Applications Of Derivatives Question 419
Question: The function $ f(x)=\frac{ln(\pi +x)}{ln(e+x)} $ is
[IIT 1995]
Options:
A) Increasing on $ [ 0,,\infty ) $
B) Decreasing on $ [ 0,,\infty ) $
C) Decreasing on $ [ 0,\frac{\pi }{e} ) $ and increasing on $ [ \frac{\pi }{e},\infty ) $
D) Increasing on $ [ 0,\frac{\pi }{e} ) $ and decreasing on $ [ \frac{\pi }{e},\infty ) $
Show Answer
Answer:
Correct Answer: B
Solution:
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 $ x\ge 0\ ,\ \ {\because \pi >e} $
Hence $ f(x) $ is decreasing in $ [0,\infty ) $ .
BETA
