Question

If $f\left(x\right)=2^{2x-1}$ and $\varnothing \left(x\right)=-2^x+2x\log 2$ . If $f\text{'}\left(x\right)>\varnothing \text{'}\left(x\right)$, then

• A. $0<x<1$
• B. $0\le x<1$
• C. $x>0$
• D. $x\ge 0$

Answer 1

The correct option is C

$x>0$

Explanation for the correct answer.

Step 1: Find the value of $f\text{'}\left(x\right)$

$\begin{array}{rcl}f\left(x\right)& =& 2^{2x-1}\\ & =& \frac{1}{2}·2^{2x}\end{array}$

So,

$\begin{array}{rcl}f\text{'}\left(x\right)& =& \frac{1}{2}·2^{2x}·\log 2·2\\ & =& 2^{2x}\log 2\end{array}$

Step 2: Find the value of $\varphi \text{'}\left(x\right)$

$\begin{array}{rcl}\varphi \left(x\right)& =& -2^x+2x\log 2\\ & =& -2^x+2\log 2·x\end{array}$

So,

$\begin{array}{rcl}\varphi \text{'}\left(x\right)& =& -2^x\log 2+2\log 2\end{array}$

Step 3: Compare $f\text{'}\left(x\right)$ and $\varphi \text{'}\left(x\right)$.

It is given that $f\text{'}\left(x\right)>\varnothing \text{'}\left(x\right)$, so

$\begin{array}{rcl}{2}^{2x}\log 2& >& -2^x\log 2+2\log 2\\ \Rightarrow 2^{2x}& >& -2^x+2....\left(1\right)\end{array}$

Let, $2^x=t>0$, so $\left(1\right)$ will become

$\begin{array}{rcl}{t}^2+t-2& >& 0\\ & \Rightarrow & \left(t+2\right)\left(t-1\right)>0\end{array}$

That means, $t\in \left(-\infty ,-2\right)\cup \left(1,\infty \right)$.

Since, $t>0$ so $t\in \left(1,\infty \right)$.

$$\begin{gathered} \Rightarrow 1<2^x<\infty \\ \Rightarrow 0<x<\infty \end{gathered}$$

Therefore, $x>0$

Hence, option C is correct.