Question

If a function \( f: \mathbb{R} \to \mathbb{R} \) is defined by \( f(x) = x^3 - x \), then \( f \) is:

• A. \text{One-one and onto}
• B. \text{One-one but not onto}
• C. \text{Onto but not one-one}
• D. \text{Neither one-one nor onto}

Hint:

To check if a function is one-one, differentiate and check monotonicity. If it changes sign, the function is not one-one. To check onto, verify if the function covers the entire codomain.

Answer 1

The Correct Option is C

Step 1: Check for One-One (Injectivity) A function is one-one if it is monotonic or if \( f(a) = f(b) \) implies \( a = b \). \[ f(x) = x^3 - x \] Differentiate to check monotonicity: \[ f'(x) = 3x^2 - 1 \] Setting \( f'(x) = 0 \): \[ 3x^2 - 1 = 0 \Rightarrow x = \pm \frac{1}{\sqrt{3}} \] Since \( f'(x) \) changes sign, \( f(x) \) is not one-one. Step 2: Check for Onto (Surjectivity) To check onto, solve for \( y \) in terms of \( x \): \[ y = x^3 - x \] Rewriting, \[ g(x) = x^3 - x \] \[ \lim_{x \to \infty} f(x) = \infty, \quad \lim_{x \to -\infty} f(x) = -\infty \] Since \( f(x) \) covers all real values \( y \), the function is onto. Conclusion:
- \( f(x) \) is not one-one (fails injectivity test).
- \( f(x) \) is onto (covers all real numbers).
- Hence, the function is onto but not one-one.