Question

Let $f:R \rightarrow R$ be a function defined by $f\left(x\right)=-x^{3}-3x^{2}-6x+1.$ Number of integers in the solution set of $x$ satisfying the inequality $f\left(f \left(x^{3} + f \left(x\right)\right)\ge f\left(f \left(- f \left(x\right) - x^{3}\right)\right)\right)$ is

• A. $3$
• B. $4$
• C. $5$
• D. $6$

Answer 1

Answer: (A)

$f^{'}\left(x\right) < 0$
$\therefore f\left(x^{3} + f \left(x\right)\le f\left(- f \left(x\right) - x^{3}\right)\right)$
$\Rightarrow x^{3}+f\left(x\right)\geq -f\left(x\right)-x^{3}$
$\Rightarrow f\left(x\right)+x^{3}\geq 0$
$\Rightarrow 3x^{2}+6x-1\leq 0$
As $x\in Z,x\in \left\{\right.-2,-1,0\left.\right\}$