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Please find the solution inside the image. IMO option B.
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File comment: Q2
Q2_sep29.jpeg
Q2_sep29.jpeg [ 117.39 KiB | Viewed 3202 times ]

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since y is an integer 2nd statement says y is less then 2 therefore y can only be 0
if x is positive then y=2x which is not possible in this case
if x is -ve then y has to be 0
therefore 2nd statement is sufficient
B is correct
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y=|x|+x => y=0 when x<0 * y=2x when x>0
1) . Suff as x! is not defined when , x<0
2) Insuff ... as y is any integer lower than 2. Thus A.
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Consider 'A', x>0. This is the only possible case for x! + 7 to be prime. x = 3,4,5,..... satisfies the condition and y is clearly not equal to zero. So A is sufficient.
Consider B, y <2, x can be 0.5 -> y = 1, x can be -0.5 -> y= 0. So B is not sufficient.

Answer A.
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Question asks to prove whether x is 0 or negative then only y will be 0.

a) X!+ 7 is prime
If x=0, therefore 0!+7 is not prime out.
X=1, not prime,out,
X=2, not prime out,
X=3, 3!+7=13 prime,
Sufficient,
X is not negative therefore y is not zero.
b) Y<2
Means y=-1,0,1
Insufficient.
ANSWER=A
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1) x will be a integer >=0
If x = 0 then, x! + 7 = 7 which is a prime number
Therefore y = 0

If x = 3 then x! = 6, x! +7 = 13 which is a prime number. Thus, y = 6

Hence A is not sufficient

2) y < 2

If x <=0 then y = 0

If x> 0 then y = 2x <2
Thus, x <1, thus y is not equal to 0

Thus B is not sufficient

Both,
From A, x is >=0 and is an integer. From B x < 1. Therefore, x =0. Thus, y = 0. Hence, C is sufficient.

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stem is asking if x= 0 or x<0. If x<0. As y=0 only in these two cases.

From 1: x!+7 is prime, implies that x is not equal 0. Also x is not negative. so this statement is sufficient
from 2: y<2, if x=0, then y=0, if x= 0.5, then y=1, an integer. 2 different answers, Not sufficient

Answer is A
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In the solution shown above, I have a doubt.
What if in the second statement y=1 , x=1/2
It’s not mentioned that x has to be an int . Then there will be two cases
Y=1 when x=1/2
Y=0 when X<0


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