KCET · Maths · Limits
If \(\lim\limits_{x \to 3}\left(\dfrac{x^2 - ax - 3b}{x - 3}\right) = 5\), then \(a + b = \)
- A \(1\)
- B \(2\)
- C \(3\)
- D \(4\)
Answer & Solution
Correct Answer
(C) \(3\)
Step-by-step Solution
Detailed explanation
For the limit to exist and be finite, the expression must take the \(\dfrac{0}{0}\) form at \(x = 3\).
Substituting \(x = 3\) in the numerator gives:
\(3^2 - a(3) - 3b = 0\)
\(9 - 3a - 3b = 0\)
\(3(a + b) = 9\)
\(a + b = 3\)
Applying L'Hopital's rule to find the individual values:
\(\lim\limits_{x \to 3} \dfrac{2x - a}{1} = 5\)
\(6 - a = 5 \Rightarrow a = 1\)
Substituting \(a = 1\) into \(a + b = 3\) gives \(b = 2\).
Thus, \(a + b = 1 + 2 = 3\).
Answer: \(3\)
Substituting \(x = 3\) in the numerator gives:
\(3^2 - a(3) - 3b = 0\)
\(9 - 3a - 3b = 0\)
\(3(a + b) = 9\)
\(a + b = 3\)
Applying L'Hopital's rule to find the individual values:
\(\lim\limits_{x \to 3} \dfrac{2x - a}{1} = 5\)
\(6 - a = 5 \Rightarrow a = 1\)
Substituting \(a = 1\) into \(a + b = 3\) gives \(b = 2\).
Thus, \(a + b = 1 + 2 = 3\).
Answer: \(3\)
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