Question

The function of $x=\dfrac{\log \left( 1+x \right)}{x}\left( x>0 \right)$ is increasing in
a) $\left( 1,\infty \right)$
b) $\left( 0,\infty \right)$
c) $\left( 2,2e \right)$
d) $\dfrac{1}{e},2e$

Answer 1

Hint: Any function f(x) is increasing in any interval, if ${{f}^{'}}\left( x \right)>0$ for that interval, so find the derivative of the given function in the problem. Use division rule of differentiation $\left( \dfrac{u}{v} \right)$ rule given as

Complete step-by-step answer:

$\dfrac{d}{dx}\left( \dfrac{u}{v} \right)=\dfrac{v\dfrac{du}{dx}-u\dfrac{dv}{dx}}{{{v}^{2}}}$
i.e. derivative of f(x). Observe the terms ${{f}^{'}}\left( x \right)$ and relate it with the given domain for function (x > 0). Check the value of ${{f}^{'}}\left( x \right)$ for $x\to {{0}^{+}}$ and use the following results to get the answer:

$\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\log \left( 1+x \right)}{x}=1$


Now coming to the question, we are given a function as

$x-\dfrac{\log \left( 1+x \right)}{x}\Rightarrow x>0$

Hence, we need to determine the interval in which it will be increasing. So, let the given function is represented by f (x) so we get

$f\left( x \right)=x-\dfrac{\log \left( 1+x \right)}{x}...................\left( i \right)$

Where domain is defined as x > 0 i.e. $x\in \left( 0,\infty \right)$ . Now to get the interval where f (x) will be increasing, let us differentiate f (x) i.e. calculate ${{f}^{'}}\left( x \right)$ . So, we get

$\begin{align}

  & {{f}^{'}}\left( x \right)=\dfrac{d}{dx}\left( x-\dfrac{\log \left( 1+x \right)}{x} \right) \\

 & {{f}^{'}}\left( x \right)=\dfrac{dx}{dx}-\dfrac{d}{dx}\left( x-\dfrac{\log \left( 1+x \right)}{x} \right) \\

\end{align}$

We know the derivative of ${{x}^{n}}$ is given as

$\dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}}..........................\left( ii \right)$

Hence, we get ${{f}^{'}}\left( x \right)$ as

${{f}^{'}}\left( x \right)=1-\dfrac{d}{dx}\left( \dfrac{\log \left( 1+x \right)}{x} \right)...........................\left( iii \right)$

Now as the term $\dfrac{\log \left( 1+x \right)}{x}$ in the above equation is fraction, so we need to apply division rule of differentiation i.e. $\left( \dfrac{u}{v} \right)$ rule of derivative. So, we know division rule of derivative is defined as

$\dfrac{d}{dx}\left( \dfrac{u}{v} \right)=\dfrac{v\dfrac{du}{dx}-u\dfrac{dv}{dx}}{{{v}^{2}}}...............................\left( iv \right)$
Now, we can compare the term $\dfrac{d}{dx}\left( \dfrac{\log \left( 1+x \right)}{x} \right)$ from the equation (iii) and equation (iv) and suppose u = log (1 + x) and v = x. So, we get ${{f}^{'}}\left( x \right)$ as

${{f}^{1}}\left( x \right)=1-\dfrac{x\dfrac{d}{dx}\left( \log \left( 1+x \right) \right)-\log \left( 1+x \right)\dfrac{dx}{dx}}{{{x}^{2}}}$

Now, we know

$\dfrac{d}{dx}\log x=\dfrac{1}{x}......................\left( v \right)$

So, we get ${{f}^{'}}\left( x \right)$ as

$\begin{align}

  & {{f}^{1}}\left( x \right)=1-\dfrac{x\times \dfrac{1}{1+x}-\log \left( 1+x \right)\times 1}{{{x}^{2}}} \\

 & {{f}^{1}}\left( x \right)=1-\left[ \dfrac{x}{{{x}^{2}}\left( 1+x \right)}-\dfrac{\log \left( 1+x \right)}{{{x}^{2}}} \right] \\

 & {{f}^{1}}\left( x \right)=1+\dfrac{\log \left( 1+x \right)}{{{x}^{2}}}-\dfrac{1}{x\left( 1+x \right)}.........................\left( vi \right) \\

\end{align}$

Now, as we know any function is increasing if ${{f}^{1}}\left( x \right)>0$ for the given interval. It means we have to put ${{f}^{1}}\left( x \right)>0$ for getting the interval for x so that f (x) is increasing. So, write ${{f}^{1}}\left( x \right)$ as

${{f}^{1}}\left( x \right)>0.................\left( vii \right)$

Now, as the domain of the function f (x) is x > 0 or $x\in \left( 0,\infty \right)$ . It means we can put only positive values of x to the function f (x) and hence, to the ${{f}^{1}}\left( x \right)$ as well. So, we get domain of ${{f}^{1}}\left( x \right)$ as x > 0………………..(viii)

Now, observe the terms of the equation (vi). First term is ‘1’ which will always be positive for any x from the domain.

Second term $\dfrac{\log \left( 1+x \right)}{{{x}^{2}}}$ will also be positive for x > 0, as we know the range of log x for x > 0 is $\left( -\infty ,\infty \right)$ , where log x gives

$\left( -\infty ,0 \right)\to x\in \left( 0,1 \right),\left( 0,\infty \right)\to x>1$

Now we have to observe function log (1 + x) in place of log x for the same domain x > 0. As we can observe 1 + x will give $\left( 1,\infty \right)$ for x > 0 in the following way:

x > 0

Add 1 to both sides of the equation

1 + x > 1

Now as we can see log (1 + x) will receive values from $1\to \infty $ inside the bracket and hence it will not give any negative value (log x is an increasing function and log 1 = 0). Hence, log (1 + x) will also be a positive value.

