Question

For all$x\in (0,1)$

• A. $e^x<1+x$
• B. ${\log }_e\left(1+x\right)<x$
• C. $\sin x>x$
• D. ${\log }_e\left(x\right)>x$

Answer 1

The correct option is B

${\log }_e\left(1+x\right)<x$

Explanation for the correct option:

Take option $\mathbf{\left(}\mathit{B}\mathbf{\right)}\mathbf{}\mathit{l}\mathit{o}{\mathit{g}}_{\mathbf{e}}\mathbf{(}\mathbf{1}\mathbf{+}\mathit{x}\mathbf{)}\mathbf{<}\mathit{x}$

We will take$\text{'}e\text{'}$ base both the sides,

$\begin{array}{rcl}{e}^{{\log }_e(1+x)}& <& e^x\\ & \Rightarrow & 1+x<e^x\end{array}$

For all the value of$x\in (0,1)$.

$1+x<e^x$ is true, as we see in the above option.

Hence, option $\mathbf{\left(}\mathit{B}\mathbf{\right)}$ is the correct option.

Explanation for incorrect options:

$\mathbf{\left(}\mathit{A}\mathbf{\right)}\mathbf{}{\mathit{e}}^{\mathbf{x}}\mathbf{<}\mathbf{1}\mathbf{+}\mathit{x}$

we can check by putting the value of$x\in (0,1)$.

$$\begin{gathered} \Rightarrow e^x<1+x \\ \Rightarrow x=\frac{1}{2}, \\ \Rightarrow e^x=\sqrt{e}=1.65 \\ \Rightarrow 1+x=1+\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.=1.5 \end{gathered}$$

from above, its clear that$e^x>1+x$.

Therefore, option$\mathbf{\left(}\mathit{A}\mathbf{\right)}$ is not correct option.

$\mathbf{\left(}\mathit{C}\mathbf{\right)}\mathbf{}\mathit{s}\mathit{i}\mathit{n}\mathit{x}\mathbf{>}\mathit{x}$

we have $x\in (0,1)$.

for$x=\frac{\mathrm{\pi }}{4}$

$$\begin{gathered} \Rightarrow \sin \frac{\mathrm{\pi }}{4}=\frac{1}{\sqrt{2}}\approx \frac{5}{7} \\ \Rightarrow \frac{\mathrm{\pi }}{4}=\frac{22}{28}=\frac{5.5}{7} \end{gathered}$$

from above, we can see$\sin x<x$.

Therefore, option$\mathbf{\left(}\mathit{C}\mathbf{\right)}\mathbf{}\mathit{s}\mathit{i}\mathit{n}\mathit{x}\mathbf{>}\mathit{x}$ is not correct.

$\mathbf{\left(}\mathit{D}\mathbf{\right)}\mathbf{}\mathit{l}\mathit{o}{\mathit{g}}_{\mathbf{e}}\mathit{x}\mathbf{>}\mathit{x}$

we can check by putting the value of$x\in (0,1)$.

for$x=\frac{1}{3}$,

$\begin{array}{rcl}{\log }_e\frac{1}{3}& <& 0\\ \frac{1}{3}& >& 0\end{array}$

from above, we can see ${\log }_{e}x<x$,

Therefore, option $\mathbf{\left(}\mathit{D}\mathbf{\right)}\mathbf{}\mathit{l}\mathit{o}{\mathit{g}}_{\mathbf{e}}\mathit{x}\mathbf{>}\mathit{x}$ is not correct.