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

$\underset{x\to 7}{\lim }\frac{18-[1-x]}{\left[x\right]-3a}$

L.H.L. $\underset{x\to 7-}{\lim }\frac{18-[1-x]}{\left[x\right]-3a}$

$\begin{array}{l}=\frac{18-(-6)}{6-3a} \\ \\ =\frac{24}{6-3a}\end{array}$

R.H.L. $\underset{x\to 7+}{\lim }\frac{18-[1-x]}{\left[x\right]-3a}$

$\begin{array}{l}=\frac{18-(-7)}{7-3a} \\ \\ =\frac{25}{7-3a}\end{array}$

Now L.H.L. = R.H.L. 

$\begin{array}{l}\frac{24}{6-3a}=\frac{25}{7-3a}\\ \\ \Rightarrow 168-72a=150-75a\\ \\ \Rightarrow 18=-3a\\ \\ \Rightarrow a=-6\end{array}$