Let a be an integer such that lim limits_{x r | vedclass

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(C) Given the limit $\lim \limits_{x \rightarrow 7} \frac{18-[1-x]}{[x]-3a}$ exists.For $x \rightarrow 7^-$,$[x] = 6$ and $[1-x] = [1 - (7-h)] = [-6+h

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$-6$

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(C) Given the limit $\lim \limits_{x \rightarrow 7} \frac{18-[1-x]}{[x]-3a}$ exists.For $x \rightarrow 7^-$,$[x] = 6$ and $[1-x] = [1 - (7-h)] = [-6+h] = -6$ (where $h > 0$ is very small).$L.H.L. = \lim \limits_{x \rightarrow 7^-} \frac{18 - (-6)}{6 - 3a} = \frac{24}{6 - 3a}$.For $x \rightarrow 7^+$,$[x] = 7$ and $[1-x] = [1 - (7+h)] = [-6-h] = -7$.$R.H.L. = \lim \limits_{x \rightarrow 7^+} \frac{18 - (-7)}{7 - 3a} = \frac{25}{7 - 3a}$.Since the limit exists,$L.H.L. = R.H.L.$$\frac{24}{6 - 3a} = \frac{25}{7 - 3a}$.Cross-multiplying gives:$24(7 - 3a) = 25(6 - 3a)$$168 - 72a = 150 - 75a$$75a - 72a = 150 - 168$$3a = -18$$a = -6$.

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

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