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If $\sinh \left(2 \tanh ^{-1} x\right)=\frac{11}{60}$, then $x=$
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The correct answer is:
$\frac{1}{11}$
$\begin{aligned} & \text { Given, } \sinh \left(2 \tanh ^{-1} x\right)=\frac{11}{60} \\ & \frac{2 \tanh \left(\tanh h^{-1} x\right)}{1-\left(\tanh \left(\tanh ^{-1} x\right)\right)^2}=\frac{11}{60}\end{aligned}$
$\left[\because \sinh (2 x)=\frac{2 \tanh (x)}{1-\tanh ^2(x)}\right]$
$\begin{aligned} & \\ & \Rightarrow \quad \frac{2 x}{1-x^2}=\frac{11}{60} \Rightarrow 11 x^2+120 x-11=0 \\ & \Rightarrow 11 x^2+121 x-x-11=0 \Rightarrow(11 x-1)(x+11)=0 \\ & \Rightarrow \quad 11 x-1=0, x \neq-11 \Rightarrow x=\frac{1}{11}\end{aligned}$
$\left[\because \sinh (2 x)=\frac{2 \tanh (x)}{1-\tanh ^2(x)}\right]$
$\begin{aligned} & \\ & \Rightarrow \quad \frac{2 x}{1-x^2}=\frac{11}{60} \Rightarrow 11 x^2+120 x-11=0 \\ & \Rightarrow 11 x^2+121 x-x-11=0 \Rightarrow(11 x-1)(x+11)=0 \\ & \Rightarrow \quad 11 x-1=0, x \neq-11 \Rightarrow x=\frac{1}{11}\end{aligned}$
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