Let $\vec p=3ax^2\hat i-2(x-1)\hat j,\,\vec q = b(x-1)\hat i+x\hat j$ and ab <0. Then $\vec p$ and $\vec q$ are parallel for:

at least one x in (0, 1)

$\vec p=k\vec q⇒[3ax^2\hat i-2(x-1)\hat j]=k[b(x-1)\hat i+x\hat j]$ ⇒ 3ax² = kb(x - 1) and -2(x - 1) = kx 

3ax³ + 2b(x – 1)² = 0

Let f(x) = 3ax³ + 2bx² – 4bx + 2b = 0

f(0) = 2b ; f(1) = 3a + 2b – 4b + 2b = 3a

As ab < 0 ; at least one root exists in (0, 1)

f(–1) = –3a + 2b + 4b + 2b ( No conclusion);  f(2) = 24a + 2b (No conclusion)