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g(x+y)=g(x)+g(y)+3xy(x+y)AA x, y in R" a...

`g(x+y)=g(x)+g(y)+3xy(x+y)AA x, y in R" and "g'(0)=-4.`
The value of g'(1) is

A

0

B

1

C

-1

D

none of these

Text Solution

Verified by Experts

`g(x+y)=g(x)+g(y)+3x^(2)y+3xy^(2)" (1)"`
`"or "g'(x+y)=g'(x)+6yx+3y^(2)" (differentiating w.r.t. x keeping y as constant)"`
Put `x=0.` Then,
`"or "g'(y)=g'(0)+3y^(2)=-4+3y^(2)`
`"or "g'(x)=-4+3x^(2)`
`"or "g(x)=-4x+x^(3)+c`
Now, put `x=y=0 in (1)." Then "g(0)=g(0)+g(0)+0`
`"or "g(0)=0`
`"or "g(x)=x^(3)-4x`
`g(x)=0rArrx^(3)-4x=0rArrx=0,2,-2.`
Hence. three roots.
`sqrt(g(x))=sqrt(x^(3)-4x)" is defined if "x^(3)-4xge0`
`"or "x in [-2, 0]cup[2,oo).`
`"Also, "g'(x)=3x^(2)-4rArrg'(1)=-1.`
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Knowledge Check

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