A) Increasing in \[[0,\infty )\]
B) Decreasing in \[[0,\infty )\]
C) Decreasing in \[\left[ 0,\frac{\pi }{e} \right]\] & increasing in \[\left[ \frac{\pi }{e},\infty \right]\]
D) Increasing in \[\left[ 0,\frac{\pi }{e} \right]\] & decreasing in \[\left[ \frac{\pi }{e},\infty \right)\]
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 \[[0,\infty )\].
You need to login to perform this action.
You will be redirected in
3 sec
