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

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.