A) \[y(lo{{g}_{e}}2)=lo{{e}_{e}}4\]
B) y(x) is decreasing in (0, 1)
C) y(x) is decreasing in \[\left( \frac{1}{2},1 \right)\]
D) \[y(lo{{g}_{e}}2)=\frac{{{\log }_{e}}2}{4}\]
Correct Answer: C
Solution :
Given, \[\frac{dy}{dx}+\left( \frac{2x+1}{x} \right)y={{e}^{-2x}},x>0\] Now. I.F. = \[{{e}^{\int_{{}}^{{}}{\left( \frac{2x+1}{x} \right)dx}}}={{e}^{\int_{{}}^{{}}{\left( 2+\frac{1}{x} \right)dx}}}={{e}^{2x+\log x}}={{e}^{2x}}.x\] So,\[y(x{{e}^{2x}})=\int_{{}}^{{}}{{{e}^{-2x}}}.x{{e}^{2x}}dx+C\] \[\Rightarrow \]\[xy{{e}^{2x}}=\int_{{}}^{{}}{x\,dx+C}\]\[\Rightarrow \]\[2xy{{e}^{2x}}={{x}^{2}}+2C\] Since, it passes through \[\left( 1,\frac{1}{2}{{e}^{-2}} \right),\]So C = 0 \[\therefore \]\[y=\frac{x{{e}^{-2x}}}{2}\] ?(i) Differentiating (i) w.r.t, x, we get \[\frac{dy}{dx}=\frac{1}{2}{{e}^{-2x}}(-2x+1)\] \[\Rightarrow \]y (x) is decreasing in \[\left( \frac{1}{2},1 \right)\] Now,\[y(lo{{g}_{e}}2)=\frac{{{(lo{{g}_{e}}2)}^{-2(lo{{g}_{e}}2)}}}{2}=\frac{1}{8}{{\log }_{e}}2\]
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