Question

The function \[f(x) = \dfrac{{\ln \left[ {\pi + x} \right]}}{{\ln (e + x)}}\] is
A. Increasing on \[\left[ {0,\infty } \right]\]
B. Decreasing on \[\left[ {0,\infty } \right)\]
C. Decreasing on \[\left[ {0,\dfrac{\pi }{e}} \right)\] and Increasing on \[\left[ {\dfrac{\pi }{e},\infty } \right)\]
D. Increasing on \[\left[ {0,\dfrac{\pi }{e}} \right)\] and decreasing on \[\left[ {\dfrac{\pi }{e},\infty } \right)\]

Answer 1

Hint: Find the derivative of the function as it can be used to determine whether the function is increasing or decreasing on any intervals in its domain.. If \[f'(x) > 0\] at each point in an interval I, then the function is said to be increasing on I. \[f'(x) < 0\] at each point in an interval I, then the function is said to be decreasing on I.

Complete step-by-step answer:
The derivative of a function of a real variable measures the sensitivity to change of the function value with respect to a change in its argument. Derivatives are a fundamental tool of calculus.
If \[f'(x) > 0\] then \[f\] is increasing on the interval, and if \[f'(x) < 0\] then \[f\] is decreasing on the interval.
We are given the function \[f(x) = \dfrac{{\ln \left[ {\pi + x} \right]}}{{\ln (e + x)}}\]
Differentiating both side with respect to \[x\] we get ,
\[f'(x) = \dfrac{{\dfrac{1}{{\pi + x}}\ln (e + x) - \dfrac{1}{{e + x}}\ln \left( {\pi + x} \right)}}{{{{\left( {\ln (e + x)} \right)}^2}}}\]
We will divide \[f'(x)\] into different functions so as to ease our calculation.
Let \[h(x) = \dfrac{1}{{\pi + x}}\ln (e + x) - \dfrac{1}{{e + x}}\ln \left( {\pi + x} \right)\]
Let us consider \[g(x) = x\ln x\]
Differentiating both side with respect to \[x\] we get ,
\[g'(x) = x \times \dfrac{1}{x} + \ln x = 1 + \ln x\]
We know that \[g'(x) > 0\; \forall \;x \in \left( {\dfrac{1}{e},\infty } \right)\] and \[g'(x) < 0\forall x \in \left( {0,\dfrac{1}{e}} \right)\]
Now we know that \[e < \pi \]
Hence \[\forall x \in \left( {0,\infty } \right)\] \[e + x < \pi + x\]
And since \[g(x)\] is an increasing function for \[x > \dfrac{1}{e}\] . Therefore we have \[g\left( {e + x} \right) < g\left( {\pi + x} \right)\]
Since we have assumed \[g(x) = x\ln x\]
Therefore we get \[\left( {e + x} \right)\ln \left( {e + x} \right) < \left( {\pi + x} \right)\ln \left( {\pi + x} \right)\]
On rearranging the terms we get ,
\[\dfrac{{\ln \left( {e + x} \right)}}{{\left( {\pi + x} \right)}} < \dfrac{{\ln \left( {\pi + x} \right)}}{{\left( {e + x} \right)}}\]
Taking all the terms on one side we get ,
\[\dfrac{{\ln \left( {e + x} \right)}}{{\left( {\pi + x} \right)}} - \dfrac{{\ln \left( {\pi + x} \right)}}{{\left( {e + x} \right)}} < 0\]
Therefore we get \[h(x) < 0\]
Therefore \[f'(x) = \dfrac{{h(x)}}{{{{\left( {\ln (e + x)} \right)}^2}}} < 0\] \[\forall x \in \left[ {0,\infty } \right)\]
Therefore \[f(x)\] decreases for \[\left[ {0,\infty } \right)\] .
So, the correct answer is “Option B”.

Note: In determining intervals where a function is increasing or decreasing, you first find domain values where all critical points will occur; then, test all intervals in the domain of the function of these values to determine if the derivative is positive or negative.