JEE MAIN - Mathematics (2023 - 25th January Evening Shift - No. 21)
If $$\int\limits_{{1 \over 3}}^3 {|{{\log }_e}x|dx = {m \over n}{{\log }_e}\left( {{{{n^2}} \over e}} \right)} $$, where m and n are coprime natural numbers, then $${m^2} + {n^2} - 5$$ is equal to _____________.
Answer
20
Explanation
$I=\int\limits_{\frac{1}{3}}^{3}|\ln x| d x=-\int\limits_{\frac{1}{3}}^{1} \ln x d x+\int\limits_{1}^{3} \ln x d x$
$$ \begin{aligned} & \left.=-[x \ln x-x]_{\frac{1}{3}}^{1}+x \ln x-x\right]_{1}^{3} \\\\ & =-\left[(0-1)-\left(\frac{1}{3} \ln 3-\frac{1}{3}\right)\right]+[(3 \ln 3-3)-(0-1)] \\\\ & =\frac{2}{3}-\frac{1}{3} \ln 3+3 \ln 3-2 \end{aligned} $$
$$ \begin{aligned} & =\frac{8}{3} \ln 3-\frac{4}{3} \\\\ & =\frac{4}{3}(2 \ln 3-\ln e) \\\\ & =\frac{4}{3} \ln \left(\frac{3^{2}}{e}\right) \\\\ & m=4, m^{m}=3 \\\\ & m^{2}+n^{2}-5=20 \end{aligned} $$
$$ \begin{aligned} & \left.=-[x \ln x-x]_{\frac{1}{3}}^{1}+x \ln x-x\right]_{1}^{3} \\\\ & =-\left[(0-1)-\left(\frac{1}{3} \ln 3-\frac{1}{3}\right)\right]+[(3 \ln 3-3)-(0-1)] \\\\ & =\frac{2}{3}-\frac{1}{3} \ln 3+3 \ln 3-2 \end{aligned} $$
$$ \begin{aligned} & =\frac{8}{3} \ln 3-\frac{4}{3} \\\\ & =\frac{4}{3}(2 \ln 3-\ln e) \\\\ & =\frac{4}{3} \ln \left(\frac{3^{2}}{e}\right) \\\\ & m=4, m^{m}=3 \\\\ & m^{2}+n^{2}-5=20 \end{aligned} $$
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