Since x→7lim[x−3a]18−[1−x] exists (a∈I)
∴L.H.L=R.H.L
⇒h→0lim[7−h−3a]18−(−6)=h→0lim[7+h−3a]18−(−7)
⇒h→0lim(7−3a)+[−h]24=h→0lim(7−3a)+[h]25
⇒6−3a24=7−3a25 ⇒a=−6
JEE Main 2022 — Mathematics Calculus
Let a be an integer such that x→7lim[x−3a]18−[1−x] exists, where [t] is greatest integer ≤t. Then a is equal to
Held on 27 Jun 2022 · Verified 6 Jul 2026.
−2
6
−6
−7
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