Question

The solution set of $f^{\prime }\left(x\right)>g^{\prime }\left(x\right)$ where $f\left(x\right)=\frac{1}{2}\left(5^{2x+1}\right)$ & $g\left(x\right)=5^x+4x\left(\ln 5\right)$ is

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

Answer 1

Answer: (D)

$x>0$

Given, $f\left(x\right)=\frac{1}{2}\left(5^{2x+1}\right)$

$f^{\prime }\left(x\right)=5^{2x+1}\times \log 5$
$g^{\prime }\left(x\right)=5^{x}l\log 5+4\ln 5=\ln 5(5^x+4)$
$f^{\prime }\left(x\right)>g^{\prime }\left(x\right)$
$=>5\times 5^{2x}>(5^x+4)$
Let $t=5^x$
$=>5t^2-t-4>0$
$=>(t-1)(t+\frac{4}{5})>0$$=>t>1$ or $t>-\frac{4}{5}$
$=>5^x>5^0=>x>0$
so, $x>0$