Applications Of Derivatives Question 366
Question: The minimum value of $ 4e^{2x}+9{e^{-2x}} $ is
[J & K 2005]
Options:
A) 11
B) 12
C) 10
D) 14
Show Answer
Answer:
Correct Answer: B
Solution:
Let $ f(x)=4e^{2x}+9{e^{-2x}} $
$ \therefore $ $ {f}’(x)=8e^{2x}-18{e^{-2x}} $ Put $ {f}’(x)=0\Rightarrow 8e^{2x}-18{e^{-2x}}=0 $
$ e^{2x}=3/2\Rightarrow x=\log {{(3/2)}^{1/2}} $
Again $ {f}’’(x)=16e^{2x}+36{e^{-2x}}>0 $
Now $ f(\log {{(3/2)}^{1/2}})=4{e^{2.(\log {{(3/2)}^{1/2}})}}+9{e^{-2(\log {{(3/2)}^{1/2}})}} $ = $ 4\times \frac{3}{2}+9\times \frac{2}{3} $ = $ 6+6=12 $
Hence minimum value = 12.
BETA
