Question

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

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

Answer 1

Answer: (D)

$x^{12}-x^9+x^4-x+1>0$
Here, three cases arises:
Case I When $x\le 0\Rightarrow x^{12}>0,-x^9>0,x^4>0,-x>0$
$\therefore x^{12}-x^9+x^4-x+1>0,\forall x\le 0$ ...(i)
Case II When $0<x\le 1$
$x^9<x^4$ and $x<1\Rightarrow -x^9+x^4>0$ and $1-x>0$
$\therefore x^{12}-x^9+x^4-x+1>0,\forall 0<x\le 1$ ...(ii)
Case III When $x>1\Rightarrow x^{12}>x^9$ and $x^4>x$
$\therefore x^{12}-x^9+x^4-x+1>0,\forall x>1$ ...(iii)
From Eqs. (i), (ii) and (iii), the above equation holds for all $x\in$ R
Hence , option (d) is the correct answer.