If 1 / 3 3 | log e x | d x = m n log e ( n 2 e ) , where m and n are coprime natural numbers, then m 2 + n 2 - 5 is equal to ______ . [2023]

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Solution

Q.
$If \int_{1 / 3}^{3}$ $| \log_{e} x |$ $d x = \frac{m}{n} \log_{e} (\frac{n^{2}}{e}), where m and n are coprime natural numbers, then m^{2} + n^{2} - 5 is equal to$ ______ . [2023]

Ans.
(20)

$Let I = \int_{1 / 3}^{3}$ $| \log_{e} x |$ $d x = \int_{1 / 3}^{1} (- \log_{e} x) d x + \int_{1}^{3} (\log_{e} x) d x$

$= - {[x \log_{e} x - x]}_{1 / 3}^{1} + {[x \log_{e} x - x]}_{1}^{3}$

$= - [- 1 - (\frac{1}{3} \log_{e} \frac{1}{3} - \frac{1}{3})] + [3 \log_{e} 3 - 3 - (- 1)]$

$= - \frac{4}{3} + \frac{8}{3} \log_{e} 3 = \frac{4}{3} (2 \log_{e} 3 - 1) = \frac{4}{3} (\log_{e} \frac{9}{e})$

$Comparing with the given condition, we get m = 4, n = 3 .$

$Now, m^{2} + n^{2} - 5 = 16 + 9 - 5 = 20$

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