Question

Consider the function $f\left(x\right)$ which satisfying the functional equation :
$2f\left(x\right)+f(1-x)=x^2+1,\forall xϵR$ and $g\left(x\right)=3f\left(x\right)+1.$
The range of $\varphi \left(x\right)=g\left(x\right)+\frac{1}{g\left(x\right)+1}$ is :

• A. $[2,\infty )$
• B. $[1,\infty )$
• C. $[-1,\infty )$
• D. $R^{+}$

Answer 1

The correct option is B $[1,\infty )$

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

put $x=1-x$

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

now multiplying above two equations by $2$ and subtract second eqwuation from it, we get

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

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

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

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

$\varphi \left(x\right)=g\left(x\right)+\frac{1}{1+g\left(x\right)}=(x+1{)}^2+\frac{1}{(x+1{)}^2}+1$

Now ${lim}_{x\to \infty }\varphi \left(x\right)=\infty$

and${lim}_{x\to -\infty }\varphi \left(x\right)=\infty$

so, minima, $\varphi \left(x\right)=0$

${\varphi }^{\prime }\left(x\right)=2(x+1)-\frac{2(x+1)}{\left(\right(x+1{)}^2+1{)}^2}=0$

so $x=-1$ and $x=-1,\varphi \left(x\right)=1$

so range of $\varphi \left(x\right)$ is $[1,\infty )$