Statement 2:

|x| + x can never be -ve for any value of 'x'

It can take positive values or 'zero'

given 'y' is an integer and < 1. Hence y = 0

If y is an integer and y = |x| + x, is y=0? The function y =f(x) = |x| + x can also be defined by
y = 2x, if x > 0 and
y = 0 if x < 0

(1) x < 0

If x < 0, According to the function, y = 0. hence sufficient!

(2) y < 1

From the function definition, We know that 0<y<1.y is an integer between 0 and 1(0 included).So, y = 0 is the only solution. hence sufficient!

Answer D

shankar.ashwin wrote:Statement 2:

|x| + x can never be -ve for any value of 'x'

It can take positive values or 'zero'

given 'y' is an integer and < 1. Hence y = 0

i guess i overlooked the integer bit

Thanks for pointing it out

If y is an integer and y = |x| + x, is y=0?
(1) x < 0
(2) y < 1

This is a yes or no question and statement 1 everybody got as sufficient.

There is a strong contradiction for statement 2 whether it is sufficient or not.

According to me, statement 2 is sufficient but because of a different reason. According to me y can never be equal to zero from statement 2 and that is what I got...

statement 2 says y < 1.

hence can we say that |x| + x < 1, from the equation that we are given...

Now try putting all the values of x.

if x = o then the inequality does not stand. x < 0 is also not possible as it does not stand in the inequality...

According to the prephrase that we received from the question, that y = 0 only if x < 0 or x = 0...

Both the conditions do not meet. statement 2 gives us an inequality and we have to assume that it is true and from that see if our condition is met.

It does not meet and therