Question

Let \( f : \mathbb{R} \to \mathbb{R} \) be defined as \( f(x) = 10 - x^2 \), then:

• A. \( f \) is one-one and onto.
• B. \( f \) is one-one but not onto.
• C. \( f \) is neither one-one nor onto.
• D. \( f \) is onto but not one-one.

Answer 1

The Correct Option is D

The given function is:

\[ f(x) = 10 - x^2. \]

Step 1: Check for one-one.

A function \(f\) is one-one if for \(f(x_1) = f(x_2)\), we have \(x_1 = x_2\). Assume:

\[ f(x_1) = f(x_2) \implies 10 - x_1^2 = 10 - x_2^2. \]

Simplify:

\[ x_1^2 = x_2^2 \implies x_1 = \pm x_2. \]

Since \(x_1 \neq x_2\) in general, the function is not one-one.

Step 2: Check for onto.

A function \(f\) is onto if for every \(y \in \mathbb{R}\), there exists an \(x \in \mathbb{R}\) such that \(f(x) = y\). Rearrange:

\[ f(x) = 10 - x^2 \quad \text{to solve for } x: \] \[ y = 10 - x^2 \implies x^2 = 10 - y. \]

For \(x^2 \geq 0\), we require \(10 - y \geq 0\), or:

\[ y \leq 10. \]

The function \(f(x)\) maps \(x \in \mathbb{R}\) to \(y \in (-\infty, 10]\). Hence, \(f\) is onto.

Conclusion: The function \(f(x) = 10 - x^2\) is onto but not one-one.