Question

Let \(f:R→R\) be a function defined by
\(f(x) =   \begin{cases}     3e^x       & \quad \text {if}\  {x<0}\\     x^2+3x+3  & \quad \text {if}\  0≤x<1   \\ x^2-3x-3 & \quad \text {if} \ x≥1\end{cases}\)

• A. f is continuous on R\{1}
• B. f is not continuous on R
• C. f is continuous on R\{0}
• D. f is continuous on R
• E. f is not continuous on R\{0,1}

Answer 1

The Correct Option is A

Step 1: Analyze the piecewise function \( f(x) \) at critical points \( x = 0 \) and \( x = 1 \):

Step 2: Check continuity at \( x = 0 \): \[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} 3e^x = 3 \] \[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (x^2 + 3x + 3) = 3 \] \[ f(0) = 3 \] Since all three values match, \( f \) is continuous at \( x = 0 \).

Step 3: Check continuity at \( x = 1 \): \[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (x^2 + 3x + 3) = 1 + 3 + 3 = 7 \] \[ \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (x^2 - 3x - 3) = 1 - 3 - 3 = -5 \] Since \( 7 \neq -5 \), \( f \) is not continuous at \( x = 1 \).

Step 4: Verify continuity elsewhere: - For \( x < 0 \): \( 3e^x \) is continuous - For \( 0 \leq x < 1 \): \( x^2 + 3x + 3 \) is continuous - For \( x > 1 \): \( x^2 - 3x - 3 \) is continuous

Conclusion: The function is discontinuous only at \( x = 1 \), making option (D) correct: \[ \boxed{A} \text{ (f is continuous on } \mathbb{R} \setminus \{1\}) \]

Answer 2

Let's analyze the continuity of the piecewise function \( f(x) \) at \( x = 0 \) and \( x = 1 \).

1. Continuity at \( x = 0 \):

• Left-hand limit: \( \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} 3e^x = 3e^0 = 3 \)
• Right-hand limit: \( \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (x^2 + 3x + 3) = 0^2 + 3(0) + 3 = 3 \)
• Function value: \( f(0) = 0^2 + 3(0) + 3 = 3 \)

Since the left-hand limit, right-hand limit, and function value are all equal to 3, \( f(x) \) is continuous at \( x = 0 \).

2. Continuity at \( x = 1 \):

• Left-hand limit: \( \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (x^2 + 3x + 3) = 1^2 + 3(1) + 3 = 7 \)
• Right-hand limit: \( \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (x^2 - 3x - 3) = 1^2 - 3(1) - 3 = -5 \)
• Function value: \( f(1) = 1^2 - 3(1) - 3 = -5 \)

Since the left-hand limit (7) and the right-hand limit (-5) are not equal, \( f(x) \) is discontinuous at \( x = 1 \).

Conclusion:

The function \( f(x) \) is continuous everywhere except at \( x = 1 \). Therefore, the statement "f is continuous on \( \mathbb{R} \setminus \{1\} \)" is true. The other statements are false.