The diagram shows the curve y = √(1 + 4x), which intersects the x-axis at A and the y-axis at B. The normal to the curve at B meets the x-axis at C. Find [Figure 11.1]

Mark scheme

Marking points from the official mark scheme (May/June 2013). The mark or mark type is shown in brackets.

1. [1] B1 For (i): Correct differentiation of (1 + 4x)^(1/2) to (1 + 4x)^(-1/2)
2. [1] B1 For (i): Correct differentiation of (1 + 4x)^(1/2) to include the factor of 4 (i.e. dy/dx = 2(1+4x)^(-1/2))
3. [1] M1 For (i): Uses the product of gradients m1m2 = -1 to find the gradient of the normal
4. [1] M1 For (i): Correct method for forming the equation of line BC (y - y1 = m(x - x1))
5. [1] A1 For (i): Correct equation of BC (e.g., y = -1/2 x + 1)
6. [1] B1 For (ii): Finds the x-intercept A (-1/4, 0)
7. [1] B1 For (ii): Correct integration of (1 + 4x)^(1/2) to get (1 + 4x)^(3/2)/(3/2)
8. [1] B1 For (ii): Correct integration of (1 + 4x)^(1/2) to include the factor of 1/4 (i.e. 1/6 (1+4x)^(3/2))
9. [1] B1 For (ii): Correctly applies limits of integration from -1/4 to 0
10. [1] B1 For (ii): Correctly calculates the area of triangle BOC and adds to the area under the curve to find the shaded area = 7/6 (For 1 + his '1/6')