Question

The solution set of $f\text{'}\left(x\right)>g\text{'}\left(x\right)$ where $f\left(x\right)=\frac{1}{2}·5^{2x+1}$ and $g\left(x\right)=5^x+4x\log 5$ is

• A. $2$
• B. $\left(0,\infty \right)$
• C. $\sin a$
• D. None of these

Answer 1

The correct option is B

$\left(0,\infty \right)$

Explanation for the correct option:

Step 1: Find the derivative of the given functions.

In the question, two functions $f\left(x\right)=\frac{1}{2}·5^{2x+1}$ and $g\left(x\right)=5^x+4x\log 5$ and one inequality $f\text{'}\left(x\right)>g\text{'}\left(x\right)$ is given.

We know that, $\frac{d{a}^x}{dx}=a^x{\log }_{e}a$.

So, the derivative of the function $f\left(x\right)$ is as follows:

$$\begin{gathered} f\text{'}\left(x\right)=\frac{5}{2}·2·5^{2x}·\log 5 \\ \Rightarrow f\text{'}\left(x\right)=5·5^{2x}·\log 5 \end{gathered}$$

And the derivative of the function $g\left(x\right)$ is as follows:

$g\text{'}\left(x\right)=5^x\log 5+4\log 5$

Step 2: Find the solution set of the given inequality.

Since, it is given that $f\text{'}\left(x\right)>g\text{'}\left(x\right)$.

So,

$$\begin{gathered} 5·5^{2x}\log 5>5^x\log 5+4\log 5 \\ \Rightarrow 5·5^{2x}>5^x+4 \end{gathered}$$

Assume that, $5^x=v$.

$$\begin{gathered} 5v^2-v-5>0 \\ \Rightarrow (5v+4)(v-1)>0 \\ \Rightarrow t\in \left(-\infty ,-\frac{4}{5}\right)\cup \left(1,\infty \right) \end{gathered}$$

Since, $5^x$ cannot be negative.

So,

$$\begin{gathered} 5^x>1 \\ \Rightarrow x\log 5>\log 1 \\ \Rightarrow x>0 \end{gathered}$$

Therefore, the solution set of the given inequality is $\left(0,\infty \right)$.

Hence, option $B$ is the correct option.