A curve is such that dy/dx = √(4x + 1) and (2, 5) is a point on the curve.

Mark scheme

Marking points from the official mark scheme (May/June 2018). The mark or mark type is shown in brackets.

1. [B1 B1] (i) To find the equation of the curve, integrate dy/dx = (4x + 1)^(1/2). ∫(4x + 1)^(1/2) dx = (4x + 1)^(3/2) / (3/2) × (1/4) + C = (4x + 1)^(3/2) / 6 + C. B1 for (4x + 1)^(3/2) / (3/2) (without ÷ 4). B1 for ÷ 4 oe. Unsimplified OK.
2. [M1] (i) Use the point (2, 5) to find C: 5 = (4(2) + 1)^(3/2) / 6 + C. 5 = (9)^(3/2) / 6 + C. 5 = 27 / 6 + C. 5 = 4.5 + C, so C = 0.5. Uses (2, 5) in an integral (indicated by an increase in power by 1).
3. [A1] (i) The equation of the curve is y = (4x + 1)^(3/2) / 6 + ½. No isw if candidate now goes on to produce a straight line equation.
4. [M1] (ii) We are given dy/dt = 0.06. We need to find dx/dt when x = 2. Use the chain rule: dy/dt = (dy/dx) × (dx/dt). When x = 2, dy/dx = √(4(2) + 1) = √9 = 3. So, 0.06 = 3 × (dx/dt). Ignore notation. Must be 0.06÷3 for M1.
5. [A1] (ii) dx/dt = 0.06 / 3 = 0.02 units per second. Correct answer with no working scores 2/2.
6. [B1] (iii) First, find d²y/dx². dy/dx = (4x + 1)^(1/2). d²y/dx² = ½ (4x + 1)^(-1/2) × 4 = 2(4x + 1)^(-1/2).
7. [B1FT] (iii) Now, find the product (d²y/dx²) × (dy/dx): (2(4x + 1)^(-1/2)) × ((4x + 1)^(1/2)) = 2 × (4x + 1)^(-1/2 + 1/2) = 2 × (4x + 1)⁰ = 2. Since 2 is a constant, the product is constant. Must either show the algebraic product and state that it results in a constant or evaluate it as '= 2'. Must not evaluate at x =2. ft to apply only if d²y/dx² is of the form k(4x+1)^(-½).