Question

If $f : R - \{2\} \to R$ is a function defined by $f(x) = \frac{x^2 - 4}{x - 2}$ , then its range is

• A. R
• B. R - {2}
• C. R - {4}
• D. R - {-2, 2}

Answer 1

The Correct Option is C

We have,
$f(x)=\frac{x^{2}-4}{x-2}, x \neq 2 $
$f(x)=\frac{(x+2)(x-2)}{x-2}, x \neq 2 $
$f(x)=(x+2), x \neq 2$
$\therefore$ Range of $f(x)=R-\{4\}$

Answer 2

Given:
\(f(x) = \frac{x^2 - 4}{x - 2}\)

\(f(x) = \frac{(x - 2)(x + 2)}{x - 2}\)
For \(x \neq 2\), this simplifies to:
\(f(x) = x + 2\)

Finding the Range:
The simplified function \(f(x) = x + 2\) can take any real value except at x = 2. When x = 2, the original function is undefined, so \(f(x)\) cannot be 4.

Therefore, the range of f(x) is all real numbers except 4:
\(\text{Range of } f(x) = \mathbb{R} - \{4\}\)

So, the correct option is (C): \(\mathbb{R} - \{4\}\)