`f(x)` is defined for `xge0` and has a continuous derivative. It satisfies `f(0)=1,f^(')(0)=0` and `(1+f(x))f^(')(x)=1+x`. The value `f(1)` cannot be:

Options

A. a. `0`
B. b. `1.20`
C. c. `1.50`
D. d. `-2`

Answer

A, D

Explanation

`1+x` is never zero, so `1+f(x)` is never zero. It is `1` for `x=0`, so it is always positive `f(')(0)=0`. So `f^(')(x)gt0` for all `xgt0` and hence `f` is strictly increasing. So in particular, `1+f(x)ge2` for all `x`
Integrating `f^(')(x)lef^(')(0)+x/2+(x^(2))/4=x/2+(x^(2))/4`
Again integrating `f(x)lef(0)+(x^(2))/4+(x^(3))/12`
Hence `f(1)le1+1/4+1/1=4/3`