Question

If $f\left( x \right)+2f\left( 1-x \right)={{x}^{2}}+2\ \ \gamma \ \ x\in \text{R}$ then find f (x):

A. ${{\left( x-1 \right)}^{{\scriptstyle{}^{2}/{}_{3}}}}$
B. \[-{{\left( x-2 \right)}^{{\scriptstyle{}^{2}/{}_{3}}}}\]
C. ${{x}^{2}}-1$
D. ${{x}^{2}}-2$

Answer 1

Hint: Since $f(x)+2f\left( 1-x \right)={{x}^{2}}+2$ is given. We replace x by $\left( 1-x \right)$ to get another equation. We solve both the equations to calculate f (x).

Complete step by step solution: We are given that,

$f(x)+2f(1-x)={{x}^{2}}+2:$ →(1)

Replacing x by $\left( 1-x \right)$ in the above equation (1) to get;

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

Solving this equation, we get

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

Simplifying further we get,

$f(1-x)+2f\left( -x \right)={{x}^{2}}-2x+3$

We have two equations:

$f(x)+2f\left( 1-x \right)={{x}^{2}}+2$ → eq (1)
& $2f(x)+f\left( 1-x \right)={{x}^{2}}-2x+3$ → eq (2)

Subtract eq (1) from eq (2) we get

$2f(x)+f\left( 1-x \right)-f(x)-2f(1-x)={{x}^{2}}-2x+3-{{x}^{2}}-2$
$\Rightarrow f(x)-f\left( 1-x \right)=-2x+1$ → eq (3)

But we know: $2f(x)+f\left( 1-x \right)={{x}^{2}}-2x+3$ from eq (2)
$\Rightarrow f(1-x)={{x}^{2}}-2x+3-2f\left( x \right)$

Substituting this $f(1-x)={{x}^{2}}-2x+3-2f\left( x \right)$ in eq (3) to get

$\Rightarrow f(x)-\left( {{x}^{2}}-2x+3-2f(x) \right)=-2x+1$
$\Rightarrow f(x)-{{x}^{2}}+2x-3+2f(x)=-2x+1$
$\Rightarrow 3f(x)=-2x+1+{{x}^{2}}-2x+3$
$3f(x)={{x}^{2}}-4x+4$
$3f(x)={{\left( x-2 \right)}^{2}}$
$\therefore $
∴None of the option matches the answer

The answer \[f\left( x \right)=\dfrac{{{\left( x-2 \right)}^{2}}}{3}\]

Note: In these types of problem, we always replace x by some 1 – x or 1+ x or 2 – x.Whatever the question demands and then we eliminate another variable to calculate f(x).