If $\mathbf{a}=x^2 \hat{\mathbf{i}}+x \hat{\mathbf{j}}+3 \hat{\mathbf{k}}$ and $\mathbf{b}=x…

1. $x \in R^{-}$
2. $x \in(-\infty, 0) \cup(1, \infty)$
3. $x \in(-\infty,-2) \cup(2, \infty)$
4. $x \in(-2,0) \cup(2, \infty)$

Solution

Given, $ \begin{aligned} & \mathbf{a}=x^2 \hat{\mathbf{i}}+x \hat{\mathbf{j}}+3 \hat{\mathbf{k}} \\ & \mathbf{b}=x \hat{\mathbf{i}}-4 \hat{\mathbf{j}}+2 \hat{\mathbf{k}} \end{aligned} $ $ \begin{aligned} & \mathbf{a} \cdot \mathbf{b}>6 \\ & \Rightarrow\left(x^2 \hat{\mathbf{i}}+x \hat{\mathbf{j}}+3 \hat{\mathbf{k}}\right)(x \hat{\mathbf{i}}-4 \hat{\mathbf{j}}+2 \hat{\mathbf{k}})>6 \\ & \Rightarrow x^2 \cdot x+(x)(-4)+3(2)>6 \end{aligned} $ $\left[\mathbf{a} \cdot \mathbf{b}=\mathbf{a}_1 \mathbf{a}_2+\mathbf{b}_1 \mathbf{b}_2+\mathbf{c}_1 \mathbf{c}_2\right.$, where $ \left.\mathbf{a}=a_1 \hat{\mathbf{i}}+b_1 \hat{\mathbf{j}}+c_1 \hat{\mathbf{k}} \text { and } \mathbf{b}=a_2 \hat{\mathbf{i}}+b_2 \hat{\mathbf{j}}+c_2 \hat{\mathbf{k}}\right] $ $ \begin{array}{rrr} \Rightarrow & x^3-4 x+6>6 \\ \Rightarrow & x^3-4 x+6-6>0 \\ \Rightarrow & x^3-4 x>0 \\ \Rightarrow & x\left(x^2-4\right)>0 \\ \Rightarrow & x(x-2)(x+2)>0 \end{array} $ Using wavy curve method critical points are $ x=0,2,-2 $
$\therefore$ Solution is $x \in(-2,0) \cup(2, \infty)$

Asked in: AP EAMCET 2021 (24 Aug Shift 1)

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