Question

Let $f\left(x\right)$ and $g\left(x\right)$ are polynomial of degree 4 such that
$g\left(\alpha \right)=g^{\prime }\left(\alpha \right)=g^{\prime \prime }\left(\alpha \right)=0$.
If $\underset{x\to \alpha }{lim}\frac{f\left(x\right)}{g\left(x\right)}=0,$ then number of different real solutions of equation $f\left(x\right)g^{\prime }\left(x\right)+g\left(x\right)f^{\prime }\left(x\right)=0$ is equal to

Answer 1

$g\left(x\right)=a(x-\alpha {)}^3(x-\beta )$
So $f\left(x\right)=b(x-\alpha {)}^4$
$h\left(x\right)=f\left(x\right)g^{\prime }\left(x\right)+f^{\prime }\left(x\right)g\left(x\right)$
$=\frac{d}{dx}\left(f\right(x).g(x\left)\right)$
$=\frac{d}{dx}\left(ab\right(x-\alpha {)}^7(x-\beta ))$
$=ab(x-\alpha {)}^6(x-\alpha +7x-7\beta )$
$=ab(x-\alpha {)}^6(8x-\alpha -7\beta )$
So $h\left(x\right)$ has two different real solutions.