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1.Relation and Function
hard
અહી $[x]$ એ મહતમ પૃણાંક વિધેય છે. જો વાસ્તવિક વિધેય $\mathrm{f}(\mathrm{x})=\sqrt{\frac{[\mathrm{x}] \mid-2}{\sqrt{[\mathrm{x}] \mid-3}}}$ નો પ્રદેશ $(-\infty, \mathrm{a}) \cup[\mathrm{b}, \mathrm{c}) \cup[4, \infty), \mathrm{a}\,<\,\mathrm{b}\,<\,\mathrm{c}$, હોય તો $\mathrm{a}+\mathrm{b}+\mathrm{c}$ ની કિમંત મેળવો.
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
