Find the range of the real function f(x) = \(\rm \frac{x+1}{x-3}\)

Concept:

Range: The range of a function is the set of all possible values it can produce, i.e., all values of y for which x is defined.

Note:

The domain of a function f(x) is the set of all values for which the function is defined, and the range of the function is the set of all values that f takes.

Calculation:

Let, y = f(x) = \(\rm \frac{x+1}{x-3}\)

⇒y(x - 3) = x + 1

⇒yx - 3y - x = 1

⇒ x(y - 1) - 3y = 1

⇒ x(y - 1) = 1 + 3y

⇒\(\rm x = \frac{1+3y}{y-1}\)

It is clear that x is not defined when y - 1 = 0, i.e, when  y = 1

∴ Range (f) = R - {1}

Hence, option (2) is correct.

Mistake PointsIt is given in the Question that, f(x) is real function. So,

f(x) has real values for value of x other than x = 3

∴ Domain of given function = R - {3}, where R is set of all real numbers