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Question

A function f defined by f(x) = In \(\left( {\sqrt {{x^2} + 1} - x} \right)\) is

This question was previously asked in
NDA II 2019 GAT Previous Year Paper (17-Nov-2019)
The correct answer is

an odd function

Understanding Function Classification: Even or Odd?

Functions can be classified based on their symmetry properties. These classifications are known as even functions and odd functions. Understanding these classifications is crucial in calculus and function analysis.

Even Function: A function \(f(x)\) is considered an even function if, for every value of x in its domain, \(f(-x) = f(x)\). Even functions are symmetric with respect to the y-axis.

Odd Function: A function \(f(x)\) is considered an odd function if, for every value of x in its domain, \(f(-x) = -f(x)\). Odd functions are symmetric with respect to the origin.

A function can also be neither even nor odd if it does not satisfy either of these conditions. It is rare for a non-zero function to be both even and odd (only \(f(x) = 0\) satisfies both).

Analyzing the Given Function \(f(x) = \ln \left( {\sqrt {{x^2} + 1} - x} \right)\)

To determine if the given function \(f(x) = \ln \left( {\sqrt {{x^2} + 1} - x} \right)\) is even, odd, or neither, we need to evaluate \(f(-x)\) and compare it with \(f(x)\) and \(-f(x)\).

Step 1: Find \(f(-x)\)

Substitute \(-x\) for \(x\) in the function definition:

\[ f(-x) = \ln \left( {\sqrt {{(-x)}^2} + 1} - (-x) \right) \] \[ f(-x) = \ln \left( {\sqrt {{x^2} + 1} + x} \right) \]

Step 2: Compare \(f(-x)\) with \(f(x)\)

We have \(f(x) = \ln \left( {\sqrt {{x^2} + 1} - x} \right)\) and \(f(-x) = \ln \left( {\sqrt {{x^2} + 1} + x} \right)\). These two expressions are not immediately equal, so the function is likely not even.

Step 3: Manipulate \(f(-x)\) to see if it relates to \(-f(x)\)

Consider the argument of the logarithm in \(f(-x)\), which is \({\sqrt {{x^2} + 1} + x}\). Let's multiply this expression by its conjugate \({\sqrt {{x^2} + 1} - x}\). This is a common technique when dealing with expressions involving square roots and sums/differences.

\[ \left( {\sqrt {{x^2} + 1} + x} \right) \left( {\sqrt {{x^2} + 1} - x} \right) \]

Using the difference of squares formula \((a+b)(a-b) = a^2 - b^2\):

\[ = \left( {\sqrt {{x^2} + 1} } \right)^2 - x^2 \] \[ = (x^2 + 1) - x^2 \] \[ = 1 \]

So, we have \( \left( {\sqrt {{x^2} + 1} + x} \right) \left( {\sqrt {{x^2} + 1} - x} \right) = 1 \).

From this, we can express the argument of \(f(-x)\) in terms of the argument of \(f(x)\):

\[ {\sqrt {{x^2} + 1} + x} = \frac{1}{{\sqrt {{x^2} + 1} - x}} \]

We can rewrite the right side using a negative exponent:

\[ {\sqrt {{x^2} + 1} + x} = \left( {\sqrt {{x^2} + 1} - x} \right)^{-1} \]

Step 4: Substitute the manipulated expression back into \(f(-x)\)

\[ f(-x) = \ln \left( \left( {\sqrt {{x^2} + 1} - x} \right)^{-1} \right) \]

Using the logarithm property \(\ln(a^b) = b \ln(a)\):

\[ f(-x) = -1 \cdot \ln \left( {\sqrt {{x^2} + 1} - x} \right) \] \[ f(-x) = - \ln \left( {\sqrt {{x^2} + 1} - x} \right) \]

Step 5: Final Comparison

We see that the expression on the right side, \( - \ln \left( {\sqrt {{x^2} + 1} - x} \right) \), is exactly \(-f(x)\).

\[ f(-x) = -f(x) \]

This result matches the definition of an odd function.

Furthermore, the domain of \(f(x)\) requires \( {\sqrt {{x^2} + 1} - x} > 0 \). This inequality \(\sqrt{x^2+1} > x\) is true for all real \(x\), meaning the domain is \((-\infty, \infty)\), which is symmetric about the origin, as required for a function to be even or odd.

Conclusion on Function Classification

Since \(f(-x) = -f(x)\) for all \(x\) in its domain, the function \(f(x) = \ln \left( {\sqrt {{x^2} + 1} - x} \right)\) is an odd function.

Function Property Condition Result for \(f(x) = \ln \left( {\sqrt {{x^2} + 1} - x} \right)\)
Even \(f(-x) = f(x)\) \( \ln \left( {\sqrt {{x^2} + 1} + x} \right) \ne \ln \left( {\sqrt {{x^2} + 1} - x} \right) \) (Not even)
Odd \(f(-x) = -f(x)\) \( \ln \left( {\sqrt {{x^2} + 1} + x} \right) = - \ln \left( {\sqrt {{x^2} + 1} - x} \right) \) (Odd)

Revision Table: Key Function Properties

Property Definition Symmetry Example
Even Function \(f(-x) = f(x)\) Symmetric about the y-axis \(f(x) = x^2\), \(f(x) = \cos(x)\)
Odd Function \(f(-x) = -f(x)\) Symmetric about the origin \(f(x) = x^3\), \(f(x) = \sin(x)\)

Additional Information on Function Types

Beyond even and odd, functions can have other important classifications and properties:

Injective (One-to-One) Function: A function where distinct elements in the domain map to distinct elements in the codomain. If \(f(x_1) = f(x_2)\), then \(x_1 = x_2\).

Surjective (Onto) Function: A function where every element in the codomain is mapped to by at least one element in the domain. The range equals the codomain.

Bijective Function: A function that is both injective and surjective. Bijective functions have inverse functions.

Periodic Function: A function that repeats its values in regular intervals or periods. \(f(x+P) = f(x)\) for some constant period \(P\).

Monotonic Function: A function that is either entirely non-increasing or entirely non-decreasing.

These classifications help in analyzing the behavior and properties of functions, which is fundamental in calculus and higher mathematics.

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