Q95 / 180 · COMEDK 2021 · Mathematics
If $f(x)$ satisfies the relation $2 f(x)+f(1-x)=x^{2}$ for all real $x$, then $f(x)$ is
- A$\dfrac{x^{2}+2 x-1}{6}$
- Correct answerB$\dfrac{x^{2}+2 x-1}{3}$
- C$\dfrac{x^{2}+4 x-1}{3}$
- D$\dfrac{x^{2}+4 x-1}{6}$
Solution
Given the equation $2f(x) + f(1-x) = x^2$ (1).
Replacing $x$ with $1-x$ in equation (1), we get:
$2f(1-x) + f(x) = (1-x)^2 = 1 - 2x + x^2$ (2).
From (1), $f(1-x) = x^2 - 2f(x)$. Substituting this into (2):
$2(x^2 - 2f(x)) + f(x) = 1 - 2x + x^2$
$2x^2 - 4f(x) + f(x) = 1 - 2x + x^2$
$-3f(x) = 1 - 2x + x^2 - 2x^2$
$-3f(x) = -x^2 - 2x + 1$
$3f(x) = x^2 + 2x - 1$
$f(x) = \dfrac{x^2 + 2x - 1}{3}$
Answer: $\dfrac{x^{2}+2 x-1}{3}$
Replacing $x$ with $1-x$ in equation (1), we get:
$2f(1-x) + f(x) = (1-x)^2 = 1 - 2x + x^2$ (2).
From (1), $f(1-x) = x^2 - 2f(x)$. Substituting this into (2):
$2(x^2 - 2f(x)) + f(x) = 1 - 2x + x^2$
$2x^2 - 4f(x) + f(x) = 1 - 2x + x^2$
$-3f(x) = 1 - 2x + x^2 - 2x^2$
$-3f(x) = -x^2 - 2x + 1$
$3f(x) = x^2 + 2x - 1$
$f(x) = \dfrac{x^2 + 2x - 1}{3}$
Answer: $\dfrac{x^{2}+2 x-1}{3}$
Hence, the correct answer is option: 'B'.