AQA A-Level Mathematics Paper 2, June 2025: Question 10

13 marks · Medium difficulty · Multi-step Problem

Differentiate a curve given by $y = x^k \ln x$, find the $y$-coordinate of its stationary point in terms of $k$, determine $k$ from a given point, and prove the curve has no point of inflection.

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Question

Question 10 states that curve C has equation y = x^k ln x for x > 0, where k is a positive integer. Part (a) asks to show that dy/dx = x^(k-1)[A + k ln x], where A is a constant to be found (4 marks). Part (b) asks to show that the y-coordinate of the stationary point of C can be written as -1/(ke), fully justifying the answer (5 marks). Part (c) states that the stationary point has coordinates (1/e, -1/e) and asks for the value of k (1 mark). Part (d) asks to prove that C does not have a point of inflection (3 marks).
Question text

10 A curve C has equation

y = xk ln x for x > 0

where k is a positive integer.

10 (a) Show that

dy k–1

= x [A + k ln x]

dx

where A is a constant to be found.

[4 marks]

y – 1

10 (b) Hence show that the ‑coordinate of the stationary point of C can be written as

ke

Fully justify your answer.

[5 marks]

… 17

(16)

( 1 – 1 ) k

10 (c) Given that the stationary point of C has coordinates , state the value of

e e

[1 mark]

10 (d) Prove that C does not have a point of inflection.

[3 marks]

Mark scheme

Show the mark scheme Mark scheme for Question 10. Part (a): uses the product rule (M1), obtains k x^(k-1) ln x + x^k (1/x) (A1), simplifies to x^(k-1)[1 + k ln x] with complete reasoned argument (A1, R1). Part (b): equates dy/dx to 0 (M1), justifies why x cannot be 0 since x > 0 (E1), solves to find x = e^(-1/k) (A1), substitutes into y (M1), simplifies correctly to -1/(ke) (R1). Part (c): k = 1 (B1). Part (d): differentiates for k = 1 to find d^2y/dx^2 = 1/x (M1, A1), argues that 1/x is never 0 for x > 0, concluding no point of inflection exists (R1). Total 13 marks.

Q Marking instructions AO Marks Typical solution

10(a) Uses the product rule with one 3.1a M1 dy k−1 k 1

term correct = kx ln x + x

dx x

Condone vu’ – uv’ or v’u – u’v

dy k−1 k−1

k−1 k 1 1.1b A1 = kx ln x + x

Obtains kx ln x + x dx

x

k 1 k−1 1.1b A1 dy k−1

Simplifies x = x = x 1+ k ln x

x dx

Completes reasoned argument 2.1 R1

dy k−1

to show = x 1+ k ln x

dx

Argument must include

1 xk

xk or k−1

before x seen

x x

Subtotal 4

10(b) Equates the given expression 1.1a M1 xk−1 1+ k ln x = 0

dy 1+ k ln x = 0

for to zero. PI k−1

dx 1+ k ln x = 0 or x = 0

and 1

Solves to obtain a non-zero ln x = −

k

value for x 1

Gives a correct reason for 2.4 E1 −

x = e k

rejecting x = 0

Since x 0, xk−1 0

1 2.2a A1

−

Deduces x = e k k

− −

A y = e k ln e k

−

Accept x = e k

Substitutes their value for x into 1.1a M1

equation of C. y = −

Completes reasoned argument 2.1 R1 e k

to show the given result. 1

Must see = −

k k e

− 1

e k simplified to e−1 or

e

and

− 1

ln e k simplified to − before

k

the final answer – A-LEVEL MATHEMATICS – 7357/2 –

AG

Subtotal 5

10(c) States 1 1.1b B1 1

Subtotal 1

14 Q Marking instructions AO Marks Typical solution

10(d) Differentiates 3.1a M1 When k = 1

dy k −1 dy

= x 1+ k ln x =1+ ln x

dx dx

where their k > 0 d2 y

d2 y If there is a point of inflection = 0

dx2

accept 2 in terms of A and/or

dx d2 y 1

k 2 = 0

2 1.1a A1 dx x

d y 1

Obtains 2 =

dx x Therefore, C has no point of inflection.

