Find the exact coordinates of the point on the curve y = x / (1 + ln x) at which the gradient of the tangent is equal to ¼.

Mark scheme

Marking points from the official mark scheme (May/June 2019). The mark or mark type is shown in brackets.

1. [M1] Use correct quotient rule (Allow use of correct product rule on x × (1 + ln x)⁻¹)
2. [A1] Obtain correct derivative in any form (dy/dx = ((1+lnx)-x(1/x))/(1+lnx)² or ((1+lnx)-1)/(1+lnx)²)
3. [M1] Equate derivative to ¼ and obtain a quadratic in ln x or (1 + ln x) (Horizontal form. Accept ln x = -(1+lnx)²)
4. [A1] Reduce to (ln x)² - 2 ln x + 1 = 0 (or 3-term equivalent. Condone (ln x)² if later used correctly)
5. [M1] Solve a 3-term quadratic in ln x for x (Must see working if solving incorrect quadratic)
6. [A1] Obtain answer x = e (Accept e¹)
7. [A1] Obtain answer y = ½e (Exact only with no decimals seen before the exact value. Accept e/2 but not e/(1+ln e))