A) 2
B) 4
C) 6
D) 8
Correct Answer: B
Solution :
\[f(2a-x)=f(x)\] \[+\int_{1}^{2}{|1-{{x}^{2}}|dx}\] = \[-\int_{-2}^{-1}{(1-{{x}^{2}})\,dx+\int_{-1}^{1}{(1-{{x}^{2}})\,dx-\int_{1}^{2}{(1-{{x}^{2}})\,dx}}}\] = \[\frac{4}{3}+\frac{4}{3}+\frac{4}{3}\]= 4.
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