Question

The largest interval for which $x^{12}-x^9+x^4-x+1>0$ is

• A. $-4<x<0$
• B. $0<x<1$
• C. $-100<x<100$
• D. $-\infty <x<\infty$

Answer 1

The correct option is D

$-\infty <x<\infty$

Explanation for correct option

Step 1: When $x\le 0$

Given: $x^{12}-x^9+x^4-x+1$

Let us take the case $x\le 0$

• $x^{12},x^4$will be positive as they have an even power.
• $-x^9,-x$ will be also positive as both will be negative as they are multiplied by a negative symbol.

$\therefore x^{12}-x^9+x^4-x+1$ will be positive for $x\le 0$.

Step 2: When $0<x<1$

Let us take the case $0<x<1$

• $1-x$ will be positive as $1>x\Rightarrow 1-x>0$.
• Similarly, $x^4-x^9$will be positive as $x^4>x^9$ for the all numbers between $0$ and $1$.

$\therefore x^{12}-x^9+x^4-x+1$will be positive for $0<x<1$.

Step 3: When $x\ge 1$

Let us take the case $x\ge 1$

• $x^{12}-x^9$ will be positive as $x^{12}>x^9\Rightarrow x^{12}-x^9>0$.
• Similarly, $x^4-x$will be positive as $x^4>x\Rightarrow x^4-x>0$.

$\therefore x^{12}-x^9+x^4-x+1$will be positive for $x\ge 1$.

Therefore, $x^{12}-x^9+x^4-x+1$ will be greater than $0$ for all $x$ i.e.$-\infty <x<\infty$.

Hence, the correct option is option (D)