Let $a$ be an integer such that $\lim _{x \rightarrow 7} \frac{18-[1-x]}{[x-3 a]}$ exists, where $[t]$ is greatest integer $\leq t$. Then $a$ is equal to :

JEE Main · 2022 · 27 Jun 2022 · Shift 1 · Limits

Let

a

be an integer such that

lim_{x \to 7} \frac{18 - [ 1 - x ]}{[ x - 3 a ]}

exists, where

[t]

is greatest integer

\leq t

. Then

a

is equal to :

Answer: (A)

For the limit to exist, the left-hand and right-hand limits must be equal.

Left-hand limit as

x \to 7^{-}

Right-hand limit as

x \to 7^{+}

Equating LHL = RHL:

\frac{24}{6 - 3 a} = \frac{25}{7 - 3 a}

24 (7 - 3 a) = 25 (6 - 3 a)

168 - 72 a = 150 - 75 a

3 a = - 18 \Rightarrow a = - 6

JEE Main · 2022 · 27 Jun 2022 · Shift 1 · Single correct option

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