Question

Let $f\left(x\right)=(x^{12}-x^9+x^4-x+1{)}^{-1/2}$. Then domain of function $f\left(x\right)$ is

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

Answer 1

Answer: (C)

$(-\infty ,\infty )$

$f\left(x\right)=\frac{1}{(x^{12}-x^9+x^4-x+1{)}^{1/2}}$

$f\left(x\right)$ is defined for $x^{12}-x^9+x^4-x+1>0$

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

Now, for $x<1$

$g\left(x\right)=\left(x^{12}\right)+(x^4-x^9)+(1-x)>0$

(since each term in bracket is greater than zero for $x<1$)
Similarly for $x\ge 1$

$g\left(x\right)=(x^{12}-x^9)+(x^4-x)+1>0$

(Again each term in bracket is greater than zero for $x>1$)

Hence $f\left(x\right)$ is defined for all values of $x$.

$\therefore$ domain $x\in (-\infty ,\infty )$