Question

Which of the following is an even function
(a) \[\dfrac{{{a^x} + {a^{ - x}}}}{{{a^x} - {a^{ - x}}}}\]
(b) \[f(x) = \dfrac{{{a^x} + 1}}{{{a^x} - 1}}\]
(c) \[f(x) = x\dfrac{{{a^x} - 1}}{{{a^{ - x}} + 1}}\]
(d) \[f(x) = {\log _2}\left( {x + \sqrt {{x^2} + 1} } \right)\]

Answer 1

Hint: If a given function \[f(x)\] satisfied \[f( - x) = f(x)\] for all \[x\] in the domain of f, then we say that \[f(x)\] is an even function, otherwise we say that \[f(x)\] is not an even function. To check that we have to replace \[x\] by \[ - x\] in the given function \[f(x)\] . Then check if \[f( - x) = f(x)\] is satisfied or not, if it is satisfied then we say that \[f(x)\] is an even function.

Complete step-by-step answer:
Option (a)
Given \[f(x) = \dfrac{{{a^x} + {a^{ - x}}}}{{{a^x} - {a^{ - x}}}}\] -----(1)
Replace \[x\] by \[ - x\] in equation (1), we get
 \[f( - x) = \dfrac{{{a^{ - x}} + {a^{ - ( - x)}}}}{{{a^{ - x}} - {a^{ - ( - x)}}}} = - \dfrac{{{a^x} + {a^{ - x}}}}{{{a^x} - {a^{ - x}}}} = - f(x)\]
therefore \[f(x) = \dfrac{{{a^x} + {a^{ - x}}}}{{{a^x} - {a^{ - x}}}}\] is not an even function.
Hence, Option (a) is incorrect.
Option (b)
Given \[f(x) = \dfrac{{{a^x} + 1}}{{{a^x} - 1}}\] -----(2)
Replace \[x\] by \[ - x\] in equation (2), we get
 \[f( - x) = \dfrac{{{a^{ - x}} + 1}}{{{a^{ - x}} - 1}} = \dfrac{{\dfrac{{{a^x} + 1}}{{{a^x}}}}}{{\dfrac{{1 - {a^x}}}{{{a^x}}}}} = - f(x)\]
therefore \[f(x) = \dfrac{{{a^x} + 1}}{{{a^x} - 1}}\] is not an even function.
Hence, Option (b) is incorrect.
Option (c)
Given \[f(x) = x\dfrac{{{a^x} - 1}}{{{a^{ - x}} + 1}}\] ------(3)
Replace \[x\] by \[ - x\] in equation (3), we get
 \[f( - x) = \left( { - x} \right)\dfrac{{{a^{ - x}} - 1}}{{{a^{ - ( - x)}} + 1}} = - x\dfrac{{\dfrac{{1 - {a^x}}}{{{a^x}}}}}{{{a^x} + 1}} = \dfrac{1}{{{a^{2x}}}}f(x)\]
therefore \[f(x) = \dfrac{{{a^x} + {a^{ - x}}}}{{{a^x} - {a^{ - x}}}}\] is not an even function.
Hence, Option (c) is incorrect.
Option (d)
Given \[f(x) = {\log _2}\left( {x + \sqrt {{x^2} + 1} } \right)\] ------(4)
Replace \[x\] by \[ - x\] in equation (4), we get
 \[f( - x) = {\log _2}\left( { - x + \sqrt {{x^2} + 1} } \right) = {\log _2}\left( {\left( { - x + \sqrt {{x^2} + 1} } \right) \times \dfrac{{x + \sqrt {{x^2} + 1} }}{{x + \sqrt {{x^2} + 1} }}} \right) = - \log \left( {x + \sqrt {{x^2} + 1} } \right) = - f(x)\]
therefore \[f(x) = {\log _2}\left( {x + \sqrt {{x^2} + 1} } \right)\] is not an even function.
Hence, Option (d) is incorrect.
So, the correct answer is “Option d”.

Note: If a given function \[f(x)\] satisfied \[f( - x) = - f(x)\] for all \[x\] in the domain of f, then we say that \[f(x)\] is an odd function, otherwise we say that \[f(x)\] is not an odd function. To check that we have to replace \[x\] by \[ - x\] in the given function \[f(x)\] . Then check if \[f( - x) = - f(x)\] is satisfied or not, if it is satisfied then we say that \[f(x)\] is an odd function.