And the denominator ${{x}^{2}}$ will always be a positive term as square of any real number will always be a positive term. It means $\dfrac{\log \left( 1+x \right)}{{{x}^{2}}}$ will be a positive term. Now coming to the last term i.e. $\dfrac{-1}{x\left( x+1 \right)}$ it will give a negative value for x > 0. And as the terms x (x + 1) are written in denominator so x (x + 1) will increase when x will increase from$0\to \infty $ , it means the whole terms $\dfrac{1}{x\left( x+1 \right)}$ will decrease and at $x\to \infty $ the term will become 0. So, let us observe the value of ${{f}^{1}}\left( x \right)$ at $x\to {{0}^{+}}$ (we cannot x = 0 as x = 0 is not considered in the domain).

So, let us calculate limit of ${{f}^{1}}\left( x \right)$ at $x\to {{0}^{+}}$ we get

$\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\left[ 1+\dfrac{\log \left( 1+x \right)}{{{x}^{2}}}-\dfrac{1}{x\left( 1+x \right)} \right]$

So, let us replace x by 0 + h, where $h\to 0$ . So, we get the above expression as

$\begin{align}

  & =\underset{h\to 0}{\mathop{\lim }}\,\left[ 1+\dfrac{\log \left( 1+0+h \right)}{{{\left( 0+h \right)}^{2}}}-\dfrac{1}{\left( 0+h \right)\left( 1+0+h \right)} \right] \\

 & =\underset{h\to 0}{\mathop{\lim }}\,\left[ 1+\dfrac{\log \left( 1+h \right)}{{{h}^{2}}}-\dfrac{1}{h\left( 1+h \right)} \right] \\

 & =\underset{h\to 0}{\mathop{\lim }}\,1+=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{1}{h}\dfrac{\log \left( 1+h \right)}{h}-=\underset{h\to 0}{\mathop{\lim }}\,\dfrac{1}{h\left( h+1 \right)}..........................\left( \text{ix} \right) \\

\end{align}$

As, we know the value of $\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\log \left( 1+x \right)}{x}$ is given as

$\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\log \left( 1+x \right)}{x}=1............................\left( \text{x} \right)$

So, we can get the equation (ix) with the help of equation (x) as

$\begin{align}

  & =\underset{h\to 0}{\mathop{\lim }}\,1+\underset{h\to 0}{\mathop{\lim }}\,\dfrac{1}{h}\times 1-\underset{h\to 0}{\mathop{\lim }}\,\dfrac{1}{h\left( h+1 \right)} \\
 & =1+\underset{h\to 0}{\mathop{\lim }}\,\left[ \dfrac{1}{h}-\dfrac{1}{h\left( h+1 \right)} \right] \\

 & =1+\underset{h\to 0}{\mathop{\lim }}\,\left[ \dfrac{h+1-1}{h\left( h+1 \right)} \right] \\

 & =1+\underset{h\to 0}{\mathop{\lim }}\,\dfrac{1}{1+h} \\

 & =1+1 \\

 & =2 \\

\end{align}$

So, we get

$\lim x\to {{0}^{+}}{{f}^{'}}\left( x \right)=2$

So, we get that ${{f}^{'}}\left( x \right)$ will give 2 for $x\to {{0}^{+}}$ , now a per the terms of ${{f}^{'}}\left( x \right)$ the negative term $\dfrac{-1}{x\left( x+1 \right)}$ will decrease when x will increase from $0\to \infty $ . it means ${{f}^{'}}\left( x \right)$ will increase for the interval $\left( 0,\infty \right)$ and as $\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,{{f}^{'}}\left( x \right)=2$ i.e. greater than 0. Hence ${{f}^{'}}\left( x \right)$ will always be greater than 0 for $x\in \left( 0,\infty \right)$ . Hence the function $x-\dfrac{\log \left( 1+x \right)}{x}$ will increase for $x\in \left( 0,\infty \right)$ for the domain $x\in \left( 0,\infty \right)$ .

So, option (b) is correct.

Note:

$\underset{x\to 0}{\mathop{\lim }}\,\dfrac{\log \left( 1+x \right)}{x}=1$

Can be proved as, put the expression of log (1 + x) as

$\begin{align}

  & \log \left( 1+x \right)=x-\dfrac{{{x}^{2}}}{2}+\dfrac{{{x}^{3}}}{3}-\dfrac{{{x}^{4}}}{4}+.................. \\

 & \underset{x\to 0}{\mathop{\lim }}\,\dfrac{x-\dfrac{{{x}^{2}}}{2}+\dfrac{{{x}^{3}}}{3}-\dfrac{{{x}^{4}}}{4}+.................}{x} \\

 & \underset{x\to 0}{\mathop{\lim }}\,1-\dfrac{x}{2}+\dfrac{{{x}^{2}}}{3}-\dfrac{{{x}^{3}}}{4}+................. \\

 & =1-0=1 \\

\end{align}$

Don’t confuse with the term

$\dfrac{\log \left( 1+x \right)}{{{x}^{2}}}$

We can observe that log (1 + x) will be greater than 0 for x > 0. So, don’t confuse this part of the problem.