JEE Mains · Maths · STD 11 - 12. limits
Let \(\lim\limits_{x \to 2} \dfrac{(\tan(x-2))(rx^2 + (p-2)x - 2p)}{(x-2)^2} = 5\) for some \(r, p \in \mathbb{R}\). If the set of all possible values of \(q\), such that the roots of the equation \(rx^2 - px + q = 0\) lie in \((0, 2)\), be the interval \((\alpha, \beta]\), then \(4(\alpha + \beta)\) equals :
- A \(11\)
- B \(13\)
- C \(17\)
- D \(21\)
Answer & Solution
Correct Answer
(C) \(17\)
Step-by-step Solution
Detailed explanation
Given limit is \(\lim\limits_{x \to 2} \dfrac{\tan(x-2)}{x-2} \cdot \dfrac{rx^2 + (p-2)x - 2p}{x-2} = 5\) Since \(\lim\limits_{x \to 2} \dfrac{\tan(x-2)}{x-2} = 1\), we have: \(\lim\limits_{x \to 2} \dfrac{rx^2 + (p-2)x - 2p}{x-2} = 5\) For the limit to exist, the numerator must…
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