Question
For all $$x \in \left( {0,1} \right)$$
A.
$${e^x} < 1 + x$$
B.
$${\log _e}\left( {1 + x} \right) < x$$
C.
$$\sin x > x$$
D.
$${\log _e}x > x$$
Answer :
$${\log _e}\left( {1 + x} \right) < x$$
Solution :
$$\eqalign{
& {\text{Let}}\,f\left( x \right) = {e^x} - 1 - x\,{\text{then}}\,f'\left( x \right) = {e^x} - 1 > 0\,{\text{for}}\,x \in \left( {0,1} \right) \cr
& \therefore f\left( x \right)\,{\text{is}}\,{\text{an increasing function}}{\text{.}} \cr
& \therefore f\left( x \right) > f\left( 0 \right),\forall \,x\, \in \left( {0,1} \right) \cr
& \Rightarrow {e^x} - 1 - x > 0 \Rightarrow {e^x} > 1 + x \cr
& \therefore \left( {\text{a}} \right)\,{\text{does not hold}}. \cr
& \left( {\text{b}} \right)\,{\text{Let}}\,g\left( x \right) = \log \left( {1 + x} \right) - x \cr
& {\text{then}}\,g'\left( x \right) = \frac{1}{{1 + x}} - 1 = - \frac{x}{{1 + x}} < 0,\forall \,x \in \left( {0,1} \right) \cr
& \therefore g\left( x \right)\,{\text{is decreasing on}}\,\left( {0,1} \right)\,\therefore x > 0 \cr
& \Rightarrow g\left( x \right) < g\left( 0 \right) \cr
& \Rightarrow \log \left( {1 + x} \right) - x < 0 \Rightarrow \log \left( {1 + x} \right) < x \cr
& \therefore \left( {\text{b}} \right){\text{ holds}}{\text{. Similarly it can be shown that}}\,\left( {\text{c}} \right)\,{\text{and}}\,\left( {\text{d}} \right)\,{\text{do}}\,{\text{not hold}}{\text{.}} \cr} $$