For all `x epsilon(0,1)`

Options

A. `e^(x)lt1+x`
B. `log_(e)(1+x)ltx`
C. `sin x gt x`
D. `log_(e)x gt x-1`

Answer

B

Explanation

Let `f(x)=e^(x)-(1+x)`
`impliesf'(x)=e^(x)-1gt0 [ :' 0 lt x lt1]`
`:.f(x)` is increasing
`implies f(x)gtf(0)`
`impliese^(x)-(1+x)gt0impliese^(x)gt1+x [ :' x gt0]`
b. Now let `g(x)=log_(e)(1+x)-x`
`impliesg'(x)=1/(1+x)-1=-x/(1+x)lt0`
`:.g(x)` is decreasing
when `xgt0impliesg(x)ltg(0)`
`implieslog_(e)(1+x)-x lt0`
`implies log_(e)(1+x)ltx`
c. Again let `h(x)=sinx-ximpliesh'(x)=cos x -1lt0`
`h(x)` is decreasing
`:' xgt0impliesh(x)lth(0)impliessin x-x lt 0 impliessin x ltx`
d. Let `t(x)=log_(e)x-x+1`
`impliest'(x)=1/x-1=(1-x)/xgt0`
`:.t(x)` is increasing
`x lt 1=t(x)lt t (1) implieslog_(e)x=xlt-1`
`implieslog_(e)x lt x-1`