1.Relation and Function
hard

माना $[ x ]$ महत्तम पूर्णांक $\leq x$ है, जहों $x \in R$ है। यदि वास्तविक मान फलन $f(x)=\sqrt{\frac{[x] \mid-2}{[x] \mid-3}}$ का प्रांत $(-\infty, a) \cup[b, c) \cup[4, \infty), a < b < c$, है, तो $a+b+c$ का मान है

A

$-3$

B

$1$

C

$-2$

D

$8$

(JEE MAIN-2021)

Solution

For domain,

$\frac{|[x]|-2}{|[x]|-3} \geq 0$

Case $I:$ When $|[x]|-2 \geq 0$

and $|[x]|-3\,>\,0$

$\therefore x \in(-\infty,-3) \cup[4, \infty] \ldots . .(1)$

Case $II:$ When $|[x]|-2 \leq 0$

and $|[x]|-3\,<\,0$

$\therefore \mathrm{x} \in[-2,3) \quad \ldots(2)$

So, from $(1)$ and $(2)$

We get

Domain of function

$=(-\infty,-3) \cup[-2,3) \cup[4, \infty)$

$\therefore(a+b+c)=-3+(-2)+3=-2(a\,<\,b\,<\,c)$

Standard 12
Mathematics

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