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 + *2*__ydy __= pxy + px² + __d____y__ . 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+4x²**2**)dx =36x-12x²+4/3X³+C**. *

*∫(36x *- 12x³+ 4/3x³ + c)dx = 18x² - 4x³ + 1/3X^{4} + cx + k

Let the required equation be y. Then, y = 18x² - 4x³ + 1/3X^{4} + 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 = 1/3X^{4} - 4x³ + *18x² *- 27x