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

Options

A. `e^(x) lt 1+x`
B. `log_e (1+x) lt x `
C. `sin x gt x `
D. `log_(epsi) x gt x `

Answer

C

Explanation

Option (a) Let `f(x) = e^(x) - 1 - x `
then ` f '(x) = e ^(x) - 1 gt 0, AA x in (0, 1)`
`rArr f(x) ` increase in `(0, 1)`
`rArr f(x) gt f(0) ` for ` 0 lt x lt 1 `
`rArr e^(x) - 1 - x gt 0 or e ^(x) gt 1 + x ` for ` 0 lt x lt 1 `
Option (b) Let `g(x) = log_e (1 + x) - x , 0 lt x lt 1 `
`g ' (x) = (1)/( 1+ x) - 1 = - (x) /( 1 + x ) lt 0 `for ` 0 lt x lt 1 `
`rArr g (x)` decreases for ` 0 lt x lt 1 `
`rArr g (x) lt g(0) ` for ` 0 lt x lt 1 `
`rArr log _e ( 1 + x) - x lt 0` for ` 0 lt x lt 1 `
or `log _e (1 + x ) lt x ` for ` 0 lt x lt 1 `
Therefore, Option (b) is the answer.
Option `sin x gt x `
Let `h(x) = sin x - x `
` h'(x) = cos x - 1 `
For ` x in (0, 1), cos x - 1 lt 0`
`rArr h (x) ` is decreasing function.
`rArr " " h (x) lt h (0)`
`rArr " " sin x - x lt 0`
`rArr sin x lt ` which is not true.
Option (d) `p (x) = log x - x `
`p' (x) = (1)/(x) - 1 gt 0, AA x in (0, 1)`
Therefore, `p'(x)` is an increasing function.
`rArr p (0) lt p (x) lt p (1)`
`rArr -oo lt logx - x lt - 1 `
`rArr log x - x lt 0`
`rArr log x lt x `
Therefore, option (d) is not the answer.