Question

Identify even, odd or NENO

f(x) =ln[x+(x^2-1) ]

• A. Even
• B. Odd
• C. NENO

Answer 1

Answer: (C)

NENO

Answer 2

To determine if a function $f(x)$ is even, odd, or neither (NENO), we must first examine its domain.

A function is classified as even or odd only if its domain is symmetric about the origin. This means that if $x$ is in the domain, then $-x$ must also be in the domain.

Let's find the domain of the given function $f(x) = \ln[x + \sqrt{x^2-1}]$.

1. Condition for the square root: For $\sqrt{x^2-1}$ to be a real number, the term inside the square root must be non-negative: $x^2 - 1 \ge 0$ $x^2 \ge 1$ This implies $x \le -1$ or $x \ge 1$. So, $x \in (-\infty, -1] \cup [1, \infty)$.

2. Condition for the logarithm: For $\ln(Y)$ to be defined, its argument $Y$ must be positive: $x + \sqrt{x^2-1} > 0$

Let's analyze this condition based on the possible values of $x$:

• Case 1: $x \ge 1$ If $x \ge 1$, then $x$ is positive. Also, $\sqrt{x^2-1}$ is non-negative. Therefore, their sum $x + \sqrt{x^2-1}$ will be positive. For example, if $x=1$, $1 + \sqrt{1^2-1} = 1+0 = 1 > 0$. If $x > 1$, then $x > 0$ and $\sqrt{x^2-1} > 0$, so $x + \sqrt{x^2-1} > 0$. Thus, for $x \in [1, \infty)$, the function $f(x)$ is defined.

• Case 2: $x \le -1$ Let $x = -y$, where $y \ge 1$. Then the argument of the logarithm becomes $-y + \sqrt{(-y)^2-1} = -y + \sqrt{y^2-1}$. We need to check if $-y + \sqrt{y^2-1} > 0$, which is equivalent to $\sqrt{y^2-1} > y$. Since $y \ge 1$, $y$ is positive. $\sqrt{y^2-1}$ is non-negative. Squaring both sides of the inequality $\sqrt{y^2-1} > y$: $y^2 - 1 > y^2$ $-1 > 0$ This inequality is false. In fact, for $y \ge 1$, we know that $y^2-1 < y^2$. Taking the square root of both non-negative sides, we get $\sqrt{y^2-1} < \sqrt{y^2} = |y|$. Since $y \ge 1$, $|y|=y$. So, $\sqrt{y^2-1} < y$. This implies $\sqrt{y^2-1} - y < 0$. Therefore, for $x \le -1$ (i.e., $y \ge 1$), the term $x + \sqrt{x^2-1}$ is always negative. Thus, $\ln[x + \sqrt{x^2-1}]$ is undefined for $x \in (-\infty, -1]$.

Combining both cases, the domain of $f(x)$ is $[1, \infty)$.

Check for symmetry of the domain: The domain of $f(x)$ is $D = [1, \infty)$. This domain is not symmetric about the origin. For instance, $2$ is in the domain ($2 \in D$), but $-2$ is not in the domain ($-2 \notin D$).

Since the domain of $f(x)$ is not symmetric about the origin, the function cannot be classified as even or odd. It is neither even nor odd (NENO).