Question

If \[f:R\to R\] such that \[f\left( x \right)=\ln \left( x+\sqrt{{{x}^{2}}+1} \right),\] another function g(x) is defined such that \[gof\left( x \right)=x\forall x\in R,\] then g(2) is:
\[\left( a \right)\dfrac{{{e}^{2}}+{{e}^{-2}}}{2}\]
\[\left( b \right){{e}^{2}}\]
\[\left( c \right)\dfrac{{{e}^{2}}-{{e}^{-2}}}{2}\]
\[\left( d \right){{e}^{-2}}\]

Answer 1

Hint: In this question, we are given a function f(x) and we are given that g (f(x)) = x. Using this we have to find g(x) for this we will assume f(x) as y. So we need to get g(y) = x. Now we need to find the value of x in terms of y so we will find it by solving f(x) = y. At last, we will put the value of y as 2 in the found g(y) to get our final answer.

Complete step-by-step answer:
Here we are given \[f\left( x \right)=\ln \left( x+\sqrt{{{x}^{2}}+1} \right).\] Also it is given that \[gof\left( x \right)=x\forall x\in R,\] which means g(f(x)) = x. Now, let us suppose that f(x) = y. So, we get g(y) = x. Now, x becomes a function of y, so we need to find it. As we know that f(x) = y, so we get,
\[\ln \left( x+\sqrt{{{x}^{2}}+1} \right)=y\]
Now using this we will find the value of x in terms of y. Now using this we will find the value of x in terms of y. Taking the exponential on both the sides, we get,
\[\Rightarrow {{e}^{\ln \left( x+\sqrt{{{x}^{2}}+1} \right)}}={{e}^{y}}\]
As we know that e and ln (natural log) cancels out, so we get,
\[\Rightarrow x+\sqrt{{{x}^{2}}+1}={{e}^{y}}\]
Taking x to the other sides, we get,
\[\Rightarrow \sqrt{{{x}^{2}}+1}={{e}^{y}}-x\]
Now squaring both the sides, we get,
\[\Rightarrow {{x}^{2}}+1={{\left( {{e}^{y}}-x \right)}^{2}}\]
As we know that \[{{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab,\] hence we get,
\[\Rightarrow {{x}^{2}}+1={{e}^{2y}}+{{x}^{2}}-2x{{e}^{y}}\]
Cancelling \[{{x}^{2}}\] from both the sides, we get,
\[\Rightarrow 1={{e}^{2y}}-2x{{e}^{y}}\]
Rearranging the terms, we get,
\[\Rightarrow 2x{{e}^{y}}={{e}^{2y}}-1\]
Dividing both the sides by \[2{{e}^{y}},\] we get,
\[\Rightarrow x=\dfrac{{{e}^{y}}\left( {{e}^{y}}-\dfrac{1}{{{e}^{y}}} \right)}{{{e}^{y}}\left( 2 \right)}\]
Since, \[\dfrac{1}{x}={{x}^{-1}},\] so we get,
\[\Rightarrow x=\dfrac{{{e}^{y}}-{{e}^{-y}}}{2}\]
Now we know that g(y) = x, so we get,
\[g\left( y \right)=\dfrac{{{e}^{y}}-{{e}^{-y}}}{2}\]
Now we need to find the value of g(2). So putting y = 2 in g(y), we get,
\[\Rightarrow g\left( 2 \right)=\dfrac{{{e}^{2}}-{{e}^{-2}}}{2}\]
Hence, g(2) is equal to \[\dfrac{{{e}^{2}}-{{e}^{-2}}}{2}.\]

So, the correct answer is “Option (c)”.

Note: Students should carefully find the value of x in terms of y. We have simplified \[\dfrac{{{e}^{2y}}-1}{2{{e}^{y}}}\] to match our answer with the options. Students should note that only natural log (ln) cancels out with e. Students can make mistakes in signs, so take care of that. Here, got (x) means g(f(x)).