Question

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

• A. -6
• B. -2
• C. 2
• D. 6

Answer 1

Answer: (A)

$\underset{x\to 7}{\lim }\frac{18-[1-x]}{[x-3a]}\text{ exist and }a\in I$
$=\underset{x\to 7}{\lim }\frac{17-[-x]}{[x]-3a}\text{ exist }$
$\text{ RHL }=\underset{x\to 7^{+}}{\lim }\frac{17-[-x]}{[x]-3a}=\frac{25}{7-3a}\left[a\ne \frac{7}{3}\right]$
$LHL=\underset{x\to 7^{-}}{\lim }\frac{17-[-x]}{[x]-3a}=\frac{24}{6-3a}[a\ne 2]$
For limit to exist
$\text{ LHL }=\text{ RHL }$
$\frac{25}{7-3a}=\frac{24}{6-3a}$
$\Rightarrow \frac{25}{7-3a}=\frac{8}{2-a}$
$\therefore a=-6$