Question:
The function
\[
f(x)=\sin\left(\log\left(x+\sqrt{x^2+1}\right)\right)
\]
is
The function
\[
f(x)=\sin\left(\log\left(x+\sqrt{x^2+1}\right)\right)
\]
is
Show Hint
Remember the identity
\[
(x+\sqrt{x^2+1})(\sqrt{x^2+1}-x)=1.
\]
It frequently appears in logarithmic and hyperbolic-function problems and is extremely useful for checking parity.
Updated On: Jun 10, 2026
- An even function
- An odd function
- Neither even nor odd
- A periodic function
Show Solution
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The Correct Option is B
Solution and Explanation
Concept:
To determine whether a function is even or odd, we evaluate \(f(-x)\).
\[
f(-x)=f(x)
\]
implies the function is even.
\[
f(-x)=-f(x)
\]
implies the function is odd.
We use the important identity
\[
(x+\sqrt{x^2+1})(\sqrt{x^2+1}-x)=1
\]
which plays a crucial role in simplifying logarithmic expressions.
Step 1: Compute \(f(-x)\) Given \[ f(x)=\sin\left(\log\left(x+\sqrt{x^2+1}\right)\right) \] Replacing \(x\) by \(-x\), \[ f(-x) = \sin\left( \log\left( -x+\sqrt{x^2+1} \right) \right) \]
Step 2: Use the standard identity Since \[ -x+\sqrt{x^2+1} = \sqrt{x^2+1}-x \] and \[ (x+\sqrt{x^2+1})(\sqrt{x^2+1}-x)=1 \] we obtain \[ \sqrt{x^2+1}-x = \frac1{x+\sqrt{x^2+1}} \] Therefore, \[ f(-x) = \sin\left( \log\frac1{x+\sqrt{x^2+1}} \right) \]
Step 3: Apply logarithmic property Using \[ \log\frac1a=-\log a \] we get \[ f(-x) = \sin\left( -\log(x+\sqrt{x^2+1}) \right) \]
Step 4: Use odd property of sine Since \[ \sin(-\theta)=-\sin\theta \] we have \[ f(-x) = -\sin\left( \log(x+\sqrt{x^2+1}) \right) \] \[ f(-x)=-f(x) \] Hence the function is odd. \[ \boxed{\text{The function is odd}} \]
Step 1: Compute \(f(-x)\) Given \[ f(x)=\sin\left(\log\left(x+\sqrt{x^2+1}\right)\right) \] Replacing \(x\) by \(-x\), \[ f(-x) = \sin\left( \log\left( -x+\sqrt{x^2+1} \right) \right) \]
Step 2: Use the standard identity Since \[ -x+\sqrt{x^2+1} = \sqrt{x^2+1}-x \] and \[ (x+\sqrt{x^2+1})(\sqrt{x^2+1}-x)=1 \] we obtain \[ \sqrt{x^2+1}-x = \frac1{x+\sqrt{x^2+1}} \] Therefore, \[ f(-x) = \sin\left( \log\frac1{x+\sqrt{x^2+1}} \right) \]
Step 3: Apply logarithmic property Using \[ \log\frac1a=-\log a \] we get \[ f(-x) = \sin\left( -\log(x+\sqrt{x^2+1}) \right) \]
Step 4: Use odd property of sine Since \[ \sin(-\theta)=-\sin\theta \] we have \[ f(-x) = -\sin\left( \log(x+\sqrt{x^2+1}) \right) \] \[ f(-x)=-f(x) \] Hence the function is odd. \[ \boxed{\text{The function is odd}} \]
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