• # question_answer If the value of y (greater than 1) satisfying the equation $\int\limits_{1}^{y}{x\,\ell n\,x\,dx=\frac{1}{4}}$ can be expressed in the form of ${{e}^{\frac{m}{n}}}$ , where m and n are relative prime then $(m+n)$ is equal to [Note : e denotes Napier's constant] A)  1                     B)  2 C)  3                                 D)  4

$\int\limits_{1}^{y}{x\,\ln \,xdx\,=\frac{{{y}^{2}}}{2}\,\ln y-\frac{1}{4}\,{{y}^{2}}+\frac{1}{4}}$ $\left[ \left. \ln \,x.\frac{{{x}^{2}}}{2} \right|_{1}^{y}-\frac{1}{2}\,\int\limits_{1}^{y}{xdx} \right]$ $\therefore \,\,\frac{{{y}^{2}}}{2}\,\ln y-\frac{1}{4}{{y}^{2}}=0;\,\,{{y}^{2}}\left[ \frac{\ln \,y}{2}\,-\frac{1}{4} \right]=0$ $\Rightarrow \,\,y={{e}^{1/2}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\Rightarrow \,\,1+2=3$