# Let X0 be the point of local maxima of

Question:

Let $x_{0}$ be the point of local maxima of $\mathrm{f}(\mathrm{x})=\overrightarrow{\mathrm{a}} \cdot(\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}), \quad$ where $\quad \overrightarrow{\mathrm{a}}=\mathrm{x} \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}$,

$\overrightarrow{\mathrm{b}}=-2 \hat{\mathrm{i}}+x \hat{\mathrm{j}}-\hat{\mathrm{k}}$ and $\overrightarrow{\mathrm{c}}=7 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+x \hat{\mathrm{k}}$. Then the value of $\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}$ at $x=x_{0}$ is :

1. $-30$

2. 14

3. $-4$

4. $-22$

Correct Option:

Solution:

$\mathrm{f}(\mathrm{x})=\overrightarrow{\mathrm{a}} \cdot(\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}})=\left|\begin{array}{ccc}\mathrm{x} & -2 & 3 \\ -2 & \mathrm{x} & -1 \\ 7 & -2 & \mathrm{x}\end{array}\right|=\mathrm{x}^{3}-27 \mathrm{x}+26$

$\mathrm{f}^{\prime}(\mathrm{x})=3 \mathrm{x}^{2}-27=0 \Rightarrow \mathrm{x}=\pm 3$

and $f^{\prime \prime}(-3)<0$

$\Rightarrow$ local maxima at $x=x_{0}=-3$

Thus, $\overrightarrow{\mathrm{a}}=-3 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}$,

$\overrightarrow{\mathrm{b}}=-2 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}-\hat{\mathrm{k}}$

and $\overrightarrow{\mathrm{c}}=7 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}-3 \hat{\mathrm{k}}$

$\Rightarrow \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{a}}=9-5-26=-22$