A) 2
B) \[k\]
C) \[\frac{1}{2}\]
D) 1
Correct Answer: C
Solution :
\[{{I}_{1}}=\int_{1-k}^{k}{xf\{x(1-x)\}}dx\] \[=\int_{1-k}^{k}{(1-x)f[(1-x)\{1-(1-x)\}]}dx\] [Put\[x=1-x\]] \[=\int_{1-k}^{k}{(1-x)f\{x(1-x)\}}dx\] \[=\int_{1-k}^{k}{f\{x(1-x)\}}dx-\int_{1-k}^{k}{xf\{x(1-x)\}}dx\] \[={{I}_{2}}-{{I}_{1}}\] \[\therefore \] \[2{{I}_{1}}={{I}_{2}}\] \[\Rightarrow \] \[\frac{{{I}_{1}}}{{{I}_{2}}}=\frac{1}{2}\]You need to login to perform this action.
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