Application Of Derivatives Ques 38
38. The set of all $x$ for which $\log (1+x) \leq x$ is equal to ….. .
(1987, 2M)
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Answer:
Correct Answer: 38.$x \in (- \frac{1}{2}, 0 ) \cup (\frac {1}{2}), x \in (- \infty , - \frac{1}{2}) \cup (0, \frac{1}{2}) $
Solution:
Formula:
Increasing and decreasing of a function:
- Let $f(x)=\log (1+x)-x$
| $\Rightarrow$ | $f^{\prime}(x)=\frac{1}{1+x}$ | $-\frac{x}{1+x}$ |
|---|---|---|
| $\Rightarrow$ | $f^{\prime}(x)>0$ | |
| when | $-1<x<0$ | |
| and | $f^{\prime}(x)<0$ | |
| when | $x>0$ |
$\therefore f(x)$ is increasing for $-1<x<0$.
$ \begin{array}{lrl} \Rightarrow & f(x) & <f(0) \\ \Rightarrow & \log (1+x) & <x \end{array} $
Again, $f(x)$ is decreasing for $x>0$.
$ \begin{array}{ll} \Rightarrow & f(x)<f(0) \\ \Rightarrow & \log (1+x)<x \\ \therefore & \log (1+x) \leq x, \forall x>-1 \end{array} $
BETA
