Question

Let $\overrightarrow{p}=3a{x}^2\hat{i}-2(x-1)\hat{j}$ and $\overrightarrow{q}=b(x-1)\hat{i}+x\hat{j}$. If $ab<0$ then $\overrightarrow{p}$ and $\overrightarrow{q}$ are parallel for

• A. atleast one $x$ is $(0,1)$
• B. atleast one $x$ is $(-1,0)$
• C. atleast one $x$ is $(1,2)$
• D. none of these

Answer 1

The correct option is A atleast one $x$ is $(0,1)$

$\overrightarrow{p},\overrightarrow{q}$ are parallel if $\overrightarrow{p}\times \overrightarrow{q}=0$ and $\overrightarrow{p}\times \overrightarrow{q}$
$=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 3a{x}^2& -2(x-1)& 0\\ b(x-1)& x& 0\end{array}\right|$
$=\hat{i}(0-0)-\hat{j}(0-0)+\hat{k}(3a{x}^3+2b{(x-1)}^2)$
$=\hat{k}(3a{x}^3+2b{(x-1)}^2)$
$|\overrightarrow{p}\times \overrightarrow{q}|=3a{x}^3+2{(x-1)}^{2}b=f\left(x\right)$
$f\left(0\right)=2b$ , $f\left(1\right)=3a$
$f\left(0\right).f\left(1\right)=2b\times 3a=6ab<0$ (given)