DO NOT USE A CALCULATOR IN THIS QUESTION. [Figure 12.1] The diagram shows part of the curve y=1/(2x+1) and part of the line 5y = x-1. The curve meets the y-axis at point A. The line meets the x-axis at point B. The line and curve intersect at point C.

Mark scheme

Marking points from the official mark scheme (May/June 2021). The mark or mark type is shown in brackets.

1. (a)(i) B1 A(0,1) and B(1, 0)
2. (a)(ii) B2 B1 for [y =] 1/(2(2)+1) and [y =](2-1)/5 oe evaluates both expressions as 1/5
3. (a)(ii) Alternative 1: B2 B1 for 1/(2x+1) = (x-1)/5 or 5 = (2x+1)(x-1) oe to get x = 2 or 1/5 = 1/(2x+1) oe to get x = 2
4. (a)(ii) Alternative 2: B2 B1 for (2x + 1)(x – 1) = 5 or 2x²- x-6=0 solves or factorises to get (2x + 3)(x-2) and states x = 2 OR shows 2(2²) - 2-6=0 oe
5. (a)(ii) Alternative 3: B2 B1 for (2x + 1)(x – 1) = 5 oe and shows (2 × 2 + 1)(2 - 1) = 5
6. (b) B1 (1/2) × 1 × 0.2 oe or (2²/5) - (2/5) - (1²/5) - (1/5) oe
7. (b) B2 B1 for (1/2)ln(2x+1) [+c] oe or (1/2)ln(x+0.5) [+c] oe or kln(2x+1) or kln(x+0.5), k ≠ 0.5 or 0
8. (b) M1 FT their F(x) providing at least B1 for integration of curve awarded F(2) - F(0) - their 0.1
9. (b) A1 0.5ln5-0.1 or exact equivalent