Question

If $f\left(x\right)=\frac{5^x}{ln5}+\frac{7^x}{ln7}$, then maximum value of slope of tangent to f(x) in the interval [1, 2].

• A. 12
• B. 64
• C. 74
• D. 10

Answer 1

The correct option is C 74

$f\left(x\right)=\frac{5^x}{\ln 5}+\frac{7^x}{\ln 7}$

$f^{\prime }\left(x\right)=5^x\frac{\ln 5}{\ln 5}+7^x\frac{\ln 7}{\ln 7}=5^x+7^x$

$f^{\prime \prime }\left(x\right)=5^x\ln 5+7^x\ln 7$

$f^{\prime \prime }\left(x\right)>0$ in $[1,2],$ hence $f^{\prime }\left(x\right)$ is increasing in $[1,2]$

Minimum value $=f^{\prime }\left(2\right)=5^2+7^2=25+49=74$