Question

If $f\left(x\right)=30-2x-x^3$, then the number of positive integral value(s) of $x$ satisfying $f\left(f\right(f\left(x\right)\left)\right)>f\left(f\right(-x\left)\right)$ is
(correct answer + 1, wrong answer - 0.25)

• A. $3$
• B. $4$
• C. $1$
• D. $2$

Answer 1

The correct option is D $2$

$f\left(x\right)=30-2x-x^3$
$\Rightarrow f^{\prime }\left(x\right)=-2-3x^2<0\Rightarrow f^{\prime }\left(x\right)<0$
So $f\left(x\right)$ is decreasing function for all $x\in R$
Now,
$f\left(f\right(f\left(x\right)\left)\right)>f\left(f\right(-x\left)\right)\Rightarrow f\left(f\right(x\left)\right)<f(-x)\Rightarrow f\left(x\right)>-x\Rightarrow 30-2x-x^3+x>0\Rightarrow x^3+x-30<0$
By observation, $x=3$ is one root
$\Rightarrow (x-3)(x^2+3x+10)<0\Rightarrow x<3(\because x^2+3x+10>0)$

Hence, number of positive integral values of $x$ is $2$.