Let `f(x)=int_(1)^(x)(3^(t))/(1+t^(2))dt`, where `xgt0`, Then

Options

A. for `0lt alpha lt beta, f(alpha)lt f(beta)`
B. for `0lt alpha lt beta, f(alpha)gt f(beta)`
C. `f(x)+pi//4lt tan^(-1)x AAxge1`
D. `f(x)+pi//4gt tan^(-1)x AAxge1`

Answer

A, D

Explanation

`f'(x)=(3^(x))/(1+x^(2)) gt 0AAx gt0`
or `f'(x)=(3^(x))/(1+x^(2))gt 1/(1+x^(2)) AAxge1`
`:.int_(1)^(x)f'(x)dx gtint_(1)^(x)1/(1+x^(2))dx`
or `f(x)gt tan^(-1)x-tan^(-1)` or `f(x)+pi//4 gt tan^(-1)x`