Question

If $\left|\text{f}\left(\text{x}\right)\text{+}\text{6}\text{-}{\text{x}}^{\text{2}}\right|\text{=}\left|\text{f}\left(\text{x}\right)\right|\text{+}\left|\text{4}\text{-}{\text{x}}^{\text{2}}\right|\text{+}\text{2,}$ then $\text{f}\left(\text{x}\right)$ is necessarily non-negative in

• A. $[-2,2]$
• B. $\left(-\infty ,-2\right)\cup \left(2,\infty \right)$
• C. $[-\sqrt{6},\sqrt{6}]$
• D. none of these

Answer 1

Answer: (A)

$[-2,2]$

$|f\left(x\right)+6-x^2|=|f\left(x\right)|+|4-x^2|+2$

Above equation can also write it as,

$\Rightarrow$ $|f\left(x\right)+4-x^2+2|=|f\left(x\right)|+|4-x^2|+|2|$

We know, $|a+b+c|=|a|+|b|+|c|$, this is only possible if $a,b,c$ are of same sign.

Here, $a=f\left(x\right),b=4-x^2$ and $c=2$

So, here $f\left(x\right)$ and $2$ are non-negative terms.

$4-x^2$ term also be non-negative

$\therefore$ $4-x^2\ge 0$

Multiplying by $(-)$ sign we get,

$\Rightarrow$ $x^2-4\le 0$

$\Rightarrow$ $(x+2)(x-2)\le 0$

$\therefore$ $x\in [-2,2]$

$\therefore$ $f\left(x\right)$ is necessarily non-negative in $x\in [-2,2]$