A continuous function `f(x)` satisfies the relation `f(x)=e^x+int_0^1 e^xf(t)dt` then `f(1)=`

Options

A. `f(0)lt0`
B. `f(x)` is a decreasing function
C. `f(x)` is increasing function
D. `int_(0)^(1)f(x)dxgt0`

Answer

A, B

Explanation

`f(x)=e^(x)+int_(0)^(1)e^(x)f(t)dt-e^(x)+ke^(x)`, where `k=int_(0)^(1)f(t)dt`
`:. k=int_(0)^(1)(e^(t)+ke^(t))dt=e+ke-1-k`
`:.k=(e-1)/(2-e)`
Thus `f(x)=e^(x)(1+(e-1)/(2-e))=(e^(x))/(2-e)`
obviously,`f(0)=1/(2-e)lt0`
Also `f'(x)=(e^(x))/(2-e)lt 0` for `AAxepsilonR`.
Hence `f(x)` is a decreasing function.
Also`int_(0)^(1)f(x)dx=int_(0)^(1)(e^(x))/(2-e)dx`
`=[(e^(x))/(2-e)]_(0)^(1)`
`=(e-1)/(2-e) lt0`