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Question: If \[g:D \to R\] be a function such that \[g\left( x \right) = \underbrace {\ln \ln \ln ......\ln }_...

If g:D→Rg:D \to R be a function such that g(x)=ln⁡ln⁡ln⁡......ln⁡⏟ntimes(4(x2+x+1)+sin⁡(πx))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) , (n∈N)\left( {n \in N} \right) then the least value of nn for which gg becomes onto, is (where set DD contains the value of xx for which g(x)g\left( x \right) is defined)
A) 1
B) 2
C) 3
D) 4

Explanation

Solution

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(x)=ln⁡ln⁡ln⁡......ln⁡⏟ntimes(4(x2+x+1)+sin⁡(πx))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(x)=(x2+x+1)f\left( x \right) = \left( {{x^2} + x + 1} \right) …….. (1)\left( 1 \right)
We can see that the given function is always greater than zero.
⇒(x2+x+1)>0\Rightarrow \left( {{x^2} + x + 1} \right) > 0
Now, we will differentiate the function in equation (1)\left( 1 \right) with respect to xx. Therefore, we get
⇒f′(x)=ddx(x2+x+1)\Rightarrow f'\left( x \right) = \dfrac{d}{{dx}}\left( {{x^2} + x + 1} \right)
⇒f′(x)=2x+1\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.
⇒f′(x)=2x+1=0\Rightarrow f'\left( x \right) = 2x + 1 = 0
On further simplification, we get
⇒2x=−1\Rightarrow 2x = - 1
Dividing both sides by 2, we get
⇒x=−12\Rightarrow x = \dfrac{{ - 1}}{2}
Now, we will substitute this value obtained in equation 1.
⇒f(−12)=((−12)2+(−12)+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.
⇒f(−12)=(14−12+1)\Rightarrow f\left( { - \dfrac{1}{2}} \right) = \left( {\dfrac{1}{4} - \dfrac{1}{2} + 1} \right)
On adding and subtracting these numbers, we get
⇒f(−12)=34\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
⇒34≤(x2+x+1)\Rightarrow \dfrac{3}{4} \le \left( {{x^2} + x + 1} \right)
On cross multiplying the terms, we get
⇒3≤4(x2+x+1)\Rightarrow 3 \le 4\left( {{x^2} + x + 1} \right)
Now, we will add sin⁡(πx)\sin \left( {\pi x} \right) on right hand side and sin⁡(π(−12))\sin \left( {\pi \left( {\dfrac{{ - 1}}{2}} \right)} \right) on right hand side.
⇒3+sin⁡(π(−12))≤4(x2+x+1)+sin⁡(πx)\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
⇒3−1≤4(x2+x+1)+sin⁡(πx)\Rightarrow 3 - 1 \le 4\left( {{x^2} + x + 1} \right) + \sin \left( {\pi x} \right)
On subtracting the terms, we get
⇒2≤4(x2+x+1)+sin⁡(πx)\Rightarrow 2 \le 4\left( {{x^2} + x + 1} \right) + \sin \left( {\pi x} \right)
We can write it as
⇒2≤4(x2+x+1)+sin⁡(πx)≤∞\Rightarrow 2 \le 4\left( {{x^2} + x + 1} \right) + \sin \left( {\pi x} \right) \le \infty …………. (2)\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(x)g\left( x \right) should be equal to the co-domain of g(x)g\left( x \right).
Let 4(x2+x+1)+sin⁡(πx)=y4\left( {{x^2} + x + 1} \right) + \sin \left( {\pi x} \right) = y
Therefore, the function g(x)g\left( x \right) becomes,
⇒g(x)=ln⁡ln⁡ln⁡......ln⁡⏟ntimes(y)\Rightarrow g\left( x \right) = \underbrace {\ln \ln \ln ......\ln }_{n{\rm{ times}}}\left( y \right)
According to question,
−∞<g(x)<∞- \infty < g\left( x \right) < \infty
Now, we will substitute the value of g(x)g\left( x \right) here.
−∞<ln⁡ln⁡ln⁡......ln⁡⏟ntimes(y)<∞- \infty < \underbrace {\ln \ln \ln ......\ln }_{n{\rm{ times}}}\left( y \right) < \infty
We know the property of logarithm that
⇒ln⁡x=y ⇒x=ey\begin{array}{l} \Rightarrow \ln x = y\\\ \Rightarrow x = {e^y}\end{array}
Now, we will use this property here.
e−∞<ln⁡ln⁡ln⁡......ln⁡⏟ntimes(y)<e∞{e^{ - \infty }} < \underbrace {\ln \ln \ln ......\ln }_{n{\rm{ times}}}\left( y \right) < {e^\infty }
We know that the value of e−∞=0{e^{ - \infty }} = 0 and e∞=∞{e^\infty } = \infty
Therefore, we get
0<ln⁡ln⁡ln⁡......ln⁡⏟n−1times(y)<∞0 < \underbrace {\ln \ln \ln ......\ln }_{n - 1{\rm{ times}}}\left( y \right) < \infty
Again using the same property, we get
e0<ln⁡ln⁡ln⁡......ln⁡⏟n−2times(y)<e∞{e^0} < \underbrace {\ln \ln \ln ......\ln }_{n - 2{\rm{ times}}}\left( y \right) < {e^\infty }
We know that the value of e0=1{e^0} = 1 and e∞=∞{e^\infty } = \infty
Therefore, we get
1<ln⁡ln⁡ln⁡......ln⁡⏟n−2times(y)<∞1 < \underbrace {\ln \ln \ln ......\ln }_{n - 2{\rm{ times}}}\left( y \right) < \infty
Again using the same property, we get
e1<ln⁡ln⁡ln⁡......ln⁡⏟n−3times(y)<e∞{e^1} < \underbrace {\ln \ln \ln ......\ln }_{n - 3{\rm{ times}}}\left( y \right) < {e^\infty }
We know that the value of e>2e > 2 and e∞=∞{e^\infty } = \infty
Therefore, we get
2<ln⁡ln⁡ln⁡......ln⁡⏟n−3times(y)<∞2 < \underbrace {\ln \ln \ln ......\ln }_{n - 3{\rm{ times}}}\left( y \right) < \infty ………. (3)\left( 3 \right)
On comparing equation 2 and equation 3, we can see that the minimum value of nn 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.