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Question: Which of the following is an even function...

Which of the following is an even function

A

f(x)=ax+1ax−1f(x) = \frac{a^{x} + 1}{a^{x} - 1}

B

f(x)=x(ax−1ax+1)f(x) = x\left( \frac{a^{x} - 1}{a^{x} + 1} \right)

C

f(x)=ax−a−xax+a−xf(x) = \frac{a^{x} - a^{- x}}{a^{x} + a^{- x}}

D

f(x)=sin⁡xf(x) = \sin x

Answer

f(x)=x(ax−1ax+1)f(x) = x\left( \frac{a^{x} - 1}{a^{x} + 1} \right)

Explanation

Solution

In option (1), f(−x)=a−x+1a−x−1=1+ax1−ax=−ax+1ax−1=−f(x)f( - x) = \frac{a^{- x} + 1}{a^{- x} - 1} = \frac{1 + a^{x}}{1 - a^{x}} = - \frac{a^{x} + 1}{a^{x} - 1} = - f(x)

So, It is an odd function.

In option (2),

f(−x)=(−x)a−x−1a−x+1=−x(1−ax)1+ax=x(ax−1)(ax+1)=f(x)f( - x) = ( - x)\frac{a^{- x} - 1}{a^{- x} + 1} = - x\frac{(1 - a^{x})}{1 + a^{x}} = x\frac{(a^{x} - 1)}{(a^{x} + 1)} = f(x) So, It is an even function.

In option (3), f(−x)=a−x−axa−x+ax=−f(x)f( - x) = \frac{a^{- x} - a^{x}}{a^{- x} + a^{x}} = - f(x) So, It is an odd function.

In option (4), f(−x)=sin⁡(−x)=−sin⁡x=−f(x)f( - x) = \sin( - x) = - \sin x = - f(x) So, It is an odd function.