The function f(x) = frac{ln(pi+x)}{ln(e+x)} i | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let $f(x) = \frac{\ln(\pi+x)}{\ln(e+x)}$.Using the quotient rule,$f'(x) = \frac{\ln(e+x) \cdot \frac{1}{\pi+x} - \ln(\pi+x) \cdot \frac{1}{e+x}}{[

Vedclass · Accepted Answer

decreasing on $(0, \infty)$.

Vedclass · Answer

(C) Let $f(x) = \frac{\ln(\pi+x)}{\ln(e+x)}$.Using the quotient rule,$f'(x) = \frac{\ln(e+x) \cdot \frac{1}{\pi+x} - \ln(\pi+x) \cdot \frac{1}{e+x}}{[\ln(e+x)]^2}$.Simplifying the numerator,we get $f'(x) = \frac{(e+x)\ln(e+x) - (\pi+x)\ln(\pi+x)}{(\pi+x)(e+x)[\ln(e+x)]^2}$.Consider the function $g(t) = t \ln(t)$. Its derivative is $g'(t) = 1 + \ln(t)$. For $t > 1$,$g'(t) > 0$,so $g(t)$ is an increasing function.Since $\pi > e > 1$,for $x > 0$,we have $\pi+x > e+x > e > 1$.Because $g(t)$ is increasing,$g(\pi+x) > g(e+x)$,which implies $(\pi+x)\ln(\pi+x) > (e+x)\ln(e+x)$.Therefore,$(e+x)\ln(e+x) - (\pi+x)\ln(\pi+x) Since the denominator is always positive for $x > 0$,$f'(x) Thus,$f(x)$ is decreasing on $(0, \infty)$.

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

Similar Questions

If $f(x) = \frac{\lambda \sin x + 6 \cos x}{2 \sin x + 3 \cos x}$ is a strictly increasing function,then .......

The function $f(x) = \log(1+x) - \frac{2x}{2+x}$ is increasing on

If a function $f(x)$ is increasing in an interval $x \in [a, b]$,then which of the following will always be correct?

The number of real solutions of $x^{7}+5x^{3}+3x+1=0$ is equal to............

Which of the following functions is a strictly increasing function?

Vedclass Test Series

Exam Paper Generator

Online Exam Module