Question:
\[ f(x)= \begin{cases}{\left[x^2\right]-\left[-x^2\right],} & x \neq 3 \\ k & , x=3\end{cases} \]
is continuous at \(x=3\), then \(\mathrm{k}=\) where \([\cdot]\) is greatest integer function
\[ f(x)= \begin{cases}{\left[x^2\right]-\left[-x^2\right],} & x \neq 3 \\ k & , x=3\end{cases} \]
is continuous at \(x=3\), then \(\mathrm{k}=\) where \([\cdot]\) is greatest integer function
Show Hint
Check the left and right limits at x = 3 using [a] - [-a] = 2[a] + 1 for non-integer a.
Updated On: Nov 13, 2026
- 0
- 1
- -1
- Does not exist
Show Solution
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The Correct Option is D
Solution and Explanation
Step 1: Understanding the Concept
For non-integer \(a\), \([a]+[-a]=-1\), so \([a]-[-a]=2[a]+1\).
Step 2: Left limit
As \(x\to3^-\), \(x^2\to9^-\), so \([x^2]=8\) and the expression is \(2(8)+1=17\).
Step 3: Right limit
As \(x\to3^+\), \(x^2\to9^+\), so \([x^2]=9\) and the expression is \(2(9)+1=19\).
Step 4: Conclusion
The two limits differ (17 and 19), so no value of \(k\) makes \(f\) continuous.
Final Answer:
No such value of k exists. \[ \boxed{\text{(D) }\text{Does not exist}} \]
For non-integer \(a\), \([a]+[-a]=-1\), so \([a]-[-a]=2[a]+1\).
Step 2: Left limit
As \(x\to3^-\), \(x^2\to9^-\), so \([x^2]=8\) and the expression is \(2(8)+1=17\).
Step 3: Right limit
As \(x\to3^+\), \(x^2\to9^+\), so \([x^2]=9\) and the expression is \(2(9)+1=19\).
Step 4: Conclusion
The two limits differ (17 and 19), so no value of \(k\) makes \(f\) continuous.
Final Answer:
No such value of k exists. \[ \boxed{\text{(D) }\text{Does not exist}} \]
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