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 number of real roots of the equation $g\left(x\right)=0$ is

• A. $2$
• B. $0$
• C. $1$
• D. $3$

Answer 1

The correct option is D $3$

Substituting $y=0$, we get
$g\left(x\right)=g\left(0\right)+g\left(x\right)$
$\Rightarrow g\left(0\right)=0$
Let $y=c$ , where $c$ is a constant. On substituting we get,
$g(x+c)=g\left(x\right)+g\left(c\right)+3xc(x+c)$
Differentiating with respect to $x$, we get,
$g^{\prime }(x+c)=g^{\prime }\left(x\right)+6cx+3c^2$
Now, if we consider $c\to 0$, and take limits on both sides and divide by $c$, we get
${lim}_{c\to 0}\frac{g^{\prime }(x+c)-g^{\prime }\left(x\right)}{c}=\underset{c\to 0}{lim}6x+3c$
$\Rightarrow g^{\prime \prime }\left(x\right)=6x$
$\Rightarrow g^{\prime }\left(x\right)=3x^2+k$ , where k is a constant.
$g^{\prime }\left(0\right)=-4$
$\Rightarrow k=-4$
$\Rightarrow g^{\prime }\left(x\right)=3x^2-4$
$\Rightarrow g\left(x\right)=x^3-4x+m$ , where m is a constant
$g\left(0\right)=0$
$\Rightarrow g\left(x\right)=x^3-4x=x(x^2-4)$
$g\left(x\right)=0$, will have three real roots