A curve has equation y = (2x-5)/(x-1) - 12x.

Mark scheme

Marking points from the official mark scheme (May/June 2016). The mark or mark type is shown in brackets.

1. [M1A1] dy/dx = (2(x-1) - (2x-5)) / (x-1)² - 12 Allow slips in du/dx and dv/dx but must be explicit. Allow (x-1)² = x² - 2x + 1
2. [B1] -12 isw
3. [B1] -24x + 14
4. [M1] dy/dx = (x-1)(-24x+14)-(-12x²+14x-5) / (x-1)²
5. [A1FT] FT on their derivative of 3 term quadratic
6. [M1] d²y/dx² = k(x-1)⁻³ No additional terms
7. [A1] k = -6 isw
8. [M1] their 3/(x-1)² - 12 = 0 and find a value for x
9. [A1] x = 0.5 and x = 1.5 12x²-24x+9=0 oe (2x-3)(2x-1)=0 oe
10. [A1] y = 2 and y = -22 if A0 A0 then A1 for a correct (x, y) pair
11. [B1] -6/(-0.5)³ > 0 therefore min when x = 0.5 oe or [-6/(0.5)³ =] 48 therefore min when x = 0.5 oe
12. [B1] -6/(0.5)³ < 0 therefore max when x = 1.5 oe or [-6/(1.5)³ =] -48 therefore max when x = 1.5 oe M1A1 is possible from other methods