Applications Of Derivatives Question 40
Question: $ f(x)=\frac{\log (\pi +x)}{\log (e+x)} $ is
Options:
A) Increasing in $ [0,\infty ) $
B) Decreasing in $ [0,\infty ) $
C) Decreasing in $ [ 0,\frac{\pi }{e} ] $ & increasing in $ [ \frac{\pi }{e},\infty ] $
D) Increasing in $ [ 0,\frac{\pi }{e} ] $ & decreasing in $ [ \frac{\pi }{e},\infty ) $
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Answer:
Correct Answer: B
Solution:
[b] We have $ e<\pi $ and $ f’(x)=\frac{\frac{1}{\pi +x}\log (e+x)-\frac{1}{e+x}\log (\pi +x)}{{{{\log (e+x)}}^{2}}} $
$ =\frac{(e+x)\log (e+x)-(\pi +x)\log (\pi +x)}{(\pi +x)(e+x){{{\log (e+x)}}^{2}}} $ In $ [0,\infty ) $ , denominator > 0 and numerator < 0, Since, $ e+x<\pi +x $ .
Hence, f(x) is decreasing in $ [0,\infty ) $ .
BETA
