Question

Which of the following is an even function
a.\[x\left( {\dfrac{{{a^x} - 1}}{{{a^x} + 1}}} \right)\]
b. $ \tan x $
c. $ \dfrac{{{a^x} - {a^{ - x}}}}{2} $
d. $ \dfrac{{{a^x} + 1}}{{{a^x} - 1}} $

Answer 1

Hint:In this question knowing the concept of even and odd function is which says that when a function f (x) = f (-x) for all x then the function is called even function but when for all x f (x) = - f (-x) then the function is said to be odd function, use this information to approach towards the solution.

Complete step-by-step answer:
We know that a function is said to be even function when f (x) = f (-x) for all x
So now checking all given function which satisfies the above statement
In 1st option we have function f (x) = \[x\left( {\dfrac{{{a^x} - 1}}{{{a^x} + 1}}} \right)\]so for f (-x) function will be:
f (-x) = \[ - x\left( {\dfrac{{{a^{ - x}} - 1}}{{{a^{ - x}} + 1}}} \right)\]
 $ \Rightarrow $ f (-x) = \[ - x\left( {\dfrac{{\dfrac{1}{{{a^x}}} - 1}}{{\dfrac{1}{{{a^x}}} + 1}}} \right)\]
 $ \Rightarrow $ f (-x) = \[ - x\left( {\dfrac{{1 - {a^x}}}{{1 + {a^x}}}} \right)\]
 $ \Rightarrow $ f (-x) = \[x\left( {\dfrac{{{a^x} - 1}}{{{a^x} + 1}}} \right)\]
Since the function f (x) = f(-x) for all x therefore function \[x\left( {\dfrac{{{a^x} - 1}}{{{a^x} + 1}}} \right)\] is a even function
For function f (x) = $ \tan x $ so f (-x) will be:
f (-x) = $ \tan \left( { - x} \right) $
Since we know that tan (-x) = - tan x
Therefore f (-x) = $ - \tan x $
Since f (x) is not equal to f (-x) for all x therefore function $ \tan x $ is not an even function
In 3rd option we have function f (x) = $ \dfrac{{{a^x} - {a^{ - x}}}}{2} $ so f (-x) will be given by
f (-x) = $ \dfrac{{{a^{ - x}} - {a^{ - \left( { - x} \right)}}}}{2} $
 $ \Rightarrow $ f (-x) = $ \dfrac{{{a^{ - x}} - {a^x}}}{2} $
 $ \Rightarrow $ f (-x) = $ - \left( {\dfrac{{{a^x} - {a^{ - x}}}}{2}} \right) $
Since f (x) is not equal to f (-x) the function $ \dfrac{{{a^x} - {a^{ - x}}}}{2} $ is not an even function
Now in the 4th option we have function f (x) = $ \dfrac{{{a^x} + 1}}{{{a^x} - 1}} $ so f (-x) will be:
f (-x) = $ \dfrac{{{a^{ - x}} + 1}}{{{a^{ - x}} - 1}} $
 $ \Rightarrow $ f (-x) = \[\dfrac{{\dfrac{1}{{{a^x}}} + 1}}{{\dfrac{1}{{{a^x}}} - 1}}\]
 $ \Rightarrow $ f (-x) = \[\dfrac{{1 + {a^x}}}{{1 - {a^x}}}\]
 $ \Rightarrow $ f (-x) = \[ - \left( {\dfrac{{{a^x} + 1}}{{{a^x} - 1}}} \right)\]
Since f (x) is not equal to f (-x) therefore function $ \dfrac{{{a^x} + 1}}{{{a^x} - 1}} $ is not an even function
Hence option A is the correct option.


Note: In the above question we found lot about the even and odd functions but in the whole solution we didn’t discussed about the some properties of even and odd function, so let’s find out what are those properties, the even functions behaves like functions\[{x^2}\],\[{x^4}\], etc. because in such functions the nature (positive or negative) of the result doesn’t depend upon the nature of input (negative or positive input) because the result of input is always positive there are some more examples of even functions like cos x, $ \left( {{x^2} + {x^4}} \right) $ and etc. but in odd functions they operate like x, $ {x^3} $ , etc. here the nature of result of function does depend upon the nature of input some more function like odd functions are sin x, $ \left( {{x^5} + 1} \right) $ and etc.