mathematics AS help needed !!

[mania _ manal]

someone plx help me with octnov 2012 pp12 question 10 will be highly appreciated....

 

[syed1995]

someone plx help me with octnov 2012 pp12 question 10 will be highly appreciated....

i)
dy/dx = x+4x^-2
∫ x+(4x^-2) dx
y= (x^2)/2 - (4x^-1) + c
8=(4^2)/2 -4(4^-1) + c
c = 8 - (8 - 1)
c = +1

y = (x^2)/2 - 4/x + 1


ii)

This is a tricky question .. they have asked the minimum value of the *gradient* and not the minimum/maximum value of y..

so that means for this the initial equation will be the gradient and not the curve.. we will differentiate the gradient to find the stationary value of gradient..

dy/dx = x+4x^-2
d2y/dx2 = 1 - 8x^-3
d2y/dx2 = 0
1-8x^-3 = 0
8x^-3 = 1
x^3 = 8
x = 2 Answer.. Hence Proved.

when x =2 gradient = 2 + 4(2^-2) = 3

now to find whether it is minimum or maximum ..

d3y/dx3 = 24x^-4
= 24(2^-4)
= 1.5

since d3y/dx3 > 0 .. the point is minimum..

 

[mania _ manal]

ohh k thnks alooot i got
one more question if its okay wid ya octnov 2010 pp41 question 6 last part ??