Q. 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) $\frac{x^{2} - 3x + 1}{6}$
Explanation
(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
$\ \ \ \ \ 3f(x) = x^{2} + 2x - 1 \Rightarrow f(x) = \frac{x^{2} - 2x - 1}{3}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.