VITEEE 2015 Paper

$\phi_m$ (m) = 3 + 2m - $7m^2 + 2m^3$ $\Phi_2$ (m) = -14 m + $7n^2$ $\Phi_3'$ (m) = 2 - 14m + $6m^2$ Now, putting $\phi_m$ (m) = 0 , we have 3 + 2m - $7m^2 + 2m^3$ = 0 ⇒ (1 - m) (1 + 2m) (3 - m) = 0 ⇒ 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 + $6m^2$) + (- 1 4m + $7m^2$) = 0 ⇒ c = $\frac{14m-7n^2}{2-14m+6m^2}$ 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 - $\frac{3}{2}$