Further Mathematics Paper 2, Nov/Dec. 2010
 Questions: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 Main
Weakness/Remedies
Strength
Question 10

(a) If x² + y² = py (1 + X²), where p is a constant, find dy .
dx

(b) A curve cuts the x-axis at the points where x = 0 and x = 3. If d²y = 4(3-x)²
find the equation of the curve.                                               dx
²

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Observation

This question was reportedly attempted by majority of the candidates and they performed well in it.
In part (a), candidates were able to differentiate X² + y² = py(1+X)² with respect to x as follows:
2x + 2ydy = pxy + px² + dy . Bringing like terms together we have
dx                           dx
2 dy
(2y - p - px² ) dy = 2pxy - 2x⇒ dy = 2pxy - 2x
dx                         dx
2y - P - px²

In part (b) majority of candidates were able to integrate d²y = 4(3-x)² twice with respect to x,
dx
²
substitute the values x = 0 and x=3 correctly to obtain the required equation as follows:
4(3 - X)² = 4(9 - 6x + X²) = 36 - 24x + 4X²
∫4(3-x)² = ∫(36-24x+4x²2)dx =36x-12x²+4/3X³+C.
∫(36x - 12x³+ 4/3x³ + c)dx = 18x² - 4x³ + 1/3X4 + cx + k
Let the required equation be y. Then, y = 18x² - 4x³ + 1/3X4 + cx + k
When x = 0, y= 18(0)² - 4(0)³ + c(O) + 1/3 (0)4 + k= O. k = O.
Whenx=3, 162-108+3c+27=03c=-81. c=-27.
Hence, Y = 1/3X4 - 4x³ + 18x² - 27x