O-LevelAdditional MathematicsDifferentiation and integration4037/11 · May/June 2011 · Q65 Marks
The curves y = x² and 3y = −2x² + 20x – 20 meet at the point A. [Figure 6.1]
- (i)Show that the x-coordinate of A is 2.[1]
- (ii)Show that the gradients of the two curves are equal at A.[3]
- (iii)Find the equation of the tangent to the curves at A.[1]
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Mark scheme
Marking points from the official mark scheme (May/June 2011). The mark or mark type is shown in brackets.
- [1] B1 for substituting y=x² into 3y = -2x² + 20x – 20 to get 3x² = -2x² + 20x – 20, then simplifying to 5x² - 20x + 20 = 0, which gives x² - 4x + 4 = 0, or (x-2)² = 0, hence x=2 (AG).
- [1] B1 for finding dy/dx for y=x² as 2x, and at x=2, gradient = 4.
- [1] M1 for attempting to differentiate the second curve 3y = -2x² + 20x – 20 (or y = -2/3 x² + 20/3 x - 20/3) to get dy/dx = -4/3 x + 20/3.
- [1] A1 for finding the gradient of the second curve at x=2 as 4. Hence gradients are equal (AG).
- [1] B1 for y-4 = 4(x-2) or y = 4x-4.
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