MCQ
If $\int \limits_{\frac{1}{3}}^3\left|\log _e x\right| d x=\frac{m}{n} \log _e\left(\frac{n^2}{e}\right)$, where $m$ and $n$ are coprime natural numbers, then $m ^2+ n ^2-5$ is equal to $............$.
- ✓$20$
- B$21$
- C$22$
- D$24$
✓
Answer
Correct option: A.
$20$
a
$\int \limits_{\frac{1}{3}}^3|\operatorname{nx}| dx =\int \limits_{\frac{1}{3}}^1(-\ell nx ) dx +\int_1^3(\ell nx ) dx$
$\int \limits_{\frac{1}{3}}^3|\operatorname{nx}| dx =\int \limits_{\frac{1}{3}}^1(-\ell nx ) dx +\int_1^3(\ell nx ) dx$
$=-[ x \ell nx - x ]_{\ell / 3}^1+[ x \ell nx - x ]_1^3$
$=-\left[-1-\left(\frac{1}{3} \ell \ln \frac{1}{3}-\frac{1}{3}\right)\right]+[3 \ln 3-3-(-1)]$
$=\left[-\frac{2}{3}-\frac{1}{3} \ln \frac{1}{3}\right]+[3 \ln 3-2] ~\\ =-\frac{4}{3}+\frac{8}{3} \ln 3$
$=\frac{4}{3}(2 \ell n 3-1)$
$=\frac{4}{3}\left(\ln \frac{9}{ e }\right)$
$\therefore m =4, n =3$
$\text { Now }, m ^2+ n ^2-5=16+9-5=20$
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