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

Let $x_0$ be the point of local minima of $\mathrm{f}(x)=\overline{\mathrm{a}} \cdot(\overline{\mathrm{b}} \times \overline{\mathrm{c}})$ where $\overline{\mathrm{a}}=x \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}$, $\overline{\mathrm{b}}=-2 \hat{\mathrm{i}}+x \hat{\mathrm{j}}-\hat{\mathrm{k}}, \quad \overline{\mathrm{c}}=7 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+x \hat{\mathrm{k}}$, then value of $\overline{\mathrm{a}} \cdot \overline{\mathrm{b}}$ at $x=x_0$ is
  1. 15
  2. -15
  3. 12
  4. -12

Solution

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

Asked in: MHT CET 2023 (14 May Shift 2)

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