The function `f(x)=(ln(pi+x))/(ln(e+x))` is

Options

A. increasing in `(0,oo)`
B. decreasing oin `(0,oo)`
C. increasing in `(0,pi//e),decreasing in(pi//e,oo)`
D. decreasing in `(0,pi//e)` increasing in `(pi//e,oo)`

Answer

2

Explanation

`f(X)=(log(e+x)xx(1)/(pi+x)-log(pi+x)(1)/(e+x))/(log(e+x))^(2)`
`=(log(e+x)xx(e+x)-(pi+x)log(pi+x))/(pi+x)(e+x)(log(e+x))/6(2)`
Since log function is an increasing function and `eltpi`
`log(e+x)ltlog(pi+x)`
Thus `(e+x)log(e+x)lt(e+x)log(pi+x)lt(log(pi+x)`
for all `x gt 0`
Thus f(x)=0 ltrbgt Therefore f(X) decrease on `(0,oo)`