JEE MAIN - Mathematics (2021 - 27th August Morning Shift - No. 14)
If A = {x $$\in$$ R : |x $$-$$ 2| > 1},
B = {x $$\in$$ R : $$\sqrt {{x^2} - 3} $$ > 1},
C = {x $$\in$$ R : |x $$-$$ 4| $$\ge$$ 2} and Z is the set of all integers, then the number of subsets of the
set (A $$\cap$$ B $$\cap$$ C)c $$\cap$$ Z is ________________.
B = {x $$\in$$ R : $$\sqrt {{x^2} - 3} $$ > 1},
C = {x $$\in$$ R : |x $$-$$ 4| $$\ge$$ 2} and Z is the set of all integers, then the number of subsets of the
set (A $$\cap$$ B $$\cap$$ C)c $$\cap$$ Z is ________________.
Answer
256
Explanation
A = ($$-$$$$\infty$$, 1) $$\cup$$ (3, $$\infty$$)
B = ($$-$$$$\infty$$, $$-$$2) $$\cup$$ (2, $$\infty$$)
C = ($$-$$$$\infty$$, 2] $$\cup$$ [6, $$\infty$$)
So, A $$\cap$$ B $$\cap$$ C = ($$-$$$$\infty$$, $$-$$2) $$\cup$$ [6, $$\infty$$)
z $$\cap$$ (A $$\cap$$ B $$\cap$$ C)' = {$$-$$2, $$-$$1, 0, $$-$$1, 2, 3, 4, 5}
Hence, no. of its subsets = 28 = 256.
B = ($$-$$$$\infty$$, $$-$$2) $$\cup$$ (2, $$\infty$$)
C = ($$-$$$$\infty$$, 2] $$\cup$$ [6, $$\infty$$)
So, A $$\cap$$ B $$\cap$$ C = ($$-$$$$\infty$$, $$-$$2) $$\cup$$ [6, $$\infty$$)
z $$\cap$$ (A $$\cap$$ B $$\cap$$ C)' = {$$-$$2, $$-$$1, 0, $$-$$1, 2, 3, 4, 5}
Hence, no. of its subsets = 28 = 256.
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