Question

Solve $|x^2-1+\sin x|=|x^2-1|+|\sin x|$, where $x\in [-2\pi ,2\pi ]$

• A. $x\in [-3\pi ,-\pi ]\cup [-2,0]\cup [1,\pi ]$
• B. $x\in [-2\pi ,-\pi ]\cup [-1,0]\cup [1,\pi ]$
• C. $x\in [-4\pi ,-\pi ]\cup [1,0]\cup [-1,\pi ]$
• D. None of these

Answer 1

Answer: (B)

$x\in [-2\pi ,-\pi ]\cup [-1,0]\cup [1,\pi ]$

Let $f\left(x\right)=x^2-1,g\left(x\right)=\sin x$
$\therefore$ Using $|f\left(x\right)+g\left(x\right)|=|f\left(x\right)|+|g\left(x\right)|$, which is true, if $f\left(x\right)\cdot g\left(x\right)\ge 0$
$\therefore (x^2-1)\cdot \sin x\ge 0$ on $[-2\pi ,2\pi ]$
Using number line rule,
$\Rightarrow x\in [-2\pi ,-\pi ]\cup [-1,0]\cup [1,\pi ]$