Question

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

• A. -6
• B. -2
• C. 2
• D. 6

Answer 1

Answer: (A)

-6

Answer 2

$lim _{x→7} \frac{18−[1−x]}{[x−3a]}$

exist & $a∈l$.

$lim_{x→7} \frac{17−[−x]}{[x−3a]}$

exist

$RHL$ = $lim_{x→7} \frac{17−[−x]}{[x−3a]}$ = $\frac{25}{7-3a} [a ≠\frac{7}{3}]$

$LHL$ = $lim_{x→7-} \frac{17−[−x]}{[x−3a]}$

= $\frac{24}{6-3a} [ a≠2]$

For limit to exist

$LHL = RHL$

$\frac{25}{7−3a}=\frac{24}{6−3a}$

$⇒\frac{25}{7−3a}=\frac{8}{2−a}$

$∴ a = -6$

Hence, the correct option is (A): $-6$