Question

Let $p=3a{x}^{2}i-2(x-1)j,q=b(x-1)i+xj$ and $ab<0$. Then $p$ and $q$ are parallel for atleast one $x$ in

• A. $(0,1)$
• B. $(-1,0)$
• C. $(1,2)$
• D. $Noneofthese$

Answer 1

The correct option is A $(0,1)$

Solution:-

$p=3a{x}^2\hat{i}-2(x-1)\hat{j}$

$q=b(x-1)+xj$

As we know that two vectors are parallel in their cross product is zero-

$\therefore p\times q=0$

$\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 3a{x}^2& -2(x-1)& 0\\ b(x-1)& x& 0\end{array}\right|=0$

$\Rightarrow k(3a{x}^3-(-2b){(x-1)}^2)=0$

$\Rightarrow 3a{x}^3+2b(x^2+1-2x)=0$

$f\left(x\right)=3a{x}^3+2b{x}^2-4bx+2b$

Now $f\left(0\right)=2b;f\left(1\right)=3aandf\left(0\right)f\left(1\right)<0$

This implies that f(x) has atleast one root in (0,1)