Question

Let $f\left( x \right) = 1 - x - {x^3}$. Find all real values of $x$ satisfying the inequality, $1 - f\left( x \right) - {f^3}\left( x \right) > f\left( {1 - 5x} \right)$
A. $\left( { - 2,0} \right) \cup \left( {2,\infty } \right)$
B. $\left( {0,2} \right)$
C. $\left( { - \infty ,0} \right) \cup \left( {2,\infty } \right)$
D. $\left( {0,2} \right) \cup \left( { - 2, - \infty } \right)$

Answer 1

Here we will find the real values of $x$ which satisfy the given inequality. First, we will find the nature of the function whether it is an increasing or decreasing function. Then we will write the modified function value of the inequality. Finally, we will use the nature of the function to get our desired answer.

Complete step by step solution:
The function is given as,
$f\left( x \right) = 1 - x - {x^3}$…..$\left( 1 \right)$
So to find the nature of the function we will find its derivative.
Differentiating both sides of equation $\left( 1 \right)$ with respect to $x$, we get
$\Rightarrow f'\left( x \right) = - 1 - 3{x^2}$
As we can observe that for any value of $x$the value of the above derivative will always be less than 0. Which means it is a decreasing function.
The inequality given to us is $1 - f\left( x \right) - {f^3}\left( x \right) > f\left( {1 - 5x} \right)$.
As we can see that the left hand side is none other than the function value at $x = f\left( x \right)$ so we can write the left hand side as,
$f\left( {f\left( x \right)} \right) > f\left( {1 - 5x} \right)$
Now, as we know $f\left( x \right)$ it is a decreasing function so the left hand side value will be greater than right hand side only if the value at right hand side is greater than value at left side.
So we can write it as,
$\Rightarrow f\left( x \right) < 1 - 5x$
Substituting value of $f\left( x \right)$ from equation $\left( 1 \right)$ in above inequality, we get
$\Rightarrow 1 - x - {x^3} < 1 - 5x$
On simplifying the inequality, we get,
$\Rightarrow 1 - x - {x^3} - 1 + 5x < 0$
Adding the like terms, we get
$\begin{array}{l} \Rightarrow 4x - {x^3} < 0\\\ \Rightarrow {x^3} - 4x > 0\end{array}$
Taking $x$ common on left side, we get
$\Rightarrow x\left( {{x^2} - 4} \right) > 0$
So we can say
$\Rightarrow x > 0$ and ${x^2} - 4 > 0$
OR
$\Rightarrow x < 0$ and ${x^2} - 4 < 0$
Therefore,
If $x > 0,{x^2} - 4 > 0$ then $x \in \left( {2,\infty } \right)$
If $x < 0,{x^2} - 4 < 0$ then $x \in \left( { - 2,0} \right)$
This means $x \in \left( { - 2,0} \right) \cup \left( {2,\infty } \right)$

Hence, option (A) is correct.

Note:
A function is a relation from a set of inputs to a set of outputs where for every input we have only one output. A function can be constant, increasing or decreasing depending on the value of the variable. Inequality is used to show the non-equal comparison between two mathematical expressions. Mostly it is used to compare two numbers on the number lines by their sizes. An algebraic inequality such as $y > 6$ can be read as “$y$ is greater than 6” in this case $y$ can have many infinite values and its upper case is not given.