Question

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

• A. -6
• B. -2
• C. 2
• D. 6

Answer 1

Answer: (A)

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