Question

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

• A. $(-1,1)$
• B. $(0,1)$
• C. $(0,\infty )$
• D. $(-\infty ,\infty )$

Answer 1

Answer: (D)

$(-\infty ,\infty )$

$f\left(x\right)=x^{12}-x^9+x^4-x+1$

If $x<0$

$x^{12}>0,-x^9>0,x^4>0,-x>0,1>0$

$\therefore x^{12}-x^9+x^4-x+1>0\Rightarrow f\left(x\right)>0\forall x<0$

If $x=0$ (Or) $x=1$

$f\left(x\right)=1>0$ for $x=0$ or $1$

$0<x<1$

$f\left(x\right)=x^{12}-x^9+x^4-x+1$

$=(x^9+x)(x^3-1)+1$

$f\left(x\right)-1=(x-1)(x^2+x+1)(x^9+x)$

$\therefore -1<f\left(x\right)-1<0$

$0<f\left(x\right)<1$

If $x>1$

$x^{12}-x^9>0$

$x^4-x>0$

$1>0$

$\therefore f\left(x\right)>0$

$\therefore f\left(x\right)>0\forall xϵR$

$\therefore xϵ(-\infty ,\infty )$