Question

Let [.]denote the greatest integer function and f(x)=[x]+[2-x],-1≤x≤4.Then

• A. f is not differentiable at x=3/2
• B. f is continues at x=2
• C. f is continues at x=0
• D. f is not continues at x=1
• E. f is differentiable at x=3

Answer 1

Answer: (E)

Step 1: Analyze the function \( f(x) = [x] + |2 - x| \) where \([ \cdot ]\) is the greatest integer function and \(-1 \leq x \leq 4\).

Step 2: Evaluate option (A) - Continuity at \(x=2\):
Compute \(f(2) = [2] + |2-2| = 2 + 0 = 2\)
Left limit: \(\lim_{x\to2^-} f(x) = 1 + 0 = 1\)
Right limit: \(\lim_{x\to2^+} f(x) = 2 + 0 = 2\)
Since \(1 \neq 2\), \(f\) is not continuous at \(x=2\). Thus, (A) is false.

Step 3: Evaluate option (B) - Continuity at \(x=1\):
Compute \(f(1) = [1] + |2-1| = 1 + 1 = 2\)
Left limit: \(\lim_{x\to1^-} f(x) = 0 + 1 = 1\)
Right limit: \(\lim_{x\to1^+} f(x) = 1 + 1 = 2\)
Since \(1 \neq 2\), \(f\) is not continuous at \(x=1\). Thus, (B) is true.

Step 4: Evaluate option (C) - Continuity at \(x=0\):
Compute \(f(0) = [0] + |2-0| = 0 + 2 = 2\)
Left limit: \(\lim_{x\to0^-} f(x) = -1 + 2 = 1\)
Right limit: \(\lim_{x\to0^+} f(x) = 0 + 2 = 2\)
Since \(1 \neq 2\), \(f\) is not continuous at \(x=0\). Thus, (C) is false.

Step 5: Evaluate option (D) - Differentiability at \(x=3\):
First check continuity at \(x=3\):
Compute \(f(3) = [3] + |2-3| = 3 + 1 = 4\)
Left limit: \(\lim_{x\to3^-} f(x) = 2 + 1 = 3\)
Right limit: \(\lim_{x\to3^+} f(x) = 3 + 1 = 4\)
Since \(3 \neq 4\), \(f\) is not continuous (and thus not differentiable) at \(x=3\). Thus, (E) is false.

Conclusion: The only correct statement is \(\boxed{E}\).

Answer 2

Step 1: Analyze the function \( f(x) = \left\lfloor x \right\rfloor + |2 - x| \), where \( \left\lfloor x \right\rfloor \) represents the greatest integer function, and the domain is \( -1 \leq x \leq 4 \). 

Step 2: Evaluate option (A) - Continuity at \(x = 2\): 

- Compute \( f(2) = \left\lfloor 2 \right\rfloor + |2 - 2| = 2 + 0 = 2 \) - Left-hand limit: \( \lim_{x \to 2^-} f(x) = 1 + 0 = 1 \) - Right-hand limit: \( \lim_{x \to 2^+} f(x) = 2 + 0 = 2 \) 

- Since the left-hand and right-hand limits do not match (\(1 \neq 2\)), \( f \) is not continuous at \( x = 2 \). Therefore, option (A) is false. 

Step 3: Evaluate option (B) - Continuity at \(x = 1\): 

- Compute \( f(1) = \left\lfloor 1 \right\rfloor + |2 - 1| = 1 + 1 = 2 \) - Left-hand limit: \( \lim_{x \to 1^-} f(x) = 0 + 1 = 1 \) - Right-hand limit: \( \lim_{x \to 1^+} f(x) = 1 + 1 = 2 \) 

- Since the left-hand and right-hand limits do not match (\(1 \neq 2\)), \( f \) is not continuous at \( x = 1 \). Therefore, option (B) is true. 

Step 4: Evaluate option (C) 

- Continuity at \(x = 0\): - Compute \( f(0) = \left\lfloor 0 \right\rfloor + |2 - 0| = 0 + 2 = 2 \) - Left-hand limit: \( \lim_{x \to 0^-} f(x) = -1 + 2 = 1 \) - Right-hand limit: \( \lim_{x \to 0^+} f(x) = 0 + 2 = 2 \) 

- Since the left-hand and right-hand limits do not match (\(1 \neq 2\)), \( f \) is not continuous at \( x = 0 \). 

Therefore, option (C) is false. 

Step 5: Evaluate option (E) 

- Differentiability at \(x = 3\): 

- First, check continuity at \( x = 3 \): 

- Compute \( f(3) = \left\lfloor 3 \right\rfloor + |2 - 3| = 3 + 1 = 4 \) 

- Left-hand limit: \( \lim_{x \to 3^-} f(x) = 2 + 1 = 3 \) 

- Right-hand limit: \( \lim_{x \to 3^+} f(x) = 3 + 1 = 4 \) 

- Since the left-hand and right-hand limits do not match (\(3 \neq 4\)), \( f \) is not continuous (and thus not differentiable) at \( x = 3 \). 

Therefore, option (E) is false.