`f(x)`is defined for `xge0` & has a continuous derivative. It satisfies `f(0)=1, f^(')(0)=0` & `(1+f(x))f^('')(x)=1+x`. The values `f(1)` can't take is/are

Options

A. `2`
B. `1.75`
C. `1.5`
D. `1.35`

Answer

A, B, C, 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. Hence `f^('')(x)` is always positive `f^(')(0)=0` so `f^(')(x)gt0, AAxgt0`
Hence so in particular `1+f(x)ge2, AAx`
`f^('')(x) le (1+x)/2`
Integrating `f^(')(x)le f^(')+x/2+(x^(2))/4=x/2+(x^(2))/4`
Agai integrating
`f(x)lef(0)+(x^(2))/4+(x^(3))/12`
Hence `f(1)le 1+ 1/4+1/12=4/3`