Identify the solution set of $x_{1 / 10} (x^{2} + x + 1) > 0$ .

(- \infty, - 1)

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(0, 1)

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(1, \infty)

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(\frac{1}{10}, \infty)

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Solution

The correct option is A $(- \infty, - 1)$
Consider the graph of $x^{2} + x + 1$
It cuts the y-axis at $y = 1$ and is always postive
Therefore, $x {log}_{1 / 10} (x^{2} + x + 1) > 0$
$x {log}_{10} (x^{2} + x + 1) < 0$
Now $x^{2} + x + 1$ is always positive and for $- 1 < x < 0$ $0 < (x^{2} + x + 1) < 1$
Hence, $x {log}_{10} (x^{2} + x + 1)$ would be greater than zero
Therefore ${log}_{10} (x^{2} + x + 1) > 0$ for $x ϵ (- \infty, - 1) \cup (1, \infty)$ ...(i)
Hence, $x$ has to be less than zero for $x {log}_{10} (x^{2} + x + 1) < 0$ ...(ii)
Therefore from i and ii
$x ϵ (- \infty, - 1)$

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