Question

The domain of the function is? If $f\left( x \right)$ =${\left( {{x^{12}} - {x^9} + {x^4} - x + 1} \right)^{ - 1/2}}$.
A. (1, $+ \infty$)
B. ($- \infty$,-1)
C. (-1, 1)
D. ($- \infty$,$+ \infty$ )

Answer 1

HINT – In this question, the first thing that you should know is domain. To find the value of the domain we put the value of $x$< 1 and $x$ $\geqslant$1 in the function to check the function value i.e. negative or positive.

Complete step by step solution:
The domain of a function is the set of all possible inputs i.e. value of $x$ for the function.
In this question it is given that,
$f\left( x \right)$ =${\left( {{x^{12}} - {x^9} + {x^4} - x + 1} \right)^{ - 1/2}}$
We write above function like that,
$f\left( x \right)$ = $\dfrac{1}{{{{\left( {{x^{12}} - {x^9} + {x^4} - x + 1} \right)}^{1/2}}}}$
$f\left( x \right)$is defined for $\left( {{x^{12}} - {x^9} + {x^4} - x + 1} \right)$ > 0
Let c = $\left( {{x^{12}} - {x^9} + {x^4} - x + 1} \right)$
Now, for $x$ < 1
$g\left( x \right)$= $\left( {{x^{12}}} \right) + \left( {{x^4} - {x^9}} \right) + \left( {1 - x} \right)$ > 0
Since each term in the bracket is greater than zero for $x$ < 1.
Similarly for $x$ $\geqslant$1
$g\left( x \right)$=$\left( {{x^{12}} - {x^9}} \right) + \left( {{x^4} - x} \right) + 1$ > 0
Again each term in bracket is greater than zero for $x$ $\geqslant$1
Hence, $f\left( x \right)$ is defined for all values of $x$.
$\therefore$ domain $x \in \left( { - \infty , + \infty } \right)$
So option D is the correct option.

NOTE –The range of a set of data is the difference between the highest and lowest value in the set. To find the range of data from least to greatest. Then subtract the smallest value from the larger value in the set. Essentially, the range tells us how spread apart a group of numbers is.