If \[f(x) = \ln ({x^2} + |x| + 10)\] is a single valued real function then the range of \[f(x)\] in its natural domain will be
A. \[[0, + \infty )\]
B. \[[\ln 10, + \infty )\]
C. \[[0,10]\]
D. \[R\]

Hint: We use the definition of the modulus of x. that is if \[x \geqslant 0\]then \[|x| = x\] and if \[x < 0\] then \[|x| = - x\]. We will find the domain of the given function. By drawing the graph at 0\[x \geqslant 0\] and \[x < 0\] we will find the minimum value of the function then on further simplification we find the natural domain.

Complete step by step answer:
Given \[f(x) = \ln ({x^2} + |x| + 10)\],
Take, \[{x^2} + 7|x| + 10 > 0\] (Because it is obvious that it is greater than zero)
When, \[x \geqslant 0\] we have \[{x^2} + 7x + 10\].
When, \[x < 0\] we have \[{x^2} - 7x + 10\].
Now solving one by one we have,
\[ \Rightarrow {x^2} + 7x + 10 > 0\] (It is obvious for the values of x).
Factoring we get,
\[ \Rightarrow {x^2} + 5x + 2x + 10 > 0\]
\[ \Rightarrow x(x + 5) + 2(x + 5) > 0\]
\[ \Rightarrow (x + 5)(x + 2) > 0\]

We can say that \[x \in [0,\infty ){\text{ - - - - - - - (1)}}\].
Similarly,
\[ \Rightarrow {x^2} - 7x + 10 < 0\] (It is obvious for the values of x)
Factoring we get,
\[ \Rightarrow {x^2} - 5x - 2x + 10 < 0\]
\[ \Rightarrow x(x - 5) - 2(x - 5) < 0\]
\[ \Rightarrow (x - 5)(x - 2) < 0\]
Now,

We can tell that, \[x \in ( - \infty ,0){\text{ - - - - - (2)}}\]
If we take the intersection of (1) and (2) we get the domain \[x \in R\].
Now, draw the graph of \[{x^2} + 7x + 10\] and \[{x^2} - 7x + 10\]

We can see that f is minimum at 10. But maximum is infinity.
Now to plot the other,

We can see that f is minimum at 10 and maximum is infinity.
That is \[10 \leqslant {x^2} + 7x + 10 < \infty \]
Taking logarithm above,
\[\ln 10 \leqslant \ln ({x^2} + 7x + 10) < \ln \infty \]
\[\ln 10 \leqslant f(x) < \infty \]
Hence the natural domain is \[[\ln 10,\infty )\]

Hence the correct option is (B).

Note: If we have \[ \leqslant \] or \[ \geqslant \] we take a closed interval. If we have \[ > \]or \[ < \] we take an open interval. We can draw the graph by putting x values as 0, 1, 2, 3,… we get the value of f(x) which is treated as y values then we can plot the graph to get the minimum values as done above.