Question

Let $f:R\to R,f\left(x\right)=1-x-4x^3,$ then the number of integral value(s) of $x\in [0,4]$ satisfying the inequality $4f^3\left(x\right)+f(1-2x)+f\left(x\right)<1$ is

Answer 1

$f^{\prime }\left(x\right)=-1-12x^2<0$
$f\left(x\right)$ is decreasing function
$f\left(f\right(x\left)\right)=1-f\left(x\right)-4f^3\left(x\right)$
Inequality reduces to $f\left(f\right(x\left)\right)>f(1-2x)$
$\Rightarrow f\left(x\right)<1-2x$
$\Rightarrow 1-x-4x^3<1-2x$
$\Rightarrow 4x^3-x>0$
$\Rightarrow x(2x-1)(2x+1)>0$

$x\in (-\frac{1}{2},0)\cup (\frac{1}{2},\infty )$
So, the number of integral values is 4.