Question

The function $ f(n) = \dfrac{{\log (\pi + x)}}{{\log (e + x)}} $ is,
A.Increasing on [1,∞]
B.Decreasing on [0,∞]
C.Increasing on \[\left[ {0,\dfrac{\pi }{e}} \right]\], decreasing on \[\left[ {\dfrac{\pi }{e},0} \right]\]
D.Decreasing on \[\left[ {0,\dfrac{\pi }{e}} \right]\], increasing on \[\left[ {\dfrac{\pi }{e},0} \right]\]

Answer 1

Hint: We will carry out differentiation [f’(x)] of the given function . The function will be increasing at points where f’(x ) > 0 and decreasing where f’(x)<0.

Complete step-by-step answer:
Differentiating the given expression, we get:
 $ f'(n) = \dfrac{{\dfrac{{\log (e + x)}}{{\pi + x}} - \dfrac{{\log (\pi + x)}}{{e + x}}}}{{{{[\log (e + x)]}^2}}} $
[Using: $ \dfrac{d}{{dx}}\left( {\dfrac{u}{v}} \right) = \dfrac{{v\dfrac{{du}}{{dx}} - u\dfrac{{dv}}{{dx}}}}{{{v^2}}} $
Here, $ u = \log (\pi + x){\text{ and }}v = \log (e + x) $ ]
\[For{\text{ }}x > 0:\]
\[\pi + x > e + x \to (1)\] \[\left[ values: \pi = 3.14, e = 2.67 \right]\]
Taking reciprocals both side:
 $ \dfrac{1}{{\pi + x}} < \dfrac{1}{{e + x}} \to (2) $ (Inequality sign changes when reciprocal is taken)
Taking log both sides in (1):
 $ \log (\pi + x) > \log (e + x) \to (3) $
Multiplying (3) with (2), we get:
 $ \dfrac{{\log (\pi + x)}}{{e + x}} > \dfrac{{\log (e + x)}}{{\pi + x}} $
Therefore, f’(x) <0
Thus, the answer is B), the function is decreasing for the interval [0,∞].
So, the correct answer is “Option B”.

Note: Apply differentiation formulas according to the need and be careful while differentiating.Along with the application of formulas, the product of differentiation w.r.t. x is also considered.