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Let \[ f(x)=\begin{cases} \dfrac{\sin[x]}{([x]+1)^2}, & x>0,\\[6pt] \dfrac{\cos\left(\frac{\pi}{2}[x]\right)}{[x]}, & x<0, \end{cases} \] where \([x]\) denotes the greatest integer not exceeding \(x\). For \(f\) to be continuous at \(x=0\), the value of \(k\) must be

1. 0
2. 1
3. -1
4. indeterminate

Correct answer: 1

Solution

For the function to be continuous at x=0, the left-hand limit as x approaches 0 from the negative side must equal the right-hand limit as x approaches 0 from the positive side, both equating to the function's value at x=0. Evaluating these limits shows that they converge to 1, thus requiring k to be 1 for continuity.

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