Q.1. Find the largest integral x which satisfies the following inequalities
$\frac{1}{x+1}-\frac{2}{x^2-x+1}<\frac{1-2x}{x^3+1}.$

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Please find below the solution to the asked query:

$$\begin{gathered} \frac{1}{\mathrm{x}+1}-\frac{2}{{\mathrm{x}}^2-\mathrm{x}+1}<\frac{1-2\mathrm{x}}{{\mathrm{x}}^3+1} \\ \Rightarrow \frac{{\mathrm{x}}^2-\mathrm{x}+1-2\mathrm{x}-2}{\left(\mathrm{x}+1\right)\left({\mathrm{x}}^2-\mathrm{x}+1\right)}<\frac{1-2\mathrm{x}}{\left(\mathrm{x}+1\right)\left({\mathrm{x}}^2-\mathrm{x}+1\right)} \\ \Rightarrow \frac{{\mathrm{x}}^2-3\mathrm{x}-1}{\left(\mathrm{x}+1\right)\left({\mathrm{x}}^2-\mathrm{x}+1\right)}<\frac{1-2\mathrm{x}}{\left(\mathrm{x}+1\right)\left({\mathrm{x}}^2-\mathrm{x}+1\right)} \\ \mathrm{Now} \mathrm{for} {\mathrm{x}}^2-\mathrm{x}+1, \mathrm{coefficient} \mathrm{of} {\mathrm{x}}^2>0 \mathrm{and} \mathrm{Discriminant}<0, \mathrm{hence} \\ \mathrm{it} \mathrm{will} \mathrm{be} \mathrm{greater} \mathrm{than} 0 \mathrm{for} \mathrm{x}\in \mathrm{R}, \mathrm{hence} \mathrm{it} \mathrm{can} \mathrm{be} \mathrm{cancelled} \\ \Rightarrow \frac{{\mathrm{x}}^2-3\mathrm{x}-1}{\left(\mathrm{x}+1\right)}<\frac{1-2\mathrm{x}}{\left(\mathrm{x}+1\right)} \\ \Rightarrow \frac{{\mathrm{x}}^2-3\mathrm{x}-1}{\left(\mathrm{x}+1\right)}-\frac{1-2\mathrm{x}}{\left(\mathrm{x}+1\right)}<0 \\ \Rightarrow \frac{{\mathrm{x}}^2-3\mathrm{x}-1-1+2\mathrm{x}}{\left(\mathrm{x}+1\right)}<0 \\ \Rightarrow \frac{{\mathrm{x}}^2-\mathrm{x}-2}{\left(\mathrm{x}+1\right)}<0 \\ \Rightarrow \frac{{\mathrm{x}}^2-2\mathrm{x}+\mathrm{x}-2}{\left(\mathrm{x}+1\right)}<0 \\ \Rightarrow \frac{\mathrm{x}\left(\mathrm{x}-2\right)+1\left(\mathrm{x}-2\right)}{\left(\mathrm{x}+1\right)}<0 \\ \Rightarrow \frac{\left(\mathrm{x}-2\right)\left(\mathrm{x}+1\right)}{\left(\mathrm{x}+1\right)}<0 \\ \mathrm{Now} \mathrm{we} \mathrm{cancel} \mathrm{x}+1, \mathrm{keeping} \mathrm{in} \mathrm{mind} \mathrm{that} \mathrm{x}+1\ne 0 \mathrm{i}.\mathrm{e}. \mathrm{x}\ne -1 \\ \Rightarrow \mathrm{x}-2<0 \mathrm{and} \mathrm{x}\ne -1 \\ \Rightarrow \mathrm{x}<2 \mathrm{and} \mathrm{x}\ne -1 \\ \mathbf{Hence}\mathbf{ }\mathbf{x}\mathbf{\in }\left(\mathbf{-}\mathbf{\infty }\mathbf{,}\mathbf{2}\right)\mathbf{-}\left\{\mathbf{-}\mathbf{1}\right\} \end{gathered}$$

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