Real Numbers And Their Chain Of Inequalities

Relevant wiki: Polynomial Inequalities Problem Solving - Basic

$(A)$ , $(B)$ , $(E)$ and $(F)$ can be true. Thus, the answer is $\boxed{4}$ .

$(A)$ is true for any real number that is greater than $1$ .

$(B)$ is true for any real number that is greater than $-1$ but less than $0$ .

$(E)$ is true for any real number that is less than $-1$ .

$(F)$ is true for any real number that is greater than $0$ but less than $1$ .

To disprove why $(C)$ and $(D)$ are wrong, we divide real numbers into $7$ groups.

$(1)$ Real numbers less than $-1$ . Example: $-2$ .

$(2)$ $-1$ .

$(3)$ Real numbers greater than $-1$ but less than $0$ . Example: $-0.5$ .

$(4)$ $0$ .

$(5)$ Real numbers greater than $0$ but less than $1$ . Example: $0.5$ .

$(6)$ $1$ .

$(7)$ Real numbers greater than $1$ . Example: $2$ .

For group $(1):\quad { x }^{ 3 }\, \, { < }\, \, { x\, \, { < }\, \, x }^{ 2 }$ .

For group $(2):\quad x\, \, =\, \, { x }^{ 3 }\, \, { < }\, \, { x }^{ 2 }$ .

For group $(3):\quad x\, \, {<}\, \, { x }^{ 3 }\, \, { < }\, \, { x }^{ 2 }$ .

For group $(4):\quad x\, \,=\, \,x^2\, \,=\, \,x^3$ .

For group $(5):\quad x^3\, \,{ < }\, \,x^2\, \,{ < }\, \,x$ .

For group $(6):\quad x\, \,=\, \,x^2\, \,=\, \,x^3$ .

For group $(7) : \quad x\, \, { < }\, \, {x }^{ 2 }\, \, { < }\, \, { x }^{ 3 }$ .

$(C)$ and $(D)$ are not satisfied by none of the cases above.

The easiest way to approach this question (in my opinion) is to draw a good sketch.

From here, we see that 4 cases arise depending on the value of $x$ .

$\begin{cases} x<-1 & \Rightarrow \, x^3<x<x^2 & (E) \\ -1<x<0 & \Rightarrow \, x<x^3<x^2 & (B) \\ 0<x<1 & \Rightarrow \, x^3<x^2<x & (F) \\ x>1 & \Rightarrow \, x<x^2<x^3 & (A) \end{cases}$

Thus there are $\large \color{#20A900}{\boxed{4}}$ inequalities which can be true.

We have $6$ cases for this inequality:

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Case 1 : For $x > 1$ , we have

$x<x^2<x^3$

Example: $x=2,\;x^2=4,\;x^3=8\;,2<4<8$

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Case 2 : For $0<x<1$ , we have

$x^3<x^2<x$

Example: $x=0.1,\;x^2=0.01,\;x^3=0.001,\;0.001<0.01<0.1$

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Case 3 : For $x=0,1$ , we have

$x=x^2=x^3$

Proof: $x=0,\;x^2=0,\;x^3=0\quad\quad\quad x=1,\;x^2=1,\;x^3=1$

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Case 4 : For $x=-1$ , we have

$x=x^3<x^2$

Proof: $x=-1,\;x^2=1,\;x^3=-1,\;-1<1$

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Case 5 : For $-1<x<0$ , we have

$x<x^3<x^2$

Example: $x=-0.1,\;x^2=0.01,\;x^3=-0.001,\;-0.1<-0.001<0.01$

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Case 6 : For $x<-1$ , we have

$x^3<x<x^2$

Example: $x=-2,\;x^2=4,\;x^3=-8,\;-8<-2<4$

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Now, let's check back the given options.

$(A)$ is satisfied by Case 1

$(B)$ is satisfied by Case 5

$(C)$ is satisfied by none of the cases

$(D)$ is satisfied by none of the cases

$(E)$ is satisfied by Case 6

$(F)$ is satisfied by Case 2

Therefore, $\boxed{4}$ of these inequalities can be true

@Armain Labeeb Nice solution! +1