JEE Main 12 Jan 2019 Morning Question 15
Question: If $ x>1 $ for $ {{(2x)}^{2y}}=4{e^{2x-2y}}, $ then $ {{(1+log _{e}2x)}^{2}}\frac{dy}{dx} $ equals [JEE Main Online Paper Held On 12-Jan-2019 Morning]
Options:
A) $ \frac{x,{\log_e}2x-{\log_e}2}{x} $
B) $ {\log_e}2x $
C) $ \frac{x,{\log_e}2x+{\log_e}2}{x} $
D) $ x,{\log_e}2x $
Show Answer
Answer:
Correct Answer: A
Solution:
$ {{(2x)}^{2y}}=4{e^{2x-2y}} $
$ \Rightarrow $ $ 2y{\log_e}2x={\log_e}4+2x-2y $
$ \Rightarrow $ $ 2y{\log_e}2x=2{\log_e}2+2x-2y $
$ \Rightarrow $ $ y{\log_e}2x={\log_e}2+x-y\Rightarrow y=\frac{x+{\log_e}2}{1+{\log_e}2x} $
$ \therefore $ $ \frac{dy}{dx}=\frac{1+{\log_e}2x-(x+log _{e}2)\frac{1}{x}}{{{(1+log _{e}2x)}^{2}}} $
$ \Rightarrow $ $ \frac{dy}{dx}={{(1+log _{e}2x)}^{2}}=\frac{x{\log_e}2x-{\log_e}2}{x} $
