Question

The values of x satisfying the inequality $|x^3-1|\ge 1-x$ belong to

• A. $(-\infty ,-1]$
• B. $(-\infty ,-1]\cup [0,\infty )$
• C. $[1,\infty )$
• D. none of these

Answer 1

Answer: (B)

$|x^3-1|\ge 1-x\Rightarrow |x-1|(x^2+x+1)\ge 1-x$ $[\because x^2+x+1>0]$
Let $x<1$, then we get
$(1-x)(x^2+x+1)\ge 1-x$
$\Rightarrow (x-1)(x^2+x+1-1)\le 0$
$\Rightarrow x(x+1)(x-1)\le 0$
Solving by method of intervals we get
$x\in (-\infty ,-1]\cup [0,1]$
Let $x\ge 1$ , then, we get
$(x-1)(x^2+x+1)\ge 1-x$
$\Rightarrow (x-1)(x^2+x+2)\ge 0$
$\Rightarrow x\ge 1[\because x^2+x+2>0\forall x\in R]$
$\Rightarrow x\in [1,\infty )$
Combining the two solutions we get
$x\in (-\infty ,-1]\cup [0,1]\cup [1,\infty )$
or $x\in (-\infty ,-1]\cup [0,\infty )$