Question

Let $f\left(x\right)+(x+\frac{1}{2})f(1-x)=1$ and $g\left(x\right)=\frac{e^{-x}}{1+\left[\ln x\right]},$ where $[.]$ is the greatest integer function. Then domain of $(f+g)$ is

• A. $R-[\frac{1}{e},1)$
• B. $(0,\infty )-[-\frac{1}{2},\frac{1}{e})$
• C. $(0,\infty )-[\frac{1}{e},1)$
• D. $(0,\infty )-[-\frac{1}{2},1)$

Answer 1

The correct option is C $(0,\infty )-[\frac{1}{e},1)$

$f\left(x\right)+(x+\frac{1}{2})f(1-x)=1...\left(1\right)$
Replacing $x$ by $1-x$, we get
$f(1-x)+(\frac{3}{2}-x)f\left(x\right)=1...\left(2\right)$

From eqn(1) and (2), we have
$\frac{1-f\left(x\right)}{x+\frac{1}{2}}+(\frac{3}{2}-x)f\left(x\right)=1$
$\Rightarrow 1-f\left(x\right)+(x+\frac{1}{2})(\frac{3}{2}-x)f\left(x\right)=x+\frac{1}{2}$
$\Rightarrow f\left(x\right)=\frac{2}{1-2x}$

$\Rightarrow f\left(x\right)$ is not defined at $x=\frac{1}{2}$
Hence, domain of $f=R-\left\{\frac{1}{2}\right\}$

Now, for $g\left(x\right)$, $e^{-x}$ is defined $\forall x\in R$
But $\ln x$ is defined $\forall x\in (0,\infty )$
Also, for $g\left(x\right)$ to be defined, $\left[\ln x\right]\ne -1$, which is possible only if
$\Rightarrow -1\le \ln x<0$
$\Rightarrow e^{-1}\le x<1$
$\Rightarrow x\in [\frac{1}{e},1)$
Thus, domain of $g$ is defined as $x\in (0,\infty )-[\frac{1}{e},1)$

Hence, domain of $(f+g)$ is
$x\in (0,\infty )-[\frac{1}{e},1)$.