- A$20$
- B$21$
- C$22$
- D$24$
$=-[ x \ell nx - x ]_{\ell / 3}^1+[ x \ell nx - x ]_1^3$
$=-\left[-1-\left(\frac{1}{3} \ell \ln \frac{1}{3}-\frac{1}{3}\right)\right]+[3 \ln 3-3-(-1)]$
$=\left[-\frac{2}{3}-\frac{1}{3} \ln \frac{1}{3}\right]+[3 \ln 3-2] ~\\ =-\frac{4}{3}+\frac{8}{3} \ln 3$
$=\frac{4}{3}(2 \ell n 3-1)$
$=\frac{4}{3}\left(\ln \frac{9}{ e }\right)$
$\therefore m =4, n =3$
$\text { Now }, m ^2+ n ^2-5=16+9-5=20$
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$f(x)=\left\{\begin{array}{ll}\frac{x^{3}}{(1-\cos 2 x)^{2}} \log _{e}\left(\frac{1+2 x e^{-2 x}}{\left(1-x e^{-x}\right)^{2}}\right), & x \neq 0 \\ \,\alpha & , x=0\end{array}\right.$ જો $\mathrm{f}$ એ $\mathrm{x}=0$ આગળ સતત હોય તો $\alpha$ ની કિમંત મેળવો.