JEE MAIN - Mathematics (2021 - 20th July Morning Shift - No. 7)
Let [ x ] denote the greatest integer $$\le$$ x, where x $$\in$$ R. If the domain of the real valued function $$f(x) = \sqrt {{{\left| {[x]} \right| - 2} \over {\left| {[x]} \right| - 3}}} $$ is ($$-$$ $$\infty$$, a) $$]\cup$$ [b, c) $$\cup$$ [4, $$\infty$$), a < b < c, then the value of a + b + c is :
8
1
$$-$$2
$$-$$3
Explanation
For domain,
$${{{\left| {[x]} \right| - 2} \over {\left| {[x]} \right| - 3}}}$$ $$\ge$$ 0
Case I :
When $${\left| {[x]} \right| - 2}$$ $$\ge$$ 0
and $${\left| {[x]} \right| - 3}$$ > 0
$$\therefore$$ x $$\in$$ ($$-$$ $$\infty$$, $$-$$3) $$\cup$$ [4, $$\infty$$) ...... (1)
Case II :
When $${\left| {[x]} \right| - 2}$$ $$\le$$ 0
and $${\left| {[x]} \right| - 3}$$ < 0
$$\therefore$$ x $$\in$$ [$$-$$2, 3) ..... (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)
$$\Rightarrow$$ Option (3) is correct.
$${{{\left| {[x]} \right| - 2} \over {\left| {[x]} \right| - 3}}}$$ $$\ge$$ 0
Case I :
When $${\left| {[x]} \right| - 2}$$ $$\ge$$ 0
and $${\left| {[x]} \right| - 3}$$ > 0
$$\therefore$$ x $$\in$$ ($$-$$ $$\infty$$, $$-$$3) $$\cup$$ [4, $$\infty$$) ...... (1)
Case II :
When $${\left| {[x]} \right| - 2}$$ $$\le$$ 0
and $${\left| {[x]} \right| - 3}$$ < 0
$$\therefore$$ x $$\in$$ [$$-$$2, 3) ..... (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)
$$\Rightarrow$$ Option (3) is correct.
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