Introduction to real valued functions

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Question

$The least integral value of x satisfying the equation ∣ x^{2} - 1 + \log x | = | x^{2} - 1 | + | \log x ∣ is$

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Solution

$|x^{2} - 1 + \log x| = |x^{2} - 1| + | \log x |$
$\begin{array}{l} \Rightarrow (x^{2} - 1) \log x \geq 0 \\ We must have x > 0 \\ \Rightarrow x \in (0, \infty) \end{array}$
$\Rightarrow Least integral value of x is 1$

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