CA Foundation · Quantitative Aptitude · Sets, Relations and Functions, Limits and Continuity
The function f is defined by f(x) = (x² − 9)/(x − 3) for x ≠ 3 and f(3) = k. For what value of k is f continuous at x = 3?
k must be 6. For x not equal to 3, (x²−9)/(x−3) simplifies to x+3, so the limit at 3 is 6. Continuity requires f(3) to equal this limit, hence k = 6. Choosing 0 or 3 ignores the simplification of the factor.
- A3
- B6Correct
- C9
- D0
Explanation
For x ≠ 3, (x² − 9)/(x − 3) = (x − 3)(x + 3)/(x − 3) = x + 3. So the limit as x tends to 3 is 6. Continuity needs f(3) = this limit, so k = 6. Putting k = 3 or 0 ignores the cancellation of the common factor.
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