Question

Let $y=\sqrt{\{\left(\frac{2}{x^2-x+1}\right)-\left(\frac{1}{1+x}\right)-\left(\frac{2x+1}{x^3+1}\right)\}}$

The solution set of $x$ for which $y$ is real is the union of

• A. an open set and a closed set
• B. two open sets
• C. two closed sets
• D. a constant and an open set

Answer 1

The correct option is A an open set and a closed set

We have,
$y=\sqrt{\{\left(\frac{2}{x^2-x+1}\right)-\left(\frac{1}{1+x}\right)-\left(\frac{2x+1}{x^3+1}\right)\}}$

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

Since, $y$ takes real values

$\therefore \frac{x(1-x)}{(x+1)(x^2-x+1)}\ge 0\Rightarrow \frac{x(1-x)}{(x+1)\{{(x-\frac{1}{2})}^2+\frac{3}{4}\}}\ge 0\Rightarrow \frac{x(1-x)}{(x+1)}\ge 0\Rightarrow \frac{x(1-x)}{x+1}\le 0$

$\therefore x\in (-\infty ,-1)\cup [0,1]$