Search any question & find its solution
Question:
Answered & Verified by Expert
If $2 \cosh 2 x+10 \sinh 2 x=5$, then $x=$
Options:
Solution:
1956 Upvotes
Verified Answer
The correct answer is:
$\frac{1}{2} \log \frac{4}{3}$
Here, $2 \cosh 2 x+10 \sinh 2 x=5$
We know that,
$\cosh 2 x=\frac{1}{2}\left(e^{2 x}+e^{-2 x}\right)$
$\sinh 2 x=\frac{1}{2}\left(e^{2 x}-e^{-2 x}\right)$
So, $2\left[\frac{1}{2}\left(e^{2 x}+e^{-2 x}\right)\right]+10\left[\frac{1}{2}\left(e^{2 x}-e^{-2 x}\right)\right]=5$
$\begin{aligned} & \Rightarrow e^{2 x}+e^{-2 x}+5 e^{2 x}-5 e^{-2 x}=5 \\ & \Rightarrow 6 e^{2 x}-5-4 e^{-2 x}=0 \Rightarrow\left(3 e^{2 x}-4\right)\left(2 e^{2 x}+1\right)=0\end{aligned}$
$e^{2 x} \stackrel{4}{3}$ or $e^{2 x}=-\frac{1}{2}$
The only real solution occurs when $e^{2 x}>0$.
So, $2 x=\ln \frac{4}{3}$
$\Rightarrow \quad x=\frac{1}{2} \ln \frac{4}{3}$
We know that,
$\cosh 2 x=\frac{1}{2}\left(e^{2 x}+e^{-2 x}\right)$
$\sinh 2 x=\frac{1}{2}\left(e^{2 x}-e^{-2 x}\right)$
So, $2\left[\frac{1}{2}\left(e^{2 x}+e^{-2 x}\right)\right]+10\left[\frac{1}{2}\left(e^{2 x}-e^{-2 x}\right)\right]=5$
$\begin{aligned} & \Rightarrow e^{2 x}+e^{-2 x}+5 e^{2 x}-5 e^{-2 x}=5 \\ & \Rightarrow 6 e^{2 x}-5-4 e^{-2 x}=0 \Rightarrow\left(3 e^{2 x}-4\right)\left(2 e^{2 x}+1\right)=0\end{aligned}$
$e^{2 x} \stackrel{4}{3}$ or $e^{2 x}=-\frac{1}{2}$
The only real solution occurs when $e^{2 x}>0$.
So, $2 x=\ln \frac{4}{3}$
$\Rightarrow \quad x=\frac{1}{2} \ln \frac{4}{3}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.