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f(x) is defined for xge0 and has a conti...

`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:

A

`2`

B

`1.75`

C

`1.5`

D

`1.35`

Text Solution

Verified by Experts

The correct Answer is:
A, B, C, D
`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`
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