Question

Let $f : \mathbb{R} \to \mathbb{R}$ be defined by $f(x) = x^2 + 1$. Then the pre-images of $17$ and $-3$ respectively are:

• A. $\phi, \{4, -4\}$
• B. $\{3, -3\}, \phi$
• C. $\{4, -4\}, \phi$
• D. $\{4, -4\}, \{2, -2\}$

Answer 1

The Correct Option is C

1. Understand the problem:

We are given the function \( f(x) = x^2 + 1 \) and need to find the pre-images of 17 and -3, i.e., the set of all \( x \) such that \( f(x) = 17 \) and \( f(x) = -3 \).

2. Find the pre-image of 17:

Solve \( x^2 + 1 = 17 \):

\[ x^2 = 16 \implies x = \pm 4 \]

Thus, the pre-image is \( \{4, -4\} \).

3. Find the pre-image of -3:

Solve \( x^2 + 1 = -3 \):

\[ x^2 = -4 \]

This has no real solutions, so the pre-image is the empty set \( \phi \).

Correct Answer: (C) \(\{4, -4\}, \phi\)

Answer 2

The function is given by \( f(x) = x^2 + 1 \). We need to find the pre-images of 17 and -3.

Pre-image of 17:

We want to find \( x \) such that \( f(x) = 17 \). This means:

\[ x^2 + 1 = 17 \] \[ x^2 = 16 \] \[ x = \pm 4 \]

Therefore, the pre-image of 17 is \( \{4, -4\} \).

Pre-image of -3:

We want to find \( x \) such that \( f(x) = -3 \). This means:

\[ x^2 + 1 = -3 \] \[ x^2 = -4 \]

Since there are no real numbers whose square is negative, there are no real solutions for \( x \). 

Therefore, the pre-image of -3 is the empty set, denoted by \( \phi \).

So the pre-images of 17 and -3 are \( \{4, -4\} \) and \( \phi \), respectively.