Consider the function defined implicitly by the equation `y^3-3y+x=0` on various intervals in the real line. If `x in (-oo,-2) uu (2,oo)`, the equation implicitly defines a unique real-valued defferentiable function `y=f(x)`. If `x in (-2,2)`, the equation implicitly defines a unique real-valud diferentiable function `y-g(x)` satisfying `g_(0)=0`.
If `f(-10sqrt2)=2sqrt(2)`, then `f"(-10sqrt(2))` is equal to

Options

A. `2g(-1)`
B. 0
C. `-2g(1)`
D. `2g(1)`

Answer

D

Explanation

Let I`= int_(-1)^(1)g'(x)dx=[g(x)]_(-1)^(1)=g(1)-g(-1)`
Since, ` y^(3)-3y+x=0 " "...(i)`
and ` y=g(x)`
` therefore {g(x)}^(3)-3g(x)+x=0 " "["from Eq."(i)]`
At `x=1, {g(1)}^(3)-3g(1)+1=0 " "(ii)`
At `x=-1, {g(-1)}^(3)-3g(-1)-1=0 " "...(iii)`
On adding Eqs.(i) and (ii), we get
`{g(1)}^(3)+{g(-1)}^(3)-3{g(1)+g(-1)}=0 `
` rArr [g(1)+g(-1)][{g(1)}^(2)+{g(-1)}^(2)-g(1)g(-1)-3]=0 `
` rArr g(1)+g(-1)=0 `
` rArr g(1)=-g(-1) `
`therefore I=g(1)-g(-1) `
`=g(1)-{-g(1)}=2g(1)`