn with the curves meeting the x-axis and stopping or worse still crossing over it.

a.

There were many good attempts at this question. Care needs to be taken over graph sketching, and the existence of asymptotes or the position of intersections needs to be shown clearly. Many candidates correctly found dydx=0 at 0,1 in part (a). However, they were then misled into finding a solution curve through this point rather than graphing the points where dydx=0 as required in part (b). Part (c) was answered well with a number of correct answers. Often the curve through 0,0.75 had a flat central section and did not show a clear maximum and minimum. The asymptotes were generally poorly drawn with the curves meeting the x-axis and stopping or worse still crossing over it.

b.

There were many good attempts at this question. Care needs to be taken over graph sketching, and the existence of asymptotes or the position of intersections needs to be shown clearly. Many candidates correctly found dydx=0 at 0,1 in part (a). However, they were then misled into finding a solution curve through this point rather than graphing the points where dydx=0 as required in part (b). Part (c) was answered well with a number of correct answers. Often the curve through 0,0.75 had a flat central section and did not show a clear maximum and minimum. The asymptotes were generally poorly drawn with the curves meeting the x-axis and stopping or worse still crossing over it.

c.

Syllabus sections

Topic 5—Calculus » AHL 5.15—Slope fields
Topic 5—Calculus

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