Question

Let $x_0$ be the point of Local maxima of $f\left(x\right)=$$\overrightarrow{a}\cdot (\overrightarrow{b}\times \overrightarrow{c}),$ where $\overrightarrow{a}=x\hat{i}-2\hat{j}+3\hat{k},\overrightarrow{b}=-2\hat{i}+x\hat{j}-\hat{k}$ and $\overrightarrow{c}=7\hat{i}-2\hat{j}+x\hat{k}$. Then the value of $\overrightarrow{a}\cdot \overrightarrow{b}+\overrightarrow{b}\cdot \overrightarrow{c}+\overrightarrow{c}\cdot \overrightarrow{a}$ at $x=x_0$ is

• A. $-22$
• B. $-4$
• C. $-30$
• D. $14$

Answer 1

The correct option is A $-22$

$f\left(x\right)=\overrightarrow{a}.\overrightarrow{b}\times \overrightarrow{c}$
$\Rightarrow f\left(x\right)=\left|\begin{array}{ccc}x& -2& 3\\ -2& x& -1\\ 7& -2& x\end{array}\right|$
$\Rightarrow f\left(x\right)=x(x^2-2)+2(-2x+7)+3(4-7x)$
$\Rightarrow f\left(x\right)=x^3-2x-4x+14+12-21x$
$\Rightarrow f\left(x\right)=x^3-27x+26$
$f^{\prime }\left(x\right)=3x^2-27=0\Rightarrow x=\pm 3$
$f^{\prime \prime }\left(x\right)=6x$
At $x=3\Rightarrow f^{\prime \prime }\left(3\right)=18>0$
At $x=-3\Rightarrow f^{\prime \prime }(-3)=-18<0$
$\therefore$ Local maxima at $x=x_0=-3$
Thus,
$\overrightarrow{a}=-3\hat{i}-2\hat{j}+3\hat{k}$
$\overrightarrow{b}=-2\hat{i}-3\hat{j}-\hat{k}$
$\overrightarrow{c}=7\hat{i}-2\hat{j}-3\hat{k}$
Now, $\overrightarrow{a}\cdot \overrightarrow{b}+\overrightarrow{b}\cdot \overrightarrow{c}+\overrightarrow{c}\cdot \overrightarrow{a}$
$=(6+6-3)+(-14+6+3)+(-21+4-9)$
$=9-5-26$
$=-22$