
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.
Is the above statement true or false?

Answer
483.9k+ views
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.
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.
Complete step-by-step answer:
The below graph shown in the above question has two curves marked with red and green colour.

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.
Recently Updated Pages
Glucose when reduced with HI and red Phosphorus gives class 11 chemistry CBSE

The highest possible oxidation states of Uranium and class 11 chemistry CBSE

Find the value of x if the mode of the following data class 11 maths CBSE

Which of the following can be used in the Friedel Crafts class 11 chemistry CBSE

A sphere of mass 40 kg is attracted by a second sphere class 11 physics CBSE

Statement I Reactivity of aluminium decreases when class 11 chemistry CBSE

Trending doubts
10 examples of friction in our daily life

The correct order of melting point of 14th group elements class 11 chemistry CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

State and prove Bernoullis theorem class 11 physics CBSE

What organs are located on the left side of your body class 11 biology CBSE
