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The function f(x)=(tan |pi[x-pi]|)/(1+[x...

The function `f(x)=(tan |pi[x-pi]|)/(1+[x]^(2))`, where [x] denotes the greatest integer less than or equal to x, is

A

f(x) is discontinuous at some x

B

f(x) is continuous at all x, but the derivative f'(x) does not exist for some x

C

`f'(x)` exists for all x, but f''(x) does not exist for some x

D

f'(x) exists for all x

Text Solution

AI Generated Solution

The correct Answer is:
To analyze the function \( f(x) = \frac{\tan(\pi |x - \pi|)}{1 + [x]^2} \), where \([x]\) denotes the greatest integer less than or equal to \(x\), we will go through the following steps: ### Step 1: Understanding the Function We start with the function: \[ f(x) = \frac{\tan(\pi |x - \pi|)}{1 + [x]^2} \] Here, \(|x - \pi|\) is the absolute value of \(x - \pi\) and \([x]\) is the greatest integer function. ### Step 2: Analyzing the Numerator The numerator is \(\tan(\pi |x - \pi|)\). The tangent function is periodic with period \(\pi\): - For any integer \(n\), \(\tan(n\pi) = 0\). - Therefore, \(\tan(\pi |x - \pi|) = 0\) whenever \(|x - \pi| = n\) for any integer \(n\). ### Step 3: Analyzing the Denominator The denominator is \(1 + [x]^2\): - The greatest integer function \([x]\) is always an integer, so \([x]^2\) is non-negative. - Thus, \(1 + [x]^2 \geq 1\) for all \(x\), meaning it is always positive and never zero. ### Step 4: Simplifying the Function Since the numerator \(\tan(\pi |x - \pi|) = 0\) for all \(x\) (as \(|x - \pi|\) will take values that make the tangent zero), we can conclude: \[ f(x) = \frac{0}{1 + [x]^2} = 0 \] for all \(x\). ### Step 5: Conclusion about Continuity and Differentiability Since \(f(x) = 0\) for all \(x\): - \(f(x)\) is a constant function. - A constant function is continuous everywhere. - The derivative \(f'(x) = 0\) exists for all \(x\). - The second derivative \(f''(x) = 0\) also exists for all \(x\). ### Final Answer Thus, we conclude: - The function \(f(x)\) is continuous and differentiable everywhere, and both the first and second derivatives exist for all \(x\). ### Summary of Options 1. **Incorrect**: \(f'(x)\) does exist for all \(x\). 2. **Incorrect**: \(f(x)\) is continuous and \(f'(x)\) exists. 3. **Incorrect**: \(f''(x)\) exists for all \(x\). 4. **Correct**: \(f(x)\) is a constant function.
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