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

To determine the properties of the function \( f(x) = 10 - x^2 \), we'll analyze its behavior regarding injectivity (one-one) and surjectivity (onto).

Injectivity (One-One):

A function is one-one if different inputs produce different outputs. Suppose \( f(a) = f(b) \), then:

\(10 - a^2 = 10 - b^2\)

This simplifies to:

\(a^2 = b^2\)

Which implies:

\(a = b\) or \(a = -b\)

Since \(a = b\) is not the only possible outcome (as shown by \(a = -b\)), \(f(x)\) is not injective.

Surjectivity (Onto):

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

Given \(f(x) = 10 - x^2\), we solve for \(x^2\):

\(x^2 = 10 - y\)

This implies \(y \leq 10\) (since \(x^2 \geq 0\)). Every \(y \leq 10\) can be achieved by some real \(x\), for example, \(x = \sqrt{10 - y}\) or \(x = -\sqrt{10 - y}\).

Thus, \(f(x)\) covers all \(y \leq 10\), confirming \(f(x)\) is surjective.

Conclusion:

\(f(x)\) is onto but not one-one. Therefore, the correct answer is: \(f\) is onto but not one-one.

Answer 2

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.