Question

the function f(x)=log(x+root of x^2+1) is
(A)even fun
(B)odd func
(C)not both
(D)none
just tell the option

Answer 1

Answer:

(b) odd function

Step-by-step explanation:

Given a function,

$f(x) = log(x + \sqrt{ {x}^{2} + 1} )$

To find whether it's an even or an odd function.

We know that,

For an even function, f(-x) = f(x)

And,

For an odd function, f(-x) = -f(x)

Let's find out, f(-x).

Therefore, we will get,

$= > f( - x) = log( - x + \sqrt{ {( - x)}^{2} + 1 } ) \\ \\ = > f( - x) = log( \sqrt{ {x}^{2} + 1 } - x) \\ \\ = > f( - x) = log( \sqrt{ {x}^{2} + 1} - x \times \frac{ \sqrt{ {x}^{2} + 1 } + x }{ \sqrt{ {x}^{2} + 1 } + x} ) \\ \\ = > f( - x) = log( ({x}^{2} + 1 - {x}^{2}) \times \frac{1}{ \sqrt{ {x}^{2} + 1 } + x} ) \\ \\ = > f( - x) = log( \frac{1}{ \sqrt{ {x}^{2} + 1} + x} ) \\ \\ = > f( - x) = log(1) - log( x + \sqrt{ {x}^{2} + 1} ) \\ \\ = > f( - x) = 0 - log(x + \sqrt{ {x}^{2} + 1 } ) \\ \\ = > f( - x) = - log(x + \sqrt{ {x}^{2} + 1 } ) \\ \\ = > f( - x) = - f(x)$

Clearly, f(-x) = -f(x)

Hence, the correct answer is (b) odd function.