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Question: The function \(f(x) = \sin\left( \log(x + \sqrt{x^{2} + 1}) \right)\) is...

The function f(x)=sin⁡(log⁡(x+x2+1))f(x) = \sin\left( \log(x + \sqrt{x^{2} + 1}) \right) is

A

Even function

B

Odd function

C

Neither even nor odd

D

Periodic function

Answer

Odd function

Explanation

Solution

f(x)=sin⁡(log⁡(x+1+x2))f(x) = \sin\left( \log(x + \sqrt{1 + x^{2}}) \right)

⇒ f(−x)=sin⁡[log⁡(−x+1+x2)]f( - x) = \sin\lbrack\log( - x + \sqrt{1 + x^{2}})\rbrack

⇒f(−x)=sin⁡log⁡((1+x2−x)(1+x2+x)(1+x2+x))f( - x) = \sin{\log\left( (\sqrt{1 + x^{2}} - x)\frac{(\sqrt{1 + x^{2}} + x)}{(\sqrt{1 + x^{2}} + x)} \right)}

⇒ f(−x)=sin⁡log⁡[1(x+1+x2)]f( - x) = \sin{\log\left\lbrack \frac{1}{(x + \sqrt{1 + x^{2}})} \right\rbrack}

⇒ f(−x)=sin⁡[log⁡(x+1+x2)−1]f( - x) = \sin\left\lbrack \log(x + \sqrt{1 + x^{2}})^{- 1} \right\rbrack

⇒ f(−x)=sin⁡[−log⁡(x+1+x2)]f( - x) = \sin\left\lbrack - \log(x + \sqrt{1 + x^{2}}) \right\rbrack

⇒ f(−x)=−sin⁡[log⁡(x+1+x2)]f( - x) = - \sin\left\lbrack \log(x + \sqrt{1 + x^{2}}) \right\rbrack⇒ f(−x)=−f(x)f( - x) = - f(x)

∴ f(x)f(x) is odd function.