JEE Mains · Maths · STD 11 - 12. limits
If \(\mathop {\lim }\limits_{x \to 2} \frac{{\tan \left( {x - 2} \right)\{ {x^2} + (k - 2)x - 2k\} }}{{{x^2} - 4x + 4}} = 5\) , then \(k\) is equal to
- A \(0\)
- B \(1\)
- C \(2\)
- D \(3\)
Answer & Solution
Correct Answer
(D) \(3\)
Step-by-step Solution
Detailed explanation
\(\left( d \right)\,\,\mathop {\lim }\limits_{x \to 2} \frac{{\tan \left( {x - 2} \right)\left\{ {{x^2} + \left( {k - 2} \right)x - 2k} \right\}}}{{{x^2} - 4x + 4}} = 5\)…
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