Answer:
Option C
Explanation:
$\phi_{3}(m)=3+2m-7m^{2}+2m^{3}$
$\phi_{2}(m)=-14m+7m^{2}$
$\phi_{3}'(m)=2-14m+6m^{2}$
Now, putting $\phi_{3}(m)=0$ , we have
3+2m-7m2+2m3 =0
$\Rightarrow$ $(1-m)(1+2m)(3-m)=0$
$\Rightarrow$ $ m=-\frac{1}{2},1,3$
We know that , $c\phi_{n}'(m)+\phi_{n-1}(m)=0$ , which in the given case becomes
c(2-14m+6m2)+(-14m+7m2)=0
$\Rightarrow c=\frac{14m-7m^{2}}{2-14m+6m^{2}}$
So, when $m=-\frac{1}{2},c=-\frac{5}{6}$
when m=1, c= $-\frac{7}{6}$
when m=3, c= $-\frac{3}{2}$
Asymptotes are y= $-\frac{1}{2}$ x $-\frac{5}{6}$,
$y=x-\frac{7}{6}$and $ y=3x-\frac{3}{2}$
