Question

If \[f(x) = {2^{2x - 1}}\] , \[\phi (x) = - {2^x} + 2x\log 2\] and \[{f^{'}}(x) > {\phi ^{'}}(x)\] , then
A. \[0 < x < 1\]
B. \[0 \le x < 1\]
C. \[x > 0\]
D. \[x \ge 0\]

Answer 1

Hint:By using the differentiation formula of \[{a^u}\] with respect to \[x\], where \[a\] is a constant and \[u\] is a function of \[x\], the first derivative of both the given functions are found out and then inequality is established with the calculated values as per given data to solve for \[x\] and find the correct option.

Formula used: The differentiation formula used to solve this problem are as follows:
1. \[\dfrac{{d({a^u})}}{{dx}} = {a^u} \times \log a \times \dfrac{{d(u)}}{{dx}}\] , where \[a\] is a constant and \[u\] is a function of \[x\]
2. \[\dfrac{{d({x^n})}}{{dx}} = n{x^{n - 1}}\] , where, \[x\] belongs to any real number.

Complete step-by-step solution:
We have been given two functions \[f(x)\] and \[\phi (x)\] .
First, we will found out the first derivatives of the above functions with respect to \[x\] .
Given, \[f(x) = {2^{2x - 1}}\]
\[ \Rightarrow f(x) = \dfrac{{{2^{2x}}}}{2}\]
\[ \Rightarrow {f^{'}}{x} = \dfrac{1}{2} \times {f'}{(2^{2x})}\]
Using formula \[\dfrac{{d({a^u})}}{{dx}} = {a^u} \times \log a \times \dfrac{{d(u)}}{{dx}}\] and considering \[u\] equal to \[{2^{2x}}\] , we get
\[ \Rightarrow {f^{'}}(x) = \dfrac{1}{2} \times {2^{2x}} \times \log 2 \times \dfrac{{d(2x)}}{{dx}}\]
Now, using the formula \[\dfrac{{d({x^n})}}{{dx}} = n{x^{n - 1}}\] and considering \[n\] equal to 1, we get \[\dfrac{{d(x)}}{{dx}} = 1{x^{1 - 1}} = {x^0} = 1\]
So, \[{f^{'}}(x) = \dfrac{1}{2} \times {2^{2x}} \times \log 2 \times 2\]
\[ \Rightarrow {f^{'}}(x) = {2^{2x}} \times \log 2\]

Similarly,
\[\phi (x) = - {2^x} + 2x\log 2\]
\[ \Rightarrow {\phi ^{'}}(x) = - {2^x}\log 2 + 2\log 2\]
\[ \Rightarrow {\phi ^{'}}(x) = ( - {2^x} + 2)\log 2\]

Given, \[{f^{'}}(x) > {\phi ^{'}}(x)\]
\[ \Rightarrow {2^{2x}}\log 2 > ( - {2^x} + 2)\log 2\]
\[ \Rightarrow {2^{2x}} > - {2^x} + 2\] [Dividing both the sides by \[\log 2\] ]
\[ \Rightarrow {2^{2x}} + {2^x} - 2 > 0\]

Taking the value of \[{2^x}\] equal to \[t\] , we have
\[{t^2} + t - 2 > 0\]
\[ \Rightarrow ({t^2} + 2t) - (t + 2) > 0\]
\[ \Rightarrow (t + 2)(t - 1) > 0\]
\[ \Rightarrow t \in ( - \infty , - 2) \cup (1,\infty )\]

Substituting the value of \[t\] , we get
\[{2^x} \in ( - \infty , - 2) \cup (1,\infty )\]
But, \[{2^x} \in ( - \infty , - 2)\] is not possible as \[x \in R\]
So, \[{2^x} \in (1,\infty )\]
\[ \Rightarrow 1 < {2^x} < \infty \]
\[ \Rightarrow {2^0} < {2^x} < {2^\infty }\]
\[ \Rightarrow 0 < x < \infty \]
Thus, \[x > 0\]

Hence, option C. is the correct answer.

Note: If \[f(x) = {a^u}\] , where, \[a\] is not a constant, but, a function of \[x\] and \[u\] is also a function of \[x\] , then, \[{f^{'}}(x)\] can not be found by using the formula \[\dfrac{{d({a^u})}}{{dx}} = {a^u} \times \log a \times \dfrac{{d(u)}}{{dx}}\] . In this case, the rules of logarithmic differentiation will be applied to find the first derivative, where, the natural log of both the sides are taken first and then both sides are differentiated with respect to \[x\] .