Edexcel IALMathematicsGeneral Applications10 Marks

Figure 1 shows a sketch of the curve with equation y = 2x ln x, x > 0.

(a)Find dy/dx.[2]
(b)Show that the x-coordinate of the minimum point of the curve is 1/√e.[3]
(c)The finite region R, shown shaded in Figure 1, is bounded by the curve, the x-axis and the line x = 2. Use integration by parts to find the exact area of R.[5]

This question uses a diagram from the original paper, which isn't shown here. Open the full paper from the past papers archive to see it.

Mark scheme

Marking points from the official mark scheme. The mark or mark type is shown in brackets.

**Part (a)**
M1: Uses the product rule correctly, u = 2x, v = ln x or u = x, v = 2ln x.
A1: dy/dx = 2ln x + 2x(1/x) = 2ln x + 2.
**Part (b)**
M1: Sets dy/dx = 0.
A1: Solves 2ln x + 2 = 0 to get ln x = -1.
A1: Correctly deduces x = e^(-1) or 1/e, which is 1/√e (showing that square root is equivalent to power of -1/2). This must be clearly stated as a demonstration.
**Part (c)**
M1: Attempts integration by parts using ∫ u dv/dx dx = uv - ∫ v du/dx dx, with u = ln x and dv/dx = 2x.
A1: Correct first application: x² ln x - ∫ x² (1/x) dx or x² ln x - ∫ x dx.
A1: Correct integration: x² ln x - x² (+ C).
M1: Applies limits of 1 (or 1/√e for point where y=0) and 2 to their integrated expression.
A1: Correct exact area = 4 ln 2 - 3. (Allow 4ln2 - 4 - (1ln1-1) for limit 1 to 2, or (4ln2-4) - (1/e ln(1/e) - 1/e) for 1/e to 2).

On Oracle Prep you can write or photograph your own answer and have it marked by AI against this mark scheme, with feedback on what you missed.