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:
Answer & explanation
Correct answer: option 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)