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If f(x)={:{(3", if " 0 le x le 1),(4", i...

If `f(x)={:{(3", if " 0 le x le 1),(4", if " 1 lt x lt 3),(5", if " 3 le x le 10):}`, then

A

f is continuous on [0, 10] except at `x=1, 3`

B

f is continuous on [0, 10] except at `x=1`

C

f is continuous on [0, 10] except at `x=3`

D

f is continuous on [0, 10]

Text Solution

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The correct Answer is:
To determine the continuity of the piecewise function \[ f(x) = \begin{cases} 3 & \text{if } 0 \leq x \leq 1 \\ 4 & \text{if } 1 < x < 3 \\ 5 & \text{if } 3 \leq x \leq 10 \end{cases} \] we need to check the continuity at the points where the function changes its definition, specifically at \(x = 1\) and \(x = 3\). ### Step 1: Check continuity at \(x = 1\) 1. **Find \(f(1)\)**: Since \(1\) is included in the first case, we have: \[ f(1) = 3 \] 2. **Find the left-hand limit as \(x\) approaches \(1\)**: \[ \lim_{x \to 1^-} f(x) = 3 \quad (\text{since } f(x) = 3 \text{ for } 0 \leq x \leq 1) \] 3. **Find the right-hand limit as \(x\) approaches \(1\)**: \[ \lim_{x \to 1^+} f(x) = 4 \quad (\text{since } f(x) = 4 \text{ for } 1 < x < 3) \] 4. **Compare limits and function value**: \[ \lim_{x \to 1^-} f(x) = 3 \quad \text{and} \quad \lim_{x \to 1^+} f(x) = 4 \] Since \(3 \neq 4\), \(f(x)\) is not continuous at \(x = 1\). ### Step 2: Check continuity at \(x = 3\) 1. **Find \(f(3)\)**: Since \(3\) is included in the third case, we have: \[ f(3) = 5 \] 2. **Find the left-hand limit as \(x\) approaches \(3\)**: \[ \lim_{x \to 3^-} f(x) = 4 \quad (\text{since } f(x) = 4 \text{ for } 1 < x < 3) \] 3. **Find the right-hand limit as \(x\) approaches \(3\)**: \[ \lim_{x \to 3^+} f(x) = 5 \quad (\text{since } f(x) = 5 \text{ for } 3 \leq x \leq 10) \] 4. **Compare limits and function value**: \[ \lim_{x \to 3^-} f(x) = 4 \quad \text{and} \quad \lim_{x \to 3^+} f(x) = 5 \] Since \(4 \neq 5\), \(f(x)\) is not continuous at \(x = 3\). ### Conclusion The function \(f(x)\) is continuous on the intervals \( (0, 1) \) and \( (1, 3) \) and \( (3, 10) \), but it is not continuous at \(x = 1\) and \(x = 3\). Therefore, we conclude that: - **f is continuous on \( [0, 10] \) except at \(x = 1\) and \(x = 3\)**.
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  • If f(x)={:{(3x+5", for " 0 le x lt 3),(2x+8", for " 3 le x lt 5),(x+13", for " 5 le x le 10):} , then

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    f is continuous at `x=1`
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    f is discontinuous at `x=1`
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