Question

Consider the following two statements.
P : There exist functions f: ℝ → ℝ, g: ℝ → ℝ such that f is continuous at x = 1 and g is discontinuous at x = 1 but g ∘ f is continuous at x = 1.
Q : There exist functions f: ℝ → ℝ, g: ℝ → ℝ such that both f and g are discontinuous at x = 1 but g ∘ f is continuous at x = 1.
Which one of the following holds ?

• A. Both P and Q are true
• B. Both P and Q are false
• C. P is true but Q is false
• D. P is false but Q is true

Answer 1

The Correct Option is A

- Statement P: 
Consider the functions \( f(x) = x \) and \( g(x) = \text{where } g(x) = 1 \text{ for } x \neq 1 \text{ and } g(x) = 0 \text{ for } x = 1 \). Here, \( f \) is continuous at \( x = 1 \) and \( g \) is discontinuous at \( x = 1 \), but \( g(f(x)) = g(x) = 1 \) for all \( x \neq 1 \), and at \( x = 1 \), \( g(f(1)) = g(1) = 0 \), so the composition \( g(f(x)) \) is continuous at \( x = 1 \). 
Hence, statement P is true. 

Statement Q: Consider the functions \( f(x) = \text{where } f(x) = 1 \text{ for } x \neq 1 \text{ and } f(x) = 0 \text{ for } x = 1 \) and \( g(x) = x \). Here, both \( f \) and \( g \) are discontinuous at \( x = 1 \), but the composition \( g(f(x)) = 1 \) for all \( x \neq 1 \) and \( g(f(1)) = g(0) = 0 \), so \( g \circ f \) is continuous at \( x = 1 \). Hence, statement Q is also true. Thus, both statements P and Q are true. 
Therefore, the correct answer is (A): Both P and Q are true. 

Therefore, the correct answer is (A): Both P and Q are true.