A curve is such that its gradient at the point (x, y) is given by (5x-2)³ᐟ². The curve passes through the point (2, 32/5). Find the coordinates of the stationary point on the curve.
- Find the coordinates of the stationary point on the curve.[6]
Mark scheme
Marking points from the official mark scheme (May/June 2025). The mark or mark type is shown in brackets.
- [B1] When dy/dx=0, x=2/5 (Must come from first derivative and not the second derivative)
- [M1] (5x-2)⁵ᐟ² term (from integral of (5x-2)³ᐟ²)
- [A1] Coefficient (2/5) for y = (2/5)(5x-2)⁵ᐟ² + c
- [M1] Substitute point (2, 32/5) into y expression: 32/5 = (2/5)(5(2)-2)⁵ᐟ² + c (Dep for attempt to find c using their k(5x-2)⁵ᐟ²)
- [A1] Calculate c: c=4 (Allow an unsimplified fraction)
- [B1] Stationary point (2/5, 4) (Dep on first B mark)
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