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WMA12/01 Oct 2024, Question 4: Differentiating Polynomials

Pearson Edexcel International AS Level Mathematics · Pure Mathematics 2 (WMA12) · Calculator · 7 marks

The question, in short

For the curve y = 4x^(1/2) + 9x^(−1/2) + 3 with x > 0, carry out calculus steps to obtain the gradient, locate the stationary point, classify it, and state where the curve falls.

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

Part (a) · 2 marks

Differentiate y with respect to x, writing every term in its simplest form.

  1. 1Bring down each power and reduce it by 1
  2. 24 × ½ x^(−1/2) = 2x^(−1/2)
  3. 39 × (−½) x^(−3/2) = −(9/2)x^(−3/2); the constant 3 gives 0
  4. 4dy/dx = 2x^(−1/2) − (9/2)x^(−3/2)

Answerdy/dx = 2x^(−1/2) − (9/2)x^(−3/2)

Part (b) · 2 marks

Using your derivative, work out the x coordinate where the curve is stationary.

  1. 12x^(−1/2) − (9/2)x^(−3/2) = 0
  2. 2Multiply by 2x^(3/2): 4x − 9 = 0
  3. 3x = 9/4

Answerx = 9/4

Part (ci) · 1 mark

Differentiate your gradient expression again, simplifying each term.

  1. 1Differentiate 2x^(−1/2): 2 × (−½) x^(−3/2) = −x^(−3/2)
  2. 2Differentiate −(9/2)x^(−3/2): −(9/2)(−3/2) x^(−5/2) = (27/4)x^(−5/2)
  3. 3d²y/dx² = −x^(−3/2) + (27/4)x^(−5/2)

Answerd²y/dx² = −x^(−3/2) + (27/4)x^(−5/2)

Part (cii) · 1 mark

Use the second derivative to decide what kind of stationary point C has, justifying your conclusion.

  1. 1At x = 9/4: −(9/4)^(−3/2) + (27/4)(9/4)^(−5/2)
  2. 2= −8/27 + (27/4)(32/243) = −8/27 + 8/9
  3. 3= 16/27, which is positive
  4. 4So the stationary point is a minimum

AnswerMinimum, since d²y/dx² = 16/27 > 0 at x = 9/4

Part (d) · 1 mark

Give the set of x values for which y decreases.

  1. 1y decreases where dy/dx < 0
  2. 2dy/dx < 0 when 4x − 9 < 0 with x > 0
  3. 30 < x < 9/4

Answer0 < x < 9/4

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