If log_1/3(|a| +1) > -1, then the domain of the function f(x)=​√(2x^4 + ax^3 -6x^2 -4ax -8) is a) [-1,2], b) [-2,2], c)(-∞,-2]∪[2,∞) 4) R-(-1,1)    

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$$\begin{gathered} \mathrm{We} \mathrm{have} \\ {\log }_{\frac{1}{3}}\left(\left|\mathrm{a}\right|+1\right)>-1 \\ \mathrm{Base} \mathrm{of} \log \mathrm{is} \mathrm{lying} \mathrm{between} 0 \mathrm{and} 1, \mathrm{hence} \mathrm{inequality} \mathrm{will} \mathrm{reverse} \\ \Rightarrow \left|\mathrm{a}\right|+1<{\left(\frac{1}{3}\right)}^{-1} \\ \Rightarrow \left|\mathrm{a}\right|+1<3 \\ \Rightarrow \left|\mathrm{a}\right|<2 \\ \mathrm{We} \mathrm{know} \mathrm{that} \left|\mathrm{X}\right|<\mathrm{A} \mathrm{gives} -\mathrm{A}<\mathrm{X}<\mathrm{A}, \mathrm{hence} \\ \left|\mathrm{a}\right|<2 \mathrm{gives} \\ -2<\mathrm{a}<2 \\ \mathrm{Now} \mathrm{we} \mathrm{have} \\ \mathrm{f}\left(\mathrm{x}\right)=\sqrt{2{\mathrm{x}}^4+{\mathrm{ax}}^3-6{\mathrm{x}}^2-4\mathrm{ax}-8} \\ \mathrm{Quantity} \mathrm{inside} \mathrm{root} \mathrm{must} \mathrm{be} \mathrm{positive} \\ 2{\mathrm{x}}^4+{\mathrm{ax}}^3-6{\mathrm{x}}^2-4\mathrm{ax}-8\ge 0 \\ \Rightarrow 2{\mathrm{x}}^4-6{\mathrm{x}}^2-8+{\mathrm{ax}}^3-4\mathrm{ax}\ge 0 \\ \Rightarrow 2{\mathrm{x}}^4-6{\mathrm{x}}^2-8+\mathrm{ax}\left({\mathrm{x}}^2-4\right)\ge 0 \\ \Rightarrow 2\left({\mathrm{x}}^4-3{\mathrm{x}}^2-4\right)+\mathrm{ax}\left({\mathrm{x}}^2-4\right)\ge 0 \\ \Rightarrow 2\left({\mathrm{x}}^4-4{\mathrm{x}}^2+{\mathrm{x}}^2-4\right)+\mathrm{ax}\left({\mathrm{x}}^2-4\right)\ge 0 \\ \Rightarrow 2\left\{{\mathrm{x}}^2\left({\mathrm{x}}^2-4\right)+1\left({\mathrm{x}}^2-4\right)\right\}+\mathrm{ax}\left({\mathrm{x}}^2-4\right)\ge 0 \\ \Rightarrow 2\left({\mathrm{x}}^2-4\right)\left({\mathrm{x}}^2+1\right)+\mathrm{ax}\left({\mathrm{x}}^2-4\right)\ge 0 \\ \Rightarrow \left({\mathrm{x}}^2-4\right)\left(2{\mathrm{x}}^2+2+\mathrm{ax}\right)\ge 0 \\ \Rightarrow \left({\mathrm{x}}^2-4\right)\left(2{\mathrm{x}}^2+\mathrm{ax}+2\right)\ge 0..\left(\mathrm{i}\right) \\ \mathrm{Now} \mathrm{Discriminat} \mathrm{of} 2{\mathrm{x}}^2+\mathrm{ax}+2 ={\mathrm{a}}^2-16 \\ \mathrm{Now} \mathrm{as} \mathrm{a}\in \left(-2,2\right), \mathrm{hence} \\ \mathrm{Discriminant}<0 \\ \mathrm{Also} \mathrm{coefficient} \mathrm{of} {\mathrm{x}}^2=2>0 \\ \mathrm{For} \mathrm{any} \mathrm{quadratic} \mathrm{equation}, \mathrm{if} \mathrm{coefficient} \mathrm{of} {\mathrm{x}}^2>0 \mathrm{and} \mathrm{Discriminant}<0, \\ \mathrm{then} \mathrm{quadratic} \mathrm{equation} \mathrm{is} \mathrm{positive} \mathrm{for} \mathrm{all} \mathrm{x}\in \mathrm{R} \\ \Rightarrow \left(2{\mathrm{x}}^2+\mathrm{ax}+2\right)>0 \mathrm{for} \mathrm{all} \mathrm{x}\in \mathrm{R}, \mathrm{hence} \left(\mathrm{i}\right) \mathrm{becomes} \\ \left({\mathrm{x}}^2-4\right)\ge 0 \\ \Rightarrow \left(\mathrm{x}-2\right)\left(\mathrm{x}+2\right)\ge 0 \\ \Rightarrow \mathrm{x}\in (-\infty ,2]\cup [2,\infty ) \left[\mathrm{Answer}\right] \end{gathered}$$

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