Question:
For all $x\,\in\,(0,1)$
For all $x\,\in\,(0,1)$
Updated On: Jun 14, 2022
- $e^x\,
- $\log_e (1 + x) < x $
- $\sin x > x$
- $ \log_e x > x $
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The Correct Option is B
Solution and Explanation
Let $f\left(x\right) = e^{x} - 1 - x $ then $f'\left(x\right) =e^{x } - 1 >0 $ for $x \in\left(0,1\right) $
$\therefore \, f(x)$ is an increasing function.
$\therefore \, f(x) > f(0) , \forall x \in (0,1)$
$\Rightarrow \, e^x - 1 - x > 0 \Rightarrow \, e^x > 1 + x$
$\therefore$ (a) does not hold.
(b) Let $g(x) = log (1+x) -x $
then $g'(x) = \frac{1}{1+x} - 1 = - \frac{x}{1+x} < 0 , \forall x \in (0,1)$
$\therefore \, g(x) $ is decreasing on $(0, 1) \therefore x > 0 $
$\Rightarrow \, g (x) < g (0)$
$\Rightarrow \, \log_e (1 + x) - x < 0 \, \Rightarrow \, \log_e (1 + x) < x$
$\therefore \, f(x)$ is an increasing function.
$\therefore \, f(x) > f(0) , \forall x \in (0,1)$
$\Rightarrow \, e^x - 1 - x > 0 \Rightarrow \, e^x > 1 + x$
$\therefore$ (a) does not hold.
(b) Let $g(x) = log (1+x) -x $
then $g'(x) = \frac{1}{1+x} - 1 = - \frac{x}{1+x} < 0 , \forall x \in (0,1)$
$\therefore \, g(x) $ is decreasing on $(0, 1) \therefore x > 0 $
$\Rightarrow \, g (x) < g (0)$
$\Rightarrow \, \log_e (1 + x) - x < 0 \, \Rightarrow \, \log_e (1 + x) < x$
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Application of Derivatives
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Increasing and Decreasing Function:
Consider y = f(x) be a differentiable function (whose derivative exists at all points in the domain) in an interval x = (a,b).
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