Question

If $f\left( x \right)$ satisfies the relation $2f\left( x \right) + f\left( {1 - x} \right) = {x^2}$ for all real $x$, then $f\left( x \right)$ is
1. $\dfrac{{\left[ {{x^2} + 2x - 1} \right]}}{6}$
2. $\dfrac{{\left[ {{x^2} + 2x - 1} \right]}}{3}$
3. $\dfrac{{\left[ {{x^2} + 4x - 1} \right]}}{3}$
4. $\dfrac{{\left[ {{x^2} - 3x + 1} \right]}}{6}$
5. $\dfrac{{\left[ {{x^2} + 3x - 1} \right]}}{3}$

Answer 1

Hint: In this question, we are given the relation $2f\left( x \right) + f\left( {1 - x} \right) = {x^2}$ which is satisfied by $f\left( x \right)$, and we have to find the value of $f\left( x \right)$. The first step is to put $x = 1 - x$ in the relation then multiply the given relation by $2$ and in the last subtract the required relation equation from the given one which is multiplied by $2$. Solve further and you’ll get the value of $f\left( x \right)$.

Formula Used:
Given relation $2f\left( x \right) + f\left( {1 - x} \right) = {x^2}$
Using this applied the elimination method to calculate the value of $f\left( x \right)$.

Complete step by step Solution:
Given that,
$f\left( x \right)$ satisfies the relation $2f\left( x \right) + f\left( {1 - x} \right) = {x^2} - - - - - - (1)$
Put $x = 1 - x$ in the equation (1),
Equation will be
$2f\left( {1 - x} \right) + f\left( {1 - \left( {1 - x} \right)} \right) = {\left( {1 - x} \right)^2}$
$f\left( x \right) + 2f\left( {1 - x} \right) = {\left( {1 - x} \right)^2} - - - - - \left( 2 \right)$
Multiplying equation (1) by $2$ on both sides,
$4f\left( x \right) + 2f\left( {1 - x} \right) = 2{x^2} - - - - - \left( 3 \right)$
Now, subtracting equation (2) from (3)
$3f\left( x \right) = 2{x^2} - {\left( {1 - x} \right)^2}$
$3f\left( x \right) = 2{x^2} - \left( {1 + {x^2} - 2x} \right)$
$f\left( x \right) = \dfrac{{{x^2} + 2x - 1}}{3}$

Hence, the correct option is 2.

Note: The key concept involved in solving this problem is a good knowledge of methods to find unknown terms. Students must know that there are three methods to find the unknown terms: Elimination method, Substitution method, and graphical method. Here, we have applied the elimination method, which is basically the process of eliminating one of the variables in a system of linear equations by just using any of the operation addition or subtraction methods in conjunction with the variable coefficient multiplication or division. You can solve using any of the methods that will also be correct.