Mathematics · 2022
JEE Main · 27 June 2022, Shift 1 · Q5
Let a be an integer such that lim_x → 7 (18-[1-x])/([x-3 a]) exists, where [t] is greatest integer ≤ t. Then a is equal to:
Let a be an integer such that $\displaystyle \lim _{x \rightarrow 7} \frac{18-[1-x]}{[x-3 a]}$ exists, where $\displaystyle [t]$ is greatest integer $\displaystyle \leqslant t$. Then a is equal to :
Official answer
(1)
$\displaystyle -6$
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JEE Main 2022 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.