Question

Let
ƒ : R → R
be defined as f(x) = x -1 and
g : R - { 1, -1 } → R
be defined as
g(x) = \(\frac{x²}{x² - 1}\)
Then the function fog is :

• A. One-one but not onto
• B. Onto but not one-one
• C. Both one-one and onto
• D. Neither one-one nor onto

Answer 1

The Correct Option is D

To find the composition of the functions \( f \) and \( g \), referred to as \( f \circ g \), we must evaluate \( f(g(x)) \) for the given functions \( f(x) = x - 1 \) and \( g(x) = \frac{x^2}{x^2 - 1} \).

Step-by-step Evaluation:

• Evaluate \( g(x) \):
  Given \( g(x) = \frac{x^2}{x^2 - 1} \), this function is defined for all real numbers except \( x = 1 \) and \( x = -1 \), as these values lead to a denominator of zero.
• Evaluate \( f(g(x)) \):
  Start by substituting \( g(x) \) into \( f(x) \): f(g(x)) = f\left(\frac{x^2}{x^2 - 1}\right) = \frac{x^2}{x^2 - 1} - 1
  Simplifying this expression, we get:
  • Combine the terms: \(\frac{x^2}{x^2 - 1} - 1 = \frac{x^2 - (x^2 - 1)}{x^2 - 1}\)
  • Simplify the numerator: x^2 - x^2 + 1 = 1
  • So, f(g(x)) = \frac{1}{x^2 - 1}

Analysis of the Composition \( f \circ g \):

• The function f(g(x)) = \frac{1}{x^2 - 1} is not one-one because different values of \( x \) can produce the same value of the function (e.g., \( x \) and \(-x\)).
• The function is not onto the domain of real numbers \( \mathbb{R} \) because the range of the function does not cover all real numbers—especially negative numbers are not included as the range due to the positive nature of \( \frac{1}{x^2 - 1} \) for defined values of \( x \).

Therefore, the function composition \( f \circ g \) is neither one-one nor onto.

Hence, the correct answer is: Neither one-one nor onto.