JEE Main 2022 · 27 Jun · Shift 1 · Mathematics Q5

← JEE Main 2022 · 27 Jun · Shift 1

MathematicsSingle correct+4 / -1Question 5 of 90

Let $a$ be an integer such that $\displaystyle\lim_{x\to7}\frac{18-[1-x]}{[x-3a]}$ exists, where $[t]$ is greatest integer $\le t$ . Then $a$ is equal to :

-6

-2

Answer and worked solution

Official answer: A

Numerator. As $x\to7^+$ , $[1-x]=-7$ (numerator 25); as $x\to7^-$ , $[1-x]=-6$ (numerator 24).

Denominator. With $a$ an integer, $[x-3a]$ is $7-3a$ on the right and $6-3a$ on the left.

Equal limits. $\dfrac{25}{7-3a}=\dfrac{24}{6-3a}$ gives $a=-6$ .

(A)

Topics:Limits Graphs of functions

← Previous question Next question →

JEE Main 2022 · 27 Jun · Shift 1, Mathematics question 5