A) 150
B) 100
C) 50
D) 250
Correct Answer: C
Solution :
[c] \[I=\int\limits_{0}^{100}{\left\{ x-[x] \right\}.dx}=\int\limits_{0}^{100\times 1}{\left\{ x-[x] \right\}.dx}=100\int\limits_{0}^{1}{\left\{ x-[x] \right\}.dx}\][Since, \[\left\{ x-\left[ x \right] \right\}\]is periodic function with period 1] \[100\left[ \frac{{{x}^{2}}}{2}-0 \right]_{0}^{1}=100\times \frac{1}{2}=50\] Hence, option [c] is correct.
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