If $x_0$ is the point of local minima of $\mathrm{f}(x)=\overline{\mathrm{a}} \cdot(\overline{\mathrm{b}} imes \overline{\mathrm{c}})$ where $\overline{\mathrm{a}}=x \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}$, $\overline{\mathrm{b}}=-2 \hat{\mathrm{i}}+x…

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If $x_0$ is the point of local minima of $\mathrm{f}(x)=\overline{\mathrm{a}} \cdot(\overline{\mathrm{b}} 	imes \overline{\mathrm{c}})$ where $\overline{\mathrm{a}}=x \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}$, $\overline{\mathrm{b}}=-2 \hat{\mathrm{i}}+x \hat{\mathrm{j}}-\hat{\mathrm{k}}$, $\overline{\mathrm{c}}=7 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+x \hat{\mathrm{k}}$, then value of $\overline{\mathrm{a}} \cdot \overline{\mathrm{b}}$ at $x=x_0$ is

Aicharya · Accepted Answer

$\begin{aligned} & \mathrm{f}(x)=\overline{\mathrm{a}} \cdot(\overline{\mathrm{b}} 	imes \overline{\mathrm{c}}) \ & =\left|\begin{array}{ccc}x & -2 & 3 \ -2 & x & -1 \ 7 & -2 & x\end{array}ight| \ & =x\left(x^2-2ight)+2(-2 x+7)+3(4-7 x) \ 	herefore \quad & \mathrm{f}(x)=x^3-27 x+26 \ & 	ext { Now, } \mathrm{f}^{\prime}(x)=0 \ \Rightarrow & 3 x^2-27=0 \ \Rightarrow & x^2=9 \ \Rightarrow & x= \pm 3 \ & \mathrm{f}^{\prime \prime}(x)=6 x \ & \Rightarrow \mathrm{f}^{\prime \prime}(x)=18\gt0 \ 	herefore \quad & \mathrm{f}(x) 	ext { has local minimum at } x=3 .\end{aligned}$ $\begin{aligned} 	herefore \quad \bar{a} & =3 \hat{i}-2 \hat{j}+3 \hat{k} \ \bar{b} & =-2 \hat{i}+3 \hat{j}-\hat{k} \ 	herefore \quad \bar{a} \cdot \bar{b} & =(3 \hat{i}-2 \hat{j}+3 \hat{k}) \cdot(-2 \hat{i}+3 \hat{j}-\hat{k}) \ & =3(-2)+(-2)(3)+3(-1) \ & =-15\end{aligned}$

If $x_0$ is the point of local minima of $\mathrm{f}(x)=\overline{\mathrm{a}} \cdot(\overline{\mathrm{b}}…

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