For all `x in (0,1)` `e^x<1+x` (b) `(log)_e(1+x) x` (d) `(log)_e x > x`

Options

A. `e^(x)lt1+x`
B. `loge^(1+x)ltx`
C. `sinxgtx`
D. `log_(e)xgtx`

Answer

2

Explanation

`y=e^(x) or (dy)/(dx)=e^(x)`
Then equation of the tangent at x=0 is
y-1=1(x-0)or y=x+1

Graph of `y=e^(x)` always leies above the grph of y =1 1+x
Hence `e^(x)gt1+x or x gt log_(e)(1+x)` .Hence (b) is true
option (3) is wrogns as sinx `lt x for x in (0,1)`
option (4) is wrong as `xgt log_(e)` for in (0,1)