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Question: The function \[f(x) = x - \dfrac{{\log (1 + x)}}{x}\left( {x > 0} \right)\] is increasing in: A. \...

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

Explanation

Solution

For a function to be increasing the derivative of the function with respect to the variable must be positive for all values in the domain. In simple words we can say that a function is increasing if f′(x)>0f' (x) > 0.

Complete answer:
We are given the function f(x)=x−log⁡(1+x)x(x>0)f(x) = x - \dfrac{{\log (1 + x)}}{x}\left( {x > 0} \right) . The domain of the function is for all values greater than 0.
Since, for a function to be increasing f′(x)>0f'(x) > 0. Therefore, we will start by differentiating the function with respect to x. We have,
f(x)=x−log⁡(1+x)xf(x) = x - \dfrac{{\log (1 + x)}}{x}

Differentiating the function with respect to x we get,
f′(x)=1−xddxlog⁡(1+x)−log⁡(1+x)ddx(x)x2f'(x) = 1 - \dfrac{{x\dfrac{d}{{dx}}\log (1 + x) - \log (1 + x)\dfrac{d}{{dx}}(x)}}{{{x^2}}}
Since, d[log⁡(1+x)]dx=11+x  \begin{gathered} \dfrac{{d[\log (1 + x)]}}{{dx}} = \dfrac{1}{{1 + x}} \\\ \\\ \end{gathered} we get,
f′(x)=1−x×11+x−log⁡(1+x)×(1)x2f'(x) = 1 - \dfrac{{x \times \dfrac{1}{{1 + x}} - \log (1 + x) \times (1)}}{{{x^2}}}

Simplifying the above equation we get,
f′(x)=1−1x(1+x)+log⁡(1+x)x2f'(x) = 1 - \dfrac{1}{{x(1 + x)}} + \dfrac{{\log (1 + x)}}{{{x^2}}}
Rearranging the above equation, we get,
f′(x)=1+log⁡(1+x)x2−1x(1+x)f'(x) = 1 + \dfrac{{\log (1 + x)}}{{{x^2}}} - \dfrac{1}{{x(1 + x)}}

Since, we have to find the values for x for which the function is increasing therefore the derivative of the function must be positive. Therefore,
1+log⁡(1+x)x2−1x(1+x)>01 + \dfrac{{\log (1 + x)}}{{{x^2}}} - \dfrac{1}{{x(1 + x)}} > 0

Now we are given that x is positive.
For x positive log⁡(1+x)x2\dfrac{{\log (1 + x)}}{{{x^2}}} is always positive since, numerator and denominator are both positive for x>0x > 0.
Similarly, 1x(1+x)\dfrac{1}{{x(1 + x)}} is also always positive for x>0x > 0. Since, the denominator is always positive.

Now for entire function to be positive value of 1+log⁡(1+x)x21 + \dfrac{{\log (1 + x)}}{{{x^2}}} must be always greater than the value of 1x(1+x)\dfrac{1}{{x(1 + x)}}.
Let us find the values of this derivative for h→0+h \to {0^ + }. We have,
f′(h)h→0+=1+log⁡(1+h)h2−1h(1+h)f'{(h)_{h \to {0^ + }}} = 1 + \dfrac{{\log (1 + h)}}{{{h^2}}} - \dfrac{1}{{h(1 + h)}} … (1)

We know that for h→0+h \to {0^ + } , log⁡(1+h)h=1\dfrac{{\log (1 + h)}}{h} = 1. Therefore applying this property in the given function we get,
f′(h)h→0+=1+1h−1h(1+h)f'{(h)_{h \to {0^ + }}} = 1 + \dfrac{1}{h} - \dfrac{1}{{h(1 + h)}}
Taking L.C.M. we get,
f′(h)h→0+=1+(1+h)−1h(1+h)f'{(h)_{h \to {0^ + }}} = 1 + \dfrac{{(1 + h) - 1}}{{h(1 + h)}}
f′(h)h→0+=1+hh(1+h)f'{(h)_{h \to {0^ + }}} = 1 + \dfrac{h}{{h(1 + h)}}

Cancelling the common h from denominator and numerator, we get,
f′(h)h→0+=1+1(1+h)f'{(h)_{h \to {0^ + }}} = 1 + \dfrac{1}{{(1 + h)}}
Putting h→0+h \to {0^ + }, we get the value of f′(h)h→0+f'{(h)_{h \to {0^ + }}} as a positive value.
Therefore, it is true for all x>0x > 0.

Hence, the correct answer is option (B).

Note: While calculating for the values of x for increasing or decreasing function one thing to be noted is that the values obtained after putting the derivative positive or negative one must check whether for those values the function exists or not.