Answer:
Option C
Explanation:
ϕ3(m)=3+2m−7m2+2m3
ϕ2(m)=−14m+7m2
ϕ′3(m)=2−14m+6m2
Now, putting ϕ3(m)=0 , we have
3+2m-7m2+2m3 =0
⇒ (1−m)(1+2m)(3−m)=0
⇒ m=−12,1,3
We know that , cϕ′n(m)+ϕn−1(m)=0 , which in the given case becomes
c(2-14m+6m2)+(-14m+7m2)=0
⇒c=14m−7m22−14m+6m2
So, when m=−12,c=−56
when m=1, c= −76
when m=3, c= −32
Asymptotes are y= −12 x −56,
y=x−76and y=3x−32