A) 1
B) 2
C) 3
D) 4
Correct Answer: C
Solution :
\[\int\limits_{1}^{y}{x\,\ln \,xdx\,=\frac{{{y}^{2}}}{2}\,\ln y-\frac{1}{4}\,{{y}^{2}}+\frac{1}{4}}\] \[\left[ \left. \ln \,x.\frac{{{x}^{2}}}{2} \right|_{1}^{y}-\frac{1}{2}\,\int\limits_{1}^{y}{xdx} \right]\] \[\therefore \,\,\frac{{{y}^{2}}}{2}\,\ln y-\frac{1}{4}{{y}^{2}}=0;\,\,{{y}^{2}}\left[ \frac{\ln \,y}{2}\,-\frac{1}{4} \right]=0\] \[\Rightarrow \,\,y={{e}^{1/2}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\Rightarrow \,\,1+2=3\]You need to login to perform this action.
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