A function f with domain x > 0 is such that f'(x) = 8(2x-3)^(-1/2) - 10x^3. It is given that the curve with equation y = f(x) passes through the point (1, 0).

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2024). The mark or mark type is shown in brackets.

1. [B1] -18 SOI
2. [M1] Use of m1m2=-1from f'(x) with x=1.
3. [A1] y [-0] / x-1 = 1/18 [y = 1/18 (x-1)] OE ISW
4. [B1B1] B1 for each unsimplified {}. Can be implied by equivalent simplified or partly simplified versions. [f(x)=] 8(2x-3)^(1/2) * 1/(3/2) * 1/2 - 10x^4/4 [+c]
5. [M1] Use of x=1 and y=0 in their integrated f'(x), defined as an expression with at least one correct power, which must contain + c.
6. [A1] [f(x) or y =] 3(2x-3)³ - 6x³ + 3
7. [M1*] b² – 4ac = 128² −4×125×192 and stating “< 0” OR use of the quadratic formula and stating “No solutions” OR completing the square for the given quadratic and stating positive or > 0. OR sketch of the given quadratic and stating positive.
8. [DM1] No turning points [in the original function.]
9. [A1] Decreasing because f'(any positivex value) < 0 WWW e.g. f' (1) = -18.