Consider the hyperbolic functions in Group – 1 and their definitions in Group - 2. The correct combination is Group - 1 Group - 2 P $\tanh x$ I $\frac{e^x + e^{-x}}{e^x - e^{-x}}$ Q $\coth x$ II $\frac{2}{e^x + e^{-x}}$ R $\text{sech } x$ III $\frac{2}{e^x - e^{-x}}$ S $\text{cosech } x$ IV $\frac{e^x - e^{-x}}{e^x + e^{-x}}$
To match the hyperbolic functions in Group - 1 with their definitions in Group - 2, we first recall the fundamental definitions based on exponential functions:
Using the basic definitions, we can derive the forms for the other functions:
The definition is $ \tanh x = \frac{\sinh x}{\cosh x} $. Substituting the exponential forms:
$ \tanh x = \frac{\frac{e^x - e^{-x}}{2}}{\frac{e^x + e^{-x}}{2}} = \frac{e^x - e^{-x}}{e^x + e^{-x}} $
This matches Group - 2 definition **IV**. So, P $\rightarrow$ IV.
The definition is $ \coth x = \frac{\cosh x}{\sinh x} $. Substituting the exponential forms:
$ \coth x = \frac{\frac{e^x + e^{-x}}{2}}{\frac{e^x - e^{-x}}{2}} = \frac{e^x + e^{-x}}{e^x - e^{-x}} $
This matches Group - 2 definition **I**. So, Q $\rightarrow$ I.
The definition is $ \text{sech } x = \frac{1}{\cosh x} $. Substituting the exponential form:
$ \text{sech } x = \frac{1}{\frac{e^x + e^{-x}}{2}} = \frac{2}{e^x + e^{-x}} $
This matches Group - 2 definition **II**. So, R $\rightarrow$ II.
The definition is $ \text{cosech } x = \frac{1}{\sinh x} $. Substituting the exponential form:
$ \text{cosech } x = \frac{1}{\frac{e^x - e^{-x}}{2}} = \frac{2}{e^x - e^{-x}} $
This matches Group - 2 definition **III**. So, S $\rightarrow$ III.
Combining the matches:
The correct combination is P-IV, Q-I, R – II, S – III.
Given the function $$f(x) = |x| + |x - 1|,$$ For all the real values of x, which one of the following statements is CORRECT ?