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)

$\underset{x\to 7}{\lim }\frac{18-[1-x]}{[x]-3a}$
L.H.L. $\underset{x\to 7-}{\lim }\frac{18-[1-x]}{[x]-3a}$
$=\frac{18-(-6)}{6-3a}$
$=\frac{24}{6-3a}$
$\text{ R.H.L. }\underset{x\to 7+}{\lim }\frac{18-[1-x]}{[x]-3a}$
$=\frac{18-(-7)}{7-3a}$
$=\frac{25}{7-3a}$
Now L.H.L. = R.H.L.
$\frac{24}{6-3a}=\frac{25}{7-3a}$
$\Rightarrow 168-72a=150-75a$
$\Rightarrow 18=-3a$
$\Rightarrow a=-6$