Home / IBDP Maths AA: Topic: SL 5.8: Local maximum and minimum points: IB style Questions HL Paper 3

IBDP Maths AA: Topic: SL 5.8: Local maximum and minimum points: IB style Questions HL Paper 3

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Question

The curve y = f ( x ) has a gradient function given by

\(\frac{dy}{dx}\) = x-y

The curve passes through the point ( 1, 1 ).

    1.          (i) On the same set of axes, sketch and label isoclines for \(\frac{dy}{dx}\)= -1,0 and 1, and clearly indicate the value of each y-intercept. dx

      (ii) Hence or otherwise, explain why the point ( 1, 1 ).is a local minimum. [6]

    2.  Find the solution of the differential equation \(\frac{dy}{dx}\)= x-y , which passes through the point ( 1, 1 ). Give your answer in the form y = f ( x ) . [8]

    3.  (i) Explain why the graph of y = f ( x ) . does not intersect the isocline \(\frac{dy}{dx}\)=1

                           (ii) Sketch the graph of y = f ( x )  on the same set of axes as part (a)(i). [4]

Answer/Explanation

Ans:

  1. (a)
    1. (i) attempt to find equation of isoclines by setting x – y  = −1,0,1
    3. 3 parallel lines with positive gradient
    4. y-intercept =-c for \(\frac{dy}{dx}\)= c
    5. 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 maximumof M1A0A1.
    6. (ii) at point (1, 1), \(\frac{dy}{dx}\)= 0
    7. EITHER
    8. to the left of (1, 1), the gradient is negative
    9. to the right of (1, 1), the gradient is positive
    10. OR
    11. \(\frac{d^{2}y}{d^{2}}= 1 – \frac{dy}{dx}\)
    12. \(\frac{d^{2}y}{dx^{2}}= 1 (>0)\)
    13. THEN
    14. hence (1, 1) is a local minimum
      1. Note: Award A1 for the correct RHS.
      2. substituting (1, 1) gives
      3. e = e-e+c
      4. c=e
      5. \(y= x-1+e^{1-x}\)
  3. (c)
    1. (i)
    2. METHOD 1
      1. EITHER
      2. attempt to solve for the intersection \(x-1 +e^{1-x}=x-1\)
      3. OR
      4. attempt to find the difference \(x-1 +e^{1-x}-(x-1)\)
      5. THEN
      6. \(e^{1-x}>0 \) for all x
      7. therefore the curve does not intersect the isocline
    3. METHOD 2
      1. \(y = x −1 \) is an (oblique) asymptote to the curve
      2. Note: Do not accept “the curve is parallel to y = x −1”.
      3. y = x −1 is the isocline for \(\frac{dy}{dx}= 1\)
      4. therefore the curve does not intersect the isocline
    4. METHOD 3
      1. The initial point is above 
      2. \(y = x −1 so \frac{dy}{dx}<1\)
      3. \(\Rightarrow x-y <1\)
      4. \(\Rightarrow y > x-1\)
      5. therefore the curve does not intersect the isocline
      6. ii
      8. concave up curve with minimum at approximately (1, 1)
      9. asymptote of curve is isocline y = x −1 
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