Solve the following logarithm inequality -
(e^x+ 1) (sin x - 2) / x³+1 is greater than equal to 0

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Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{Given} \mathrm{inequality} \mathrm{is} \\ \frac{\left(e^x+1\right)\left(\sin x-2\right)}{x^3+1}\ge 0 \\ \Rightarrow \frac{\left(e^x+1\right)\left(\sin x-2\right)}{x^3+1^3}\ge 0 \\ \mathrm{Using} \mathrm{identity} a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right) \mathrm{we} \mathrm{get} \\ \frac{\left(e^x+1\right)\left(\sin x-2\right)}{\left(x+1\right)\left(x^2-x+1\right)}\ge 0 ; \mathrm{inequality}\left(\mathrm{i}\right) \\ \mathrm{Now} e^x\ge 0 \mathrm{for} \mathrm{all} x\in R \\ \Rightarrow \left(e^x+1\right) \mathrm{will} \mathrm{always} \mathrm{be} \mathrm{positive} \mathrm{and} \mathrm{hence} \mathrm{can} \mathrm{be} \mathrm{shifted} \mathrm{to} \mathrm{L}.\mathrm{H}.\mathrm{S}. \mathrm{of} \mathrm{inequality}. \mathrm{So} \mathrm{inequality}\left(\mathrm{i}\right) \mathrm{becomes}, \\ \frac{\left(\sin x-2\right)}{\left(x+1\right)\left(x^2-x+1\right)}\ge 0 \\ \mathrm{Now} \\ \mathrm{sinx} \mathrm{varies} \mathrm{from} -1 \mathrm{to} 1 \\ \Rightarrow \left(\sin x-2\right) \mathrm{will} \mathrm{always} \mathrm{be} \mathrm{negative} \mathrm{and} \mathrm{hence} \mathrm{can} \mathrm{be} \mathrm{shifted} \mathrm{to} \mathrm{L}.\mathrm{H}.\mathrm{S}. \mathrm{of} \mathrm{inequality} \mathrm{but} \mathrm{since} \mathrm{it} \mathrm{is} \mathrm{negative} \mathrm{hence} \mathrm{sign} \mathrm{of} \mathrm{inequality} \mathrm{will} \mathrm{reverse}.\mathrm{So} \mathrm{inequality}\left(\mathrm{i}\right) \mathrm{becomes}, \\ \frac{1}{\left(x+1\right)\left(x^2-x+1\right)}\le 0 \\ \mathbf{Important}\mathbf{ }\mathbf{ }\mathbf{Note}\mathbf{:}\mathbf{ }\mathbf{In}\mathbf{ }\mathbf{y}\mathbf{=}{\mathbf{ax}}^{\mathbf{2}}\mathbf{+}\mathbf{bx}\mathbf{+}\mathbf{c}\mathbf{ }\mathbf{if}\mathbf{ }\mathbf{a}\mathbf{>}\mathbf{0}\mathbf{ }\mathbf{and}\mathbf{ }\mathbf{Discriminant}\left(\mathbf{D}\right)\mathbf{=}{\mathbf{b}}^{\mathbf{2}}\mathbf{-}\mathbf{4}\mathbf{ac}\mathbf{<}\mathbf{0}\mathbf{.}\mathbf{ }\mathbf{Then}\mathbf{ }{\mathbf{ax}}^{\mathbf{2}}\mathbf{+}\mathbf{bx}\mathbf{+}\mathbf{c}\mathbf{ }\mathbf{ }\mathbf{will}\mathbf{ }\mathbf{be}\mathbf{ }\mathbf{above}\mathbf{ }\mathbf{x}\mathbf{ }\mathbf{axis}\mathbf{ }\mathbf{for}\mathbf{ }\mathbf{all}\mathbf{ }\mathbf{vaues}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{x}\mathbf{.}\mathbf{ }\mathbf{Hence}\mathbf{ }\mathbf{it}\mathbf{ }\mathbf{will}\mathbf{ }\mathbf{always}\mathbf{ }\mathbf{be}\mathbf{ }\mathbf{positive}\mathbf{.} \\ \mathrm{Now} \mathrm{in} {\mathrm{x}}^2-\mathrm{x}+1 \\ \mathrm{a}=1>0 \\ \mathrm{D}={\left(-1\right)}^2-4\times 1\times 1 \\ \Rightarrow \mathrm{D}=-3<0 \\ \Rightarrow {\mathrm{x}}^2-\mathrm{x}+1 \mathrm{will} \mathrm{always} \mathrm{be} \mathrm{positive} \mathrm{a}\mathrm{n}\mathrm{d} \mathrm{h}\mathrm{e}\mathrm{n}\mathrm{c}\mathrm{e} \mathrm{c}\mathrm{a}\mathrm{n} \mathrm{b}\mathrm{e} \mathrm{s}\mathrm{h}\mathrm{i}\mathrm{f}\mathrm{t}\mathrm{e}\mathrm{d} \mathrm{t}\mathrm{o} \mathrm{L}.\mathrm{H}.\mathrm{S}. \mathrm{o}\mathrm{f} \mathrm{i}\mathrm{n}\mathrm{e}\mathrm{q}\mathrm{u}\mathrm{a}\mathrm{l}\mathrm{i}\mathrm{t}\mathrm{y}. \mathrm{S}\mathrm{o} \mathrm{i}\mathrm{n}\mathrm{e}\mathrm{q}\mathrm{u}\mathrm{a}\mathrm{l}\mathrm{i}\mathrm{t}\mathrm{y}\left(\mathrm{i}\right) \mathrm{b}\mathrm{e}\mathrm{c}\mathrm{o}\mathrm{m}\mathrm{e}\mathrm{s}, \\ \frac{1}{\mathrm{x}-1}\le 0 \\ \mathbf{\Rightarrow }\mathbf{x}\mathbf{<}\mathbf{1}\mathbf{ }\mathbf{or}\mathbf{ }\mathbf{x}\mathbf{\in }\left(\mathbf{-}\mathbf{\infty }\mathbf{,}\mathbf{ }\mathbf{1}\right)\mathbf{ }\left(\mathbf{Answer}\right)\mathbf{ }\mathbf{ }\left(\mathrm{Note}: \mathrm{Equality} \mathrm{will} \mathrm{not} \mathrm{hold} \mathrm{because} \mathrm{at} \mathrm{x}=1 \mathrm{denominator} \mathrm{will} \mathrm{become} 0\right) \end{gathered}$$

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