Question

Consider U = {x|x is a positive integer greater than 1}.

Answer 1

B) {x|x ∈ U and 2x is prime}
x is greater than 1.
the only even prime number is 2
2 times anything is even.
the minimum x value is 2. 2x is 4, therefore higher than the only even number
every value 2x will have a multiple of 2 so it will not be prime.

Answer 2

Answer:

The correct option is B.i.e., 2nd set.

Step-by-step explanation:

Given: U = {x : x is positive integer greater than 1}

               = {2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,....}      

1st Set

say, X= {x : x∈U and $\frac{1}{2}x$ is prime}

4∈U ⇒ $\frac{1}{2}4 = 2$ ∈X as 2 is prime

∴This set is not empty set.

2nd Set

say, Y={x : x∈U and 2x is prime}

This set is empty set because product of 2 and no. greater than 1 is always a composite no.

3rd Set

say, Z={x : x∈U and $\frac{1}{2}x$ can be written as fraction}

3∈U ⇒ $\frac{1}{2}3 = \frac{3}{2}$ ∈Z as $\frac{3}{2}$ is a fraction.

∴This set is not empty.

4th Set

say, S = {x : x∈U and 2x can be written as a fraction}

2∈U ⇒ 2.2 = 4 ∈S as this can be written as $\frac{4}{1}$ fraction or 3∈U ⇒ 2.3 = 6 ∈S as this can be written as $\frac{6}{1}$ fraction

∴This set is not empty.

Therefore, The correct option is B.