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0580/41 May/Jun 2021, Question 7: Stationary Points and Turning Points

Cambridge IGCSE Mathematics 0580 · Paper 41 (Extended) · Calculator · 14 marks

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The question, in short

Simplify an algebraic fraction, combine two fractions into one, and use differentiation on y = 2x³ − 4x² + 6 to find gradient and stationary points.

Summarised in our own words. Use the official paper for the exact wording and any diagrams.

Part (a) · 3 marks

Simplify the fraction (x² − 25)/(x² − x − 20) by factorising numerator and denominator.

  1. 1x² − 25 = (x − 5)(x + 5)
  2. 2x² − x − 20 = (x − 5)(x + 4)
  3. 3Cancel (x − 5): (x + 5)/(x + 4)

Answer(x + 5)/(x + 4)

Part (b) · 3 marks

Express (x + 5)/x + (x + 8)/(x − 1) as one fraction in simplest form.

  1. 1Common denominator x(x − 1)
  2. 2[(x + 5)(x − 1) + x(x + 8)]/(x(x − 1))
  3. 3= (x² + 4x − 5 + x² + 8x)/(x(x − 1))
  4. 4= (2x² + 12x − 5)/(x(x − 1))

Answer(2x² + 12x − 5)/(x(x − 1)) or (2x² + 12x − 5)/(x² − x)

Part (ci) · 2 marks

Differentiate y = 2x³ − 4x² + 6 to obtain dy/dx.

  1. 1Multiply by the power, then take 1 off the power: axⁿ → naxⁿ⁻¹
  2. 22x³ → 3 × 2x² = 6x²
  3. 3−4x² → 2 × (−4)x = −8x
  4. 46 → 0 (a number on its own differentiates to 0)
  5. 5dy/dx = 6x² − 8x

Answer6x² − 8x

Part (cii) · 2 marks

Evaluate the gradient of y = 2x³ − 4x² + 6 at x = 4.

  1. 1The gradient is the value of dy/dx
  2. 2From part (i): dy/dx = 6x² − 8x
  3. 3At x = 4: 6 × 4² − 8 × 4 = 96 − 32 = 64

Answer64

Part (ciii) · 4 marks

Find the coordinates of both stationary points on y = 2x³ − 4x² + 6.

  1. 1dy/dx = 6x² − 8x; stationary points are where dy/dx = 0
  2. 22x(3x − 4) = 0, so x = 0 or x = 4/3
  3. 3x = 0: y = 0 − 0 + 6 = 6
  4. 4x = 4/3: y = 128/27 − 192/27 + 162/27 = 98/27
  5. 5Stationary points: (0, 6) and (4/3, 98/27)

Answer(0, 6) and (4/3, 98/27)

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Also searched as: 0580/41/M/J/21 · 0580_s21_qp_41 · 0580 s21 41 · question 7.
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