CUET · MATHS · PYQ PAPER 2023
If \(f(x)=\left\{\begin{array}{ll}\frac{x^2-9}{x-3}, & x \neq 3 \\ 5, & x=3\end{array}\right.\) then \(f(x):\)
- A is continuous at \(x=3\)
- B has removable discontinuity at \(x=3\)
- C has irremovable discontinuity at \(x=3\)
- D continuous at every real number
Answer & Solution
Correct Answer
(B) has removable discontinuity at \(x=3\)
Step-by-step Solution
Detailed explanation
\(\lim_{x \to 3} f(x) = \lim_{x \to 3} \frac{(x-3)(x+3)}{x-3} = \lim_{x \to 3} (x+3) = 3+3 = 6\) \(f(3) = 5\) Since \(\lim_{x \to 3} f(x) \neq f(3)\) and \(\lim_{x \to 3} f(x)\) exists, the function has removable discontinuity at \(x=3\).
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