Question

If $f\left(x\right)+2f(1-x)=x^2+1\forall x\in R$ then range of $f$ is

• A. $(-\infty ,\frac{1}{3}]$
• B. $[-\frac{1}{3},\infty )$
• C. $[-\frac{1}{3},\frac{1}{3}]$
• D. $[\frac{1}{3},\infty )$

Answer 1

Answer: (B)

$[-\frac{1}{3},\infty )$

We have,

$\begin{array}{c}f\left(x\right)+2f\left(1-x\right)=x^2+1\\ x\to \left(1-x\right)\\ \left[f\left(1-x\right)+2f\left(x\right)={\left(1-x\right)}^2+1\right]\times 2\\ -3f\left(x\right)=x^2+1-2{\left(1-x\right)}^2-2\\ 3f\left(x\right)=2{\left(1-x\right)}^2+2-x^2-1\\ 3f\left(x\right)=2{\left(x-1\right)}^2+1-x^2\\ =2\left(x^2+1-2x\right)+1-x^2\\ 3f\left(x\right)=x^2-4x+3\\ f\left(x\right)=\frac{1}{3}\left[x^2-4x+3\right]\\ f\left(x\right)=\frac{1}{3}\left[{\left(x-2\right)}^2-1\right]\\ f\left(x\right)=\frac{1}{3}{\left(x-2\right)}^2-\frac{1}{3}\\ Rangeoff=[-\frac{1}{3},\infty )\\ Hence,\\ optionBiscorrectanswer.\end{array}$