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