Question

$g(x+y)=g\left(x\right)+g\left(y\right)+3xy(x+y)\forall x,yϵR$ and $g^{\prime }\left(0\right)=-4$.The value of $g^{\prime }\left(1\right)$ is

• A. $0$
• B. $1$
• C. $-1$
• D. none of these

Answer 1

The correct option is C $-1$

$g(x+y)=g\left(x\right)+g\left(y\right)+3x^{2}y+3x{y}^2$ (1)
or $g^{\prime }(x+y)=g^{\prime }\left(x\right)+6yx+3y^2$ (differentiating w.r.t. $x$ keeping $y$ as constant)
Put $x=0$. Then,
or $g^{\prime }\left(y\right)=g^{\prime }\left(0\right)+3y^2$
$=-4+3y^2$
or $g^{\prime }\left(y\right)=-4+3y^2$
or $g\left(x\right)=-4x+x^3+c$
Now, put $x=y=0$ in (1). Then $g\left(0\right)=g\left(0\right)+g\left(0\right)+0$
or $g\left(0\right)=0$
or $g\left(x\right)=x^3-4x$
$g\left(x\right)=0⟹x^3-4x=0⟹x=0,2,-2$.
Hence, three roots,
$\sqrt{g\left(x\right)}=\sqrt{x^3-4x}$ is defined if $x^3-4x\ge 0$ or $xϵ[-2,0]\cup [2,\infty ]$.
Also, $g^{\prime }\left(x\right)=3x^2-4⟹g^{\prime }\left(1\right)=-1$.