Completes reasoned argument 2.1 R1

by comparing with zero

x

and

concludes that there is no point

of inflection.

Subtotal 3

Question 10 Total 13

How to answer it

Differentiation & Stationary Points of y = xk ln x

What this question tests

  • Product Rule Differentiation: Correctly differentiating products involving power functions and natural logarithms ( ln x ).
  • Index & Logarithm Laws: Manipulating expressions such as xk × (1/x) = xk−1 , solving ln x = −1/k , and evaluating powers of e .
  • Rigorous Mathematical Justification: Proving statements by explicitly justifying why alternate algebraic roots (e.g., x = 0 ) must be rejected based on the domain.
  • Nature of Curves & Inflexion: Finding the second derivative ( d²y/dx² ) and proving non-existence of points of inflection.
Part (a) · 4 Marks

Show that dy/dx = xk−1 [A + k ln x]

Application of the product rule and algebraic factorisation

📐 Step-by-Step Solution

  1. Identify parts for the product rule:
    Let u = xk ⇒ du/dx = kxk−1
    Let v = ln x ⇒ dv/dx = 1/x
  2. Apply formula ( u'v + uv' ):
    dy/dx = kxk−1 ln x + xk(1/x)
  3. Simplify powers of x:
    xk × (1/x) = xk × x−1 = xk−1
    dy/dx = kxk−1 ln x + xk−1
  4. Factorise out xk−1:
    dy/dx = xk−1(1 + k ln x)
    Hence, A = 1 .

💡 Key Knowledge

  • Derivative of ln x: Always remember that d/dx(ln x) = 1/x .
  • Laws of Indices: xk / x = xk−1 . Showing this step explicitly is essential in a "show that" question.
  • Structure: Match the requested format xk−1 [A + k ln x] directly to read off the value of A .

🧠 Exam Technique & Proof Rigour

This is a "Show that" question. You cannot simply jump from the product rule line directly to the final factorised answer. The mark scheme specifically demands that you explicitly write xk × (1/x) or xk / x before writing xk−1 to secure the reasoning mark (R1).

❌ Common Errors

  • Omitting the intermediate step showing xk(1/x) = xk−1 , which forfeits the final reasoning mark.
  • Incorrectly applying the quotient rule instead of treating xk ln x as a simple product.
  • Forgetting to state the constant: A = 1 .
Mark Scheme Breakdown:
• M1 (3.1a): Uses the product rule with at least one term correct.
• A1 (1.1b): Correct unsimplified derivative: kxk−1 ln x + xk(1/x) .
• A1 (1.1b): Simplifies xk(1/x) = xk−1 .
• R1 (2.1): Completes a fully reasoned argument to reach xk−1[1 + k ln x] .
Part (b) · 5 Marks

Show that the y-coordinate of the stationary point is −1/(ke)

Setting dy/dx = 0, solving for x, justifying rejected roots, and substituting

📐 Step-by-Step Solution

  1. Set derivative equal to 0:
    xk−1(1 + k ln x) = 0
  2. Consider both factors and justify:
    Either xk−1 = 0 or 1 + k ln x = 0 .
    Given the domain is x > 0 , x ≠ 0 ⇒ xk−1 ≠ 0 .
  3. Solve for x:
    1 + k ln x = 0 ⇒ ln x = −1/k
    x = e−1/k
  4. Substitute x back into curve C:
    y = (e−1/k)k × ln(e−1/k)
  5. Simplify using index and log laws:
    (e−1/k)k = e(−1/k) × k = e−1 = 1/e
    ln(e−1/k) = −1/k
    y = (1/e) × (−1/k) = −1 / (ke) [As Required]

✅ What Full Marks Look Like

A full-mark solution explicitly:

  • States that x > 0 , hence xk−1 ≠ 0 (or x ≠ 0 ).
  • Shows x = e−1/k clearly.
  • Shows intermediate simplifications e−1 (or 1/e ) and −1/k before writing the final combined fraction.

🧠 Examiner Insight: "Fully Justify"

Whenever an exam question includes the instruction "Fully justify your answer", you are guaranteed to lose marks if you ignore possible alternate roots. Setting xk−1(1 + k ln x) = 0 produces two potential equations. Failing to mention why xk−1 = 0 gives no valid solution loses the explanation mark E1 immediately.

❌ Common Errors

  • Missing domain justification: Writing 1 + k ln x = 0 without acknowledging and rejecting xk−1 = 0 .
  • Logarithm inversion errors: Writing x = −1/(k ln) or other invalid algebraic manipulations instead of taking exponentials.
  • Skipping power laws: Jumping directly from (e−1/k)k ln(e−1/k) to −1/(ke) without showing the two separated simplified terms.
Mark Scheme Breakdown:
• M1 (1.1a): Sets dy/dx = 0 and solves to find a non-zero value for x .
• E1 (2.4): Gives correct reason for rejecting x = 0 (i.e. x > 0 so xk−1 ≠ 0 ).
• A1 (2.2a): Correctly deduces x = e−1/k .
• M1 (1.1a): Substitutes their x value back into y = xk ln x .
• R1 (2.1): Fully reasoned argument: must see (e−1/k)k simplified to e−1 (or 1/e ) AND ln(e−1/k) simplified to −1/k before concluding −1/(ke) .
Part (c) · 1 Mark

Determine the value of k

Matching given coordinates with the general stationary point

✅ Correct Answer

k = 1

Since the stationary point is given as (1/e, −1/e) :

  • Comparing y-coordinates: −1/(ke) = −1/e ⇒ k = 1
  • Comparing x-coordinates: e−1/k = e−1 ⇒ −1/k = −1 ⇒ k = 1

🧠 Exam Technique: "State"

The command word here is State. No working is required to gain the single mark; simply writing k = 1 earns full credit. However, double-checking both coordinates ensures you haven't made an arithmetic slip!

Mark Scheme Breakdown:
• B1 (1.1b): States 1 .
Part (d) · 3 Marks

Prove that C does not have a point of inflection

Finding the second derivative d²y/dx² and establishing that d²y/dx² ≠ 0

📐 Step-by-Step Proof

  1. Use the value of k = 1:
    From part (c), k = 1 .
    The first derivative simplifies to:
    dy/dx = x1−1[1 + (1)ln x] = x0(1 + ln x) = 1 + ln x
  2. Find the second derivative:
    d²y/dx² = d/dx(1 + ln x) = 1/x
  3. Analyse points of inflection condition:
    A point of inflection requires d²y/dx² = 0 .
  4. Form reasoned conclusion:
    Since x > 0 , 1/x > 0 for all x in the domain.
    Therefore, 1/x ≠ 0 for any x .
    Hence, curve C has no point of inflection.

💡 Alternative General Method (Any k > 0)

Even if differentiated generally for any k :

dy/dx = xk−1 + k xk−1 ln x

For k = 1 , it simplifies dramatically to d²y/dx² = 1/x . Always substitute earlier found values (like k = 1 ) to make subsequent calculus as simple as possible!

🧠 Completing the Reasoning

To secure the final R1 mark, examiners look for two distinct things:

  1. Direct reference to d²y/dx² = 0 as the requirement for an inflection point.
  2. An explicit statement comparing 1/x to zero (e.g., stating 1/x ≠ 0 or 1/x > 0 because x > 0 ), followed by a clear conclusion.

❌ Common Errors

  • Forgetting to substitute k = 1 from part (c) and getting tangled in unnecessary product rule algebra for general k .
  • Finding d²y/dx² = 1/x but stopping without stating that 1/x ≠ 0 . A proof requires a written conclusion!
  • Confusing points of inflection ( d²y/dx² = 0 ) with stationary points ( dy/dx = 0 ).
Mark Scheme Breakdown:
• M1 (3.1a): Differentiates dy/dx with their k > 0 .
• A1 (1.1a): Obtains d²y/dx² = 1/x .
• R1 (2.1): Completes reasoned argument by comparing 1/x with zero and explicitly concluding that there is no point of inflection.

Topics

Pure Mathematics · G: Differentiation · F: Exponentials and logarithms · A: Proof

Question and mark scheme from the AQA A-Level Mathematics examination, Paper 2, June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.