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 domain of $h\left(x\right)=\sqrt{g\left(x\right)}$ is :

• A. $R^{+}$
• B. $[0,\infty ]$
• C. $R$
• D. None of these

Answer 1

The correct option is D None of these

Given :- $2f\left(x\right)+f(1-x)=x^2+1,\forall x\in R$ --eq(1)

Put $(1-x)$ at the place of $x$, and we get

$2f(1-x)+f\left(x\right)={(1-x)}^2+1,\forall x\in R$ --eq(2)

On solving 2eq(1) - eq(2), we get

$3f\left(x\right)=2x-1,\forall x\in R$

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

So, $h\left(x\right)=\sqrt{2x},\forall x\ge 0$

Hence, the domain of $h\left(x\right)$ is [0,∞]

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