MCQ
Which of the following is not true
- A$\log (1 + x) < x\,\,\,{\rm{for}}\,\,x > 0$
- B${x \over {1 + x}} < \log (1 + x)\,\,{\rm{for}}\,\,x > {\rm{0}}$
- C${e^x} > 1 + x\,\,{\rm{for}}\,\,x > 0$
- ✓${e^x} < 1 - x\,\,{\rm{for }}\,x > {\rm{0}}$
$ = \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.
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$(A)$ $\quad a=2$ $(B)$ $a=1$ $(C)$ $L=\frac{1}{64}$ $(D)$ $L=\frac{1}{32}$