Question

The function $x-\dfrac{\log \left( 1+x \right)}{x}\left( x>0 \right)$ is increasing in
A. $\left( 1,\infty \right)$
B. $\left( 0,\infty \right)$
C. $\left( 2,2e \right)$
D. $\left( \dfrac{1}{e},2e \right)$

Answer 1

Hint : We first try to describe the relation between the slope of the curve and the characteristics of it being increasing. We find the differentiation of the curve or function by taking its slope form by differentiating it. Depending on the value of slope we get the characteristics of the function.

Complete step by step solution:
We first take the given function as $f\left( x \right)=x-\dfrac{\log \left( 1+x \right)}{x}$ .
We take differentiation of the function and find the slope of the function.
So, $\dfrac{df}{dx}={{f}^{'}}\left( x \right)$ is the slope of the function.
Now, if the slope at any fixed point is negative which means $\dfrac{df}{dx}<0$ then the function is decreasing and if $\dfrac{df}{dx}>0$ then the function is increasing.
If the changes for the whole curve happens very rapidly then the function is not monotone.
For our given function we find the slope of $f\left( x \right)=x-\dfrac{\log \left( 1+x \right)}{x}$ .
We find the slope of the function by taking $\dfrac{df}{dx}=\dfrac{d}{dx}\left[ x-\dfrac{\log \left( 1+x \right)}{x} \right]$ .
We have $\dfrac{d}{dx}\left[ x-\dfrac{\log \left( 1+x \right)}{x} \right]=1-\dfrac{\dfrac{x}{1+x}-\log \left( 1+x \right)}{{{x}^{2}}}$.
Now for the function $f\left( x \right)=x-\dfrac{\log \left( 1+x \right)}{x}$ to be increasing, $\dfrac{df}{dx}>0$ .
So, $1-\dfrac{\dfrac{x}{1+x}-\log \left( 1+x \right)}{{{x}^{2}}}>0$ which gives $\dfrac{\dfrac{x}{1+x}-\log \left( 1+x \right)}{{{x}^{2}}}<1$.
We simplify to get

& \dfrac{\dfrac{x}{1+x}-\log \left( 1+x \right)}{{{x}^{2}}}<1 \\\ & \Rightarrow \dfrac{x}{1+x}-\log \left( 1+x \right)<{{x}^{2}} \\\ & \Rightarrow \log \left( 1+x \right)>\dfrac{x}{1+x}-{{x}^{2}} \\\ \end{aligned}$$ Now we value $ x=0 $ . We take the function $$\log \left( 1+x \right)-\dfrac{x}{1+x}+{{x}^{2}}$$ and apply limit tending to 0. So, $ \underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\left[ \log \left( 1+x \right)-\dfrac{x}{1+x}+{{x}^{2}} \right]=0 $ . We again take $$G\left( x \right)=\log \left( 1+x \right)-\dfrac{x}{1+x}+{{x}^{2}}$$. Then $${{G}^{'}}\left( x \right)=\dfrac{1}{1+x}-\dfrac{1}{{{\left( 1+x \right)}^{2}}}+2x=\dfrac{x}{{{\left( 1+x \right)}^{2}}}+2x$$. Given that $ \left( x>0 \right) $ which gives $${{G}^{'}}\left( x \right)=\dfrac{x}{{{\left( 1+x \right)}^{2}}}+2x>0$$ for $ \left( x>0 \right) $ . The function is increasing for $ \left( x>0 \right) $ . The domain for increasing function will be $ \left( 0,\infty \right) $ The correct option B. **So, the correct answer is “Option B”.** **Note** : We can also find the value of $ x $ for which if we get $ {{x}_{1}}>{{x}_{2}} $ and $ f\left( {{x}_{1}} \right)>f\left( {{x}_{2}} \right) $ , the curve is increasing. If we find $ {{x}_{1}}<{{x}_{2}} $ and $ f\left( {{x}_{1}} \right)>f\left( {{x}_{2}} \right) $ , the curve is decreasing. The change of values is equal to the slope.