A) \[\frac{{{x}^{2}}+2x-1}{6}\]
B) \[\frac{{{x}^{2}}+2x-1}{3}\]
C) \[\frac{{{x}^{2}}+4x-1}{3}\]
D) \[\frac{{{x}^{2}}-3x+1}{6}\]
E) \[\frac{{{x}^{2}}+3x-1}{3}\]
Correct Answer: B
Solution :
Given, \[2f(x)+f(1-x)={{x}^{2}}\] ...(i) Replacing\[x\]by\[(1-x),\]we get \[2f(1-x)+f(x)={{(1-x)}^{2}}\] \[\Rightarrow \] \[2f(1-x)\,+f(x)=1+{{x}^{2}}-2x\] ...(ii) On multiplying Eq. (i) by 2 and subtracting from Eq. (ii), we get \[3f(x)={{x}^{2}}+2x-1\] \[\Rightarrow \] \[f(x)=\frac{{{x}^{2}}+2x-1}{3}\]
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