Question

The number of roots of the equation √2 + e^cosh^-1 x - e^sinh^-1x =0 is​

Answer 1

Answer:

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Answer 2

The number of roots of the equation $\mathsf{\sqrt{2}+e^{cosh^{-1}x}-e^{sinh^{-1}x}=0}$ is 2.

Concept:

• $\mathsf{sinh^{-1}x=log_{e}(x+\sqrt{x^{2}+1})}$

• $\mathsf{cosh^{-1}x=log_{e}(x+\sqrt{x^{2}-1})}$

• $\mathsf{e^{log_{e}x}=x}$

Step-by-step explanation:

Here, $\mathsf{\sqrt{2}+e^{cosh^{-1}x}-e^{sinh^{-1}x}=0}$

$\mathsf{\Rightarrow \sqrt{2}+e^{log_{e}(x+\sqrt{x^{2}-1})}-e^{log_{e}(x+\sqrt{x^{2}+1})}=0}$

$\mathsf{\Rightarrow \sqrt{2}+(x+\sqrt{x^{2}-1})-(x+\sqrt{x^{2}+1})=0}$

$\mathsf{\Rightarrow \sqrt{2}+x+\sqrt{x^{2}-1}-x-\sqrt{x^{2}+1}=0}$

$\mathsf{\Rightarrow \sqrt{x^{2}+1}-\sqrt{x^{2}-1}=-\sqrt{2}}$

Squaring both sides, we get

$\mathsf{x^{2}+1+x^{2}-1-2\sqrt{x^{2}+1}\sqrt{x^{2}-1}=2}$

$\mathsf{\Rightarrow 2x^{2}-2\sqrt{x^{4}-1}=2}$

$\mathsf{\Rightarrow x^{2}-1=\sqrt{x^{4}-1}}$

Again squaring both sides, we get

$\mathsf{x^{4}-2x^{2}+1=x^{4}-1}$

$\mathsf{\Rightarrow 2x^{2}-2=0}$

$\mathsf{\Rightarrow x^{2}-1=0}$

Since the highest power of $\mathsf{x}$ is 2, the given equation has two roots.

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