Question

The even function is
A. $ f(x) = \dfrac{{[{a^x} + {a^{ - x}}]}}{{[{a^x} - {a^{ - x}}]}} $
B. $ f(x) = \dfrac{{[{a^x} + 1]}}{{[{a^x} - 1]}} $
C. $ f(x) = x\left[ {\dfrac{{[{a^x} - 1]}}{{[{a^x} + 1]}}} \right] $
D. $ f(x) = {\log _2}(x + \sqrt {{x^2} + 1} ) $

Answer 1

Hint: A function, $ f(x) $ is considered to be even when $ f(x) = f( - x) $ . Therefore, to check whether the given function is even or odd, then simply replace $ x $ by $ - x $ and check the equality.
If the condition of $ f(x) = f( - x) $ is satisfied then it is an even function and if $ f(x) = - f( - x) $ , then that function is an odd function.

Complete step by step solution:
Let's use one by one to evaluate all the given functions by substituting $ - x $ in place of $ x $ .
In the (A) option,
 $ f(x) = \dfrac{{[{a^x} + {a^{ - x}}]}}{{[{a^x} - {a^{ - x}}]}} $
Now replace $ x $ by $ - x $
 $ \Rightarrow f( - x) = \dfrac{{[{a^{ - x}} + {a^x}]}}{{[{a^{ - x}} - {a^x}]}} $
 $ \Rightarrow f( - x) = \dfrac{{\left[ {\dfrac{1}{{{a^x}}} + {a^x}} \right]}}{{\left[ {\dfrac{1}{{{a^x}}} - {a^x}} \right]}} $
 $ \Rightarrow f( - x) = \dfrac{{\left[ {\dfrac{{1 + {a^{2x}}}}{{{a^x}}}} \right]}}{{\left[ {\dfrac{{1 - {a^{2x}}}}{{{a^x}}}} \right]}} $
  $ \Rightarrow f( - x) = \dfrac{{\left[ {1 + {a^{2x}}} \right]}}{{\left[ {1 - {a^{2x}}} \right]}} \ne f(x) $
Hence option (1) is NOT an even function.

Now, evaluate option (B)
Replace $ x $ by $ - x $
 $ \Rightarrow f(x) = \dfrac{{[{a^x} + 1]}}{{[{a^x} - 1]}} $
  $ \Rightarrow f( - x) = \dfrac{{[{a^{ - x}} + 1]}}{{[{a^{ - x}} - 1]}} $
 $ \Rightarrow f( - x) = \dfrac{{\left[ {\dfrac{1}{{{a^x}}} + 1} \right]}}{{\left[ {\dfrac{1}{{{a^x}}} - 1} \right]}} $
\[ \Rightarrow f( - x) = \dfrac{{\left[ {\dfrac{{1 + {a^x}}}{{{a^x}}}} \right]}}{{\left[ {\dfrac{{1 - {a^x}}}{{{a^x}}}} \right]}}\]
\[ \Rightarrow f( - x) = \dfrac{{\left[ {1 + {a^x}} \right]}}{{\left[ {1 - {a^x}} \right]}} = - f(x)\]
Hence option (2) Is an odd function.

Further, let us evaluate option (C)
Replace $ x $ by $ - x $
 $ \Rightarrow f(x) = x\left[ {\dfrac{{[{a^x} - 1]}}{{[{a^x} + 1]}}} \right] $
 $ \Rightarrow f( - x) = - x\left[ {\dfrac{{[{a^{ - x}} - 1]}}{{[{a^{ - x}} + 1]}}} \right] $

 $ \Rightarrow f( - x) = - x\left[ {\dfrac{{\left[ {\dfrac{1}{{{a^x}}} - 1} \right]}}{{\left[ {\dfrac{1}{{{a^x}}} + 1} \right]}}} \right] $
 $ \Rightarrow f( - x) = - x\left[ {\dfrac{{\left[ {\dfrac{{1 - {a^x}}}{{{a^x}}}} \right]}}{{\left[ {\dfrac{{1 + {a^x}}}{{{a^x}}}} \right]}}} \right] $
 $ \Rightarrow f( - x) = - x\left[ {\dfrac{{[1 - {a^x}]}}{{[1 + {a^x}]}}} \right] $
 $ \Rightarrow f( - x) = x\left[ {\dfrac{{[{a^x} - 1]}}{{[1 + {a^x}]}}} \right] = f(x) $
Hence, option (3) satisfies the condition of even functions, therefore, option (3) is the correct answer.

We can also check option (D) in the similar way by replacing $ x $ by $ - x $
 $ \Rightarrow f(x) = {\log _2}(x + \sqrt {{x^2} + 1} ) $
 $ \Rightarrow f( - x) = {\log _2}( - x + \sqrt {{{( - x)}^2} + 1} ) $
 $ \Rightarrow f( - x) = {\log _2}( - x + \sqrt {{{(x)}^2} + 1} ) \ne f(x) $
Hence, this function is neither even nor odd.

Note: We can also find the even and odd function by graphs. If a graph is symmetrical about the y- axis, then the function is even and if a graph is symmetrical about the origin, then the function is odd. If a graph is not symmetrical about the y-axis nor the origin, the function is neither even, nor odd.