Question

If $f\left(x\right)$ satisfies the relation $2f\left(x\right)+f(1-x)=x^2$ for all real $x$, then $f\left(x\right)$ 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

The correct option is B

$\frac{x^2+2x-1}{3}$

Explanation for the correct option.

Find the value of $f\left(x\right)$:

Given,

$2f\left(x\right)+f(1-x)=x^2\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{1}\mathbf{\right)}$

Replacing $x$ by $(1-x)$,

$$\begin{gathered} 2f(1-x)+f\left(x\right)={\left(1-x\right)}^2 \\ \Rightarrow 2f(1-x)+f\left(x\right)=1+x^2-2x\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{2}\mathbf{\right)} \end{gathered}$$

Now, multiplying equation $\left(1\right)$ by $2$.

$4f\left(x\right)+2f(1-x)=2x^2$

Subtracting this equation from equation $\left(2\right)$.

$$\begin{gathered} 3f\left(x\right)=2x^2-(1+x^2-2x) \\ \Rightarrow f\left(x\right)=\frac{x^2+2x-1}{3} \end{gathered}$$

Hence, the correct option is B.