Answer

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**Hint:**

Here we need to find the least value of the number of the time of the logarithm function used such that the given function will become onto the function. A function is said to be an onto function when the range of the given function is equal to the co-domain of the given function. So we will use this concept to solve the problem and also we will use the property of the logarithm function to solve this problem.

**Complete step by step solution:**

The given function is \[g\left( x \right) = \underbrace {\ln \ln \ln ......\ln }_{n{\rm{ times}}}\left( {4\left( {{x^2} + x + 1} \right) + \sin \left( {\pi x} \right)} \right)\].

Let \[f\left( x \right) = \left( {{x^2} + x + 1} \right)\] …….. \[\left( 1 \right)\]

We can see that the given function is always greater than zero.

\[ \Rightarrow \left( {{x^2} + x + 1} \right) > 0\]

Now, we will differentiate the function in equation \[\left( 1 \right)\] with respect to \[x\]. Therefore, we get

\[ \Rightarrow f'\left( x \right) = \dfrac{d}{{dx}}\left( {{x^2} + x + 1} \right)\]

\[ \Rightarrow f'\left( x \right) = 2x + 1\]

Now, we will equate the derivative of the given function with zero to find its minima or maxima value.

\[ \Rightarrow f'\left( x \right) = 2x + 1 = 0\]

On further simplification, we get

\[ \Rightarrow 2x = - 1\]

Dividing both sides by 2, we get

\[ \Rightarrow x = \dfrac{{ - 1}}{2}\]

Now, we will substitute this value obtained in equation 1.

\[ \Rightarrow f\left( { - \dfrac{1}{2}} \right) = \left( {{{\left( {\dfrac{{ - 1}}{2}} \right)}^2} + \left( {\dfrac{{ - 1}}{2}} \right) + 1} \right)\]

Now, we will apply the exponent on the base.

\[ \Rightarrow f\left( { - \dfrac{1}{2}} \right) = \left( {\dfrac{1}{4} - \dfrac{1}{2} + 1} \right)\]

On adding and subtracting these numbers, we get

\[ \Rightarrow f\left( { - \dfrac{1}{2}} \right) = \dfrac{3}{4}\]

We can see that this is the minimum value of the function. So we can write it as

\[ \Rightarrow \dfrac{3}{4} \le \left( {{x^2} + x + 1} \right)\]

On cross multiplying the terms, we get

\[ \Rightarrow 3 \le 4\left( {{x^2} + x + 1} \right)\]

Now, we will add \[\sin \left( {\pi x} \right)\] on right hand side and \[\sin \left( {\pi \left( {\dfrac{{ - 1}}{2}} \right)} \right)\] on right hand side.

\[ \Rightarrow 3 + \sin \left( {\pi \left( {\dfrac{{ - 1}}{2}} \right)} \right) \le 4\left( {{x^2} + x + 1} \right) + \sin \left( {\pi x} \right)\]

On further simplification, we get

\[ \Rightarrow 3 - 1 \le 4\left( {{x^2} + x + 1} \right) + \sin \left( {\pi x} \right)\]

On subtracting the terms, we get

\[ \Rightarrow 2 \le 4\left( {{x^2} + x + 1} \right) + \sin \left( {\pi x} \right)\]

We can write it as

\[ \Rightarrow 2 \le 4\left( {{x^2} + x + 1} \right) + \sin \left( {\pi x} \right) \le \infty \] …………. \[\left( 2 \right)\]

We know that a function is said to be an onto function when range of the given function is equal to the co-domain of the given function i.e.

Range of \[g\left( x \right)\] should be equal to the co-domain of \[g\left( x \right)\].

Let \[4\left( {{x^2} + x + 1} \right) + \sin \left( {\pi x} \right) = y\]

Therefore, the function \[g\left( x \right)\] becomes,

\[ \Rightarrow g\left( x \right) = \underbrace {\ln \ln \ln ......\ln }_{n{\rm{ times}}}\left( y \right)\]

According to question,

\[ - \infty < g\left( x \right) < \infty \]

Now, we will substitute the value of \[g\left( x \right)\] here.

\[ - \infty < \underbrace {\ln \ln \ln ......\ln }_{n{\rm{ times}}}\left( y \right) < \infty \]

We know the property of logarithm that

\[\begin{array}{l} \Rightarrow \ln x = y\\ \Rightarrow x = {e^y}\end{array}\]

Now, we will use this property here.

\[{e^{ - \infty }} < \underbrace {\ln \ln \ln ......\ln }_{n{\rm{ times}}}\left( y \right) < {e^\infty }\]

We know that the value of \[{e^{ - \infty }} = 0\] and \[{e^\infty } = \infty \]

Therefore, we get

\[0 < \underbrace {\ln \ln \ln ......\ln }_{n - 1{\rm{ times}}}\left( y \right) < \infty \]

Again using the same property, we get

\[{e^0} < \underbrace {\ln \ln \ln ......\ln }_{n - 2{\rm{ times}}}\left( y \right) < {e^\infty }\]

We know that the value of \[{e^0} = 1\] and \[{e^\infty } = \infty \]

Therefore, we get

\[1 < \underbrace {\ln \ln \ln ......\ln }_{n - 2{\rm{ times}}}\left( y \right) < \infty \]

Again using the same property, we get

\[{e^1} < \underbrace {\ln \ln \ln ......\ln }_{n - 3{\rm{ times}}}\left( y \right) < {e^\infty }\]

We know that the value of \[e > 2\] and \[{e^\infty } = \infty \]

Therefore, we get

\[2 < \underbrace {\ln \ln \ln ......\ln }_{n - 3{\rm{ times}}}\left( y \right) < \infty \]………. \[\left( 3 \right)\]

On comparing equation 2 and equation 3, we can see that the minimum value of \[n\] can be 3.

**Hence, the correct option is option C.**

**Note:**

To solve this question, we need to remember the properties of logarithm function and meaning of the onto function. . A function is said to be an onto function when the range of the given function is equal to the co-domain of the given function. A logarithm function is defined as a function which is inverse of the exponential function.

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