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The two curves shown in the below figure with $ \cosh \left( x \right) $ as green colored and $ {{\cosh }^{-1}}\left( x \right) $ as red colored are correctly labeled.
seo images

Is the above statement true or false?

Last updated date: 17th Jun 2024
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Hint: We know that $ \cosh \left( x \right)\And {{\cosh }^{-1}}\left( x \right) $ both are hyperbolic functions with $ \cosh \left( x \right)=\dfrac{{{e}^{x}}+{{e}^{-x}}}{2} $ and $ {{\cosh }^{-1}}\left( x \right)=\ln \left( x+\sqrt{{{x}^{2}}-1} \right) $ . Now, find the minimum value of $ \cosh \left( x \right)\And {{\cosh }^{-1}}\left( x \right) $ from the equation that we have just described by putting $ x=0 $ in these equations and then compare the minimum values from the graph shown in the above problem. If the minimum values of $ \cosh \left( x \right)\And {{\cosh }^{-1}}\left( x \right) $ from the equations are matched with the graph given above, then the curves are correctly labeled.

Complete step-by-step answer:
The below graph shown in the above question has two curves marked with red and green colour.
seo images

It is given that green color curve corresponds to $ \cosh \left( x \right) $ and red color corresponds to $ {{\cosh }^{-1}}\left( x \right) $ and we have to show whether the labeling of the curves are correct or not.
We know that $ \cosh \left( x \right)\And {{\cosh }^{-1}}\left( x \right) $ are hyperbolic functions. In the below, we are showing the hyperbolic functions in x corresponding to $ \cosh \left( x \right)\And {{\cosh }^{-1}}\left( x \right) $ .
The function of $ \cosh \left( x \right) $ is equal to:
 $ \cosh \left( x \right)=\dfrac{{{e}^{x}}+{{e}^{-x}}}{2} $ …………. Eq. (1)
The function of $ {{\cosh }^{-1}}\left( x \right) $ is equal to:
 $ {{\cosh }^{-1}}\left( x \right)=\ln \left( x+\sqrt{{{x}^{2}}-1} \right) $ ……….. Eq. (2)
Now, to check whether the curves labeled in the above question are correct or not by finding the minimum values of $ \cosh \left( x \right)\And {{\cosh }^{-1}}\left( x \right) $ .
Substituting $ x=0 $ in eq. (1) we get,
 $ \begin{align}
  & \cosh \left( 0 \right)=\dfrac{{{e}^{\left( 0 \right)}}+{{e}^{\left( 0 \right)}}}{2} \\
 & \Rightarrow \cosh \left( 0 \right)=\dfrac{1+1}{2} \\
 & \Rightarrow \cosh \left( 0 \right)=\dfrac{2}{2}=1 \\
\end{align} $
Now, at $ x=0 $ we have got the value of $ \cosh \left( x \right) $ as 1 which is the same as given in the above problem.
Hence, the green curve corresponding to $ \cosh \left( x \right) $ is correctly labeled.
If we assume that what is given in the question is right i.e. red curve corresponds to $ {{\cosh }^{-1}}\left( x \right) $ then the minimum value of this function occurs at $ x=1 $ and is 0 so let us substitute $ x=1 $ in eq. (2) to see whether the value of $ {{\cosh }^{-1}}\left( x \right) $ is coming 0 or not.
 $ \begin{align}
  & {{\cosh }^{-1}}\left( x \right)=\ln \left( x+\sqrt{{{x}^{2}}-1} \right) \\
 & \Rightarrow {{\cosh }^{-1}}\left( 1 \right)=\ln \left( 1+\sqrt{{{1}^{2}}-1} \right) \\
\end{align} $
 $ \begin{align}
  & \Rightarrow {{\cosh }^{-1}}\left( 1 \right)=\ln \left( 1+\sqrt{0} \right) \\
 & \Rightarrow {{\cosh }^{-1}}\left( 1 \right)=\ln \left( 1 \right) \\
\end{align} $
And we know that, the value of $ \ln \left( 1 \right)=0 $ so the value of the above equation becomes:
 $ {{\cosh }^{-1}}\left( 1 \right)=0 $
Hence, we have got the same minimum value of $ {{\cosh }^{-1}}\left( x \right) $ which is given in the above problem. Hence, the red curve which is labeled as $ {{\cosh }^{-1}}\left( x \right) $ is correct.
From the above, we can say that the above statement is true.

Note: Don’t confuse the functions $ \cosh \left( x \right)\And {{\cosh }^{-1}}\left( x \right) $ given in the above problem with $ \cos x\And {{\cos }^{-1}}\left( x \right) $ . These two functions are completely different from each other. The functions $ \cosh \left( x \right)\And {{\cosh }^{-1}}\left( x \right) $ are hyperbolic functions whereas $ \cos x\And {{\cos }^{-1}}\left( x \right) $ are trigonometric functions.
Usually, students ignore the “h” written with cosine and think it might be a typo but it is not so make sure you don’t repeat such mistakes.