Question

Let $f\left(x\right)=1-x-x^3.$ Find all real values of x satisfying the inequality,$1-f\left(x\right)-f^3\left(x\right)>f(1-5x)$

• A. $(-2,0)\cup (2,\infty )$
• B. $(0,2)$
• C. $(-\infty ,0)\cup (2,\infty )$
• D. $(0,2)\cup (-2,-\infty )$

Answer 1

Answer: (A)

$(-2,0)\cup (2,\infty )$

$1-f\left(x\right)-f(x{)}^2>f(1-5x)$
Or
$f\left(f\right(x\left)\right)>f(1-5x)$ ...(i)
Now
$f^{\prime }\left(x\right)=-1-3x^2$
Hence
$f^{\prime }\left(x\right)<0$ for all x. ...(ii)
Hence
$f\left(f\right(x)>f(1-5x)$
But $f\left(x\right)$ is a decreasing function (from ii).
Hence
$f\left(x\right)<1-5x$
Or
$1-x-x^3<1-5x$.
Or
$4x-x^3<0$
Or
$x^3-4x>0$
Or
$x(x^2-4)>0$
Or
$x>0$ and $x^2-4>0$ or
$x<0$ and $x^2-4<0$
Hence if $x>0$ $x^2-4>0$ implies $xϵ(2,\infty )$
And
$x<0$ and $x^2-4<0$ implies $(-2,0)$
Hence
$xϵ(-2,0)\cup (2,\infty )$.