Question

If $ f(x) $ satisfies the relation $ 2f(x)+f(1-x)={{x}^{2}} $ for all real $ x, $ then $ f(x) $ is

• 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} $

Answer 1

Answer: (B)

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} $