d
(d) \({\log _e}(1 + x) - x = {\log _e}(1 + x) - {\log _e}{e^x} = {\log _e}{{1 + x} \over {{e^x}}}\)

\( = \ln {{1 + x} \over {1 + x + {{{x^2}} \over {2\,!}} + {{{x^3}} \over {3\,!}} + ...}} < {\rm{ }}0,\,{\rm{as }}1 + x < 1 + x + {{{x^2}} \over {2\,!}} + .... + \)

\(\therefore {\log _e}(1 + x) < x\), for \(x > 0\).

\({x \over {1 + x}} - \log (1 + x) = 1 - {1 \over {1 + x}} - \log (1 + x)\)

= \(1 - \left[ {{1 \over {1 + x}} + \log (1 + x)} \right]\, < 0\), for \(x > 0\)

\(\therefore {x \over {1 + x}} < \log (1 + x)\), \(\therefore (b)\) is true

\({e^x} - (1 + x) = 1 + x + {{{x^2}} \over {2\,!}} + {{{x^3}} \over {3\,!}} + ..... - (1 + x)\)

= \({{{x^2}} \over {2\,!}} + {{{x^3}} \over {3\,!}} + ..... > 0\), for \(x > 0\)

\(\therefore {e^x} > 1 + x\), for \(x > 0\); \(\therefore (c)\) is true

\({e^x} - (1 - x) = 1 + x + {{{x^2}} \over {2\,!}} + ...... - 1 + x\)

= \(2x + {{{x^2}} \over {2\,!}} + {{{x^3}} \over {3\,!}} + ....... > 0\), for \(x > 0\)

\(\therefore {e^x} > 1 - x\), for \(x > 0\)

Thus, \({e^x} < (1 - x),\) for \(x > 0\) is not true.