The function f(x) = ln ( + x) ln (e + x) is - Vidyadip

MCQ

The function $f(x) = {{{\rm{ln}}(\pi + x)} \over {{\rm{ln}}(e + x)}}$ is

A
Increasing on $\left[ {0,\,\infty } \right)$
✓
Decreasing on $\left[ {0,\,\infty } \right)$
C
Decreasing on $\left[ {0,{\pi \over e}} \right)$ and increasing on $\left[ {{\pi \over e},\infty } \right)$
D
Increasing on $\left[ {0,{\pi \over e}} \right)$ and decreasing on $\left[ {{\pi \over e},\infty } \right)$

✓

Answer

Correct option: B.

Decreasing on $\left[ {0,\,\infty } \right)$

b
(b) Let $f(x) = \frac{{\ln (\pi + x)}}{{\ln (e + x)}}$

$\therefore f'(x) = \frac{{\ln (e + x) \times \frac{1}{{\pi + x}} - \ln (\pi + x)\frac{1}{{e + x}}}}{{{{\ln }^2}(e + x)}}$

$ = \frac{{(e + x)\ln (e + x) - (\pi + x)\ln (\pi + x)}}{{{{\ln }^2}(e + x) \times (e + x)(\pi + x)}}$

$ \Rightarrow f'(x) < 0$ for all

Hence $f(x)$ is decreasing in $[0,\infty )$.

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