Find the exact coordinates of the stationary points of the curve y = e^(3x²-1) / (1-x²).

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2023). The mark or mark type is shown in brackets.

1. [M1] Use correct product or quotient rule. Need attempt at both derivatives condone errors in chain rule.
2. [A1] Obtain correct derivative in any form, e.g. dy/dx = ( (1-x²) * e^(3x²-1) * 6x - e^(3x²-1) * (-2x) ) / (1-x²)². In quotient rule allow BOD in formula if +/- 2x seen unless clear that incorrect formula has been used. If omit denominator or forget to square or complete reversal of signs then M0 A0 M1 A1 A1 A1. If 6x(1-x²)e^(3x²-1) +2xe^(3x²-1) = 0 from the start, with no wrong formula seen, award M1A1.
3. [M1] Equate derivative (or its numerator) to zero and solve for x. 6x - 6x³ + 2x = 0 and solve. Allow for just one x value. Allow if from solution of 3 term quadratic equation, but if they get x = 0 the x must factorise out.
4. [A1] Obtain the point (0, e⁻¹) or exact equivalent. Or for all three x coordinates found 0, ±2/√3 oe and no extras but if this is the case then one pair of correct coordinates A1 and both other pairs of correct coordinates A1. Accept, e.g. x = 0, y = e⁻¹. ISW for last 3 marks.
5. [A1] Obtain the point (2/√3, e³) or exact equivalent.
6. [A1] Obtain the point (-2/√3, e³) or exact equivalent.