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,
                                                                                                      dx²
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=0⇒3c=-81. ⇒ c=-27.
Hence, Y = 1/3X⁴ - 4x³ + 18x² - 27x