If one solution of the equation $\cos \mathrm{h} x-\frac{4}{5} \sin \mathrm{h} x=1$ is $x=0$, then the other…
If one solution of the equation $\cos \mathrm{h} x-\frac{4}{5} \sin \mathrm{h} x=1$ is $x=0$, then the other solution is $\mathrm{x}=$
- $2 \log 2$
- $2 \log 5$
- $\log \left(\frac{4}{3}\right)$
- $2 \log 3$
Solution
$5 \cosh x-4 \sinh x=5$
$\begin{aligned} & \Rightarrow 5\left(\frac{\mathrm{e}^{\mathrm{n}}+\mathrm{e}^{-\mathrm{n}}}{2}\right)-4\left(\frac{\mathrm{e}^{\mathrm{x}}-\mathrm{e}^{-\mathrm{x}}}{2}\right)=5 \\ & \Rightarrow 5 \mathrm{e}^{\mathrm{n}}+5 \mathrm{e}^{-\mathrm{n}}-4 \mathrm{ex}+4 \mathrm{e}^{-\mathrm{x}}=10 \\ & \Rightarrow \mathrm{e}^{\mathrm{n}}+9 \mathrm{e}^{-\mathrm{x}}=10 \Rightarrow(\mathrm{ex})^2-10 \mathrm{e}^{\mathrm{x}}+9=0 \\ & \Rightarrow(\mathrm{ex})^2-9 \mathrm{e}^{\mathrm{x}}-\mathrm{e}^{\mathrm{x}}+9=0 \\ & \Rightarrow\left(\mathrm{e}^{\mathrm{x}}-1\right)\left(\mathrm{e}^{\mathrm{x}}-9\right)=0 \\ & \Rightarrow \mathrm{e}^{\mathrm{x}}-1=0 \Rightarrow \mathrm{e}^{\mathrm{x}}=1 \Rightarrow \mathrm{x}=0 \\ & \Rightarrow \mathrm{e}^{\mathrm{x}}-9=0 \Rightarrow \mathrm{e}^{\mathrm{x}}=9 \Rightarrow \mathrm{x}=\log 9=2 \log 3\end{aligned}$
Asked in: AP EAMCET 2022 (08 Jul Shift 1)
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