Let [x] denote the greatest integer ≤ x , where x ∈ R . If the domain of the real valued function f(x) = (|[x]| - 2)/(|[x]| - 3) is (- ∞ , a) [b, c) [4, ∞), a < b < c , then the value of a + b + c is:

Correct answer: C. - 2 We have, f(x) = (|[x]| - 2)/(|[x]| - 3) For domain, ( |[x]| - 2 |[x] | - 3 ) ≥ 0 Case I: When | [ x ] | - 2 ≥ 0 and |[x] | - 3 > 0 , then ∴ x ∈ (- ∞ , - 3) [4, ∞) . . . (1) Case II: When |[x]| - 2 ≤ 0 and |[x]| - 3 ∴ x ∈ [- 2, 3) . . . (2) So, from (1) and (2) , we get Domain of function is (- ∞ , - 3) [- 2, 3) [4, ∞) ∴ (a + b + c) = - 3 + (- 2) + 3 = - 2 (∵ a

JEE Main 202120 JulyMorning Shift

Exam: JEE Main
Year: 2021
Date: 20 July
Shift: Morning Shift
Subject: Mathematics
Chapter: Functions

JEE Main 2021 · 20 July · Morning Shift · Mathematics · Functions

Let [x] denote the greatest integer ≤ x , where x ∈ R . If the...

Let

\left[x\right]

denote the greatest integer

\le x

, where

x \in R

. If the domain of the real valued function

f\left(x\right) = \sqrt{\dfrac{\left|\left[x\right]\right| - 2}{\left|\left[x\right]\right| - 3}}

is \left(- \infty , a\right) \cup \left[b, c\right) \cup \left[4, \infty\right), a < b < c

, then the value of

a + b + c$ is:

Options

A $8$
B $1$
C $- 2$
D $- 3$

Correct answer

C. $- 2$

Step-by-step solution

We have,

$f\left(x\right) = \sqrt{\dfrac{\left|\left[x\right]\right| - 2}{\left|\left[x\right]\right| - 3}}$

For domain, $\left(\dfrac{\left|\left[x\right]\right| - 2}$ {\left|\left[x\right] $\right| - 3}\right) \ge 0$

Case $I:$ When $| [ x ] | - 2 \ge 0 and \left|\left[x\right]$ \right| - 3 > 0 $, then$

$∴ x \in \left(- \infty , - 3\right) \cup \left[4, \infty\right) . . . \left(1\right)$

Case $II:$ When \left|\left[x\right]\right| - 2 \le 0 $and \left|\left[x\right]\right| - 3 < 0$ , then $$ ∴ x \in \left[- 2, 3\right) $. . . \left(2\right)So,$ from \left(1\right) and \left(2\right) $, we getDomain of$ function is

\left(- \infty , - 3\right) \cup \left[- 2, 3\right) \cup \left[4, \infty\right)

$∴ \left(a + b + c\right)$ = - 3 + \left(- 2\right) $+ 3 = - 2$

\left(∵ a < b < c\right)

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