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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

  1. A \(2\)
  2. B \(-2\)
  3. C \(-6\)
  4. D \(6\)
Verified Solution

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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