JEE Mains · Maths · STD 11 - 12. limits
Let a be an integer such that \(\lim \limits_{x \rightarrow 7} \frac{18-[1-x]}{[x-3 a]}\) exists, where \([ t ]\) is greatest integer \(\leq t\). Then a is equal to
- A \(2\)
- B \(-2\)
- C \(-6\)
- D \(6\)
Answer & Solution
Correct Answer
(C) \(-6\)
Step-by-step Solution
Detailed explanation
\(\lim \limits_{x \rightarrow 7} \frac{18-[1-x]}{[x]-3 a}\) \(L.H.L.\) \(\lim \limits_{x \rightarrow 7-} \frac{18-[1-x]}{[x]-3 a}\) \(=\frac{18-(-6)}{6-3 a}\) \(=\frac{24}{6-3 a}\) \(R.H.L.\) \(\lim \limits_{x \rightarrow 7^{+}} \frac{18-[1-x]}{[x]-3 a}\)…
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