Question

If \( f(x) = \frac{1}{2 - x} \) and \( g(x) = \frac{1}{1 - x} \), then the point(s) of discontinuity of the function \( g(f(x)) \) is (are):

• A. \( x = 2 \)
• B. \( x = 3 \)
• C. \( x = 2, x = 3 \)
• D. \( x = 2, x = 1 \)
• E. \( x = 1, x = -2 \)

Hint:

The points of discontinuity of composite functions occur wherever the individual functions have discontinuities.

Answer 1

The Correct Option is D

We are given the composite function \( g(f(x)) \), where \( f(x) = \frac{1}{2-x} \) and \( g(x) = \frac{1}{1-x} \). 
To find the points of discontinuity of \( g(f(x)) \), we need to examine where the individual functions \( f(x) \) and \( g(x) \) are discontinuous. 
1. The function \( f(x) = \frac{1}{2 - x} \) is discontinuous where the denominator is zero, i.e., at \( x = 2 \). 
2. The function \( g(x) = \frac{1}{1 - x} \) is discontinuous where the denominator is zero, i.e., at \( x = 1 \). 
Thus, \( g(f(x)) \) is discontinuous where \( f(x) = 2 \) or \( g(x) = 1 \). 
- \( f(x) = 2 \) gives \( \frac{1}{2 - x} = 2 \), leading to \( x = 0 \), which is a valid solution. 
- \( g(x) = 1 \) gives \( \frac{1}{1 - x} = 1 \), leading to \( x = 0 \), which also affects the discontinuity. 
Hence, the discontinuities occur at \( x = 2 \) and \( x = 1 \).