Which of the following function is even funct | vedclass

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Question

(b) In $(a)$, $f( - x) = \frac{{{a^{ - x}} + 1}}{{{a^{ - x}} - 1}} = \frac{{1 + {a^x}}}{{1 - {a^x}}} = - \frac{{{a^x} + 1}}{{{a^x} - 1}} = - f(x)$

Vedclass · Accepted Answer

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

Vedclass · Answer

(b) In $(a)$, $f( - x) = \frac{{{a^{ - x}} + 1}}{{{a^{ - x}} - 1}} = \frac{{1 + {a^x}}}{{1 - {a^x}}} = - \frac{{{a^x} + 1}}{{{a^x} - 1}} = - f(x)$

So, it is an odd function.

In $(b)$, $f( - x) = ( - x)\frac{{{a^{ - x}} - 1}}{{{a^{ - x}} + 1}} = - x\frac{{1 - {a^x}}}{{1 + {a^x}}} = x\frac{{{a^x} - 1}}{{{a^x} + 1}} = f(x)$

So, it is an even function.

In $(c)$, $f( - x) = - \sin \left[ {\log (x + \sqrt {1 + {x^2}} )} \right]$

So, it is an odd function.

In $(d)$, $f( - x) = \sin ( - x) = - \sin x = - f(x)$

So, it is an odd function.

Which of the following function is even function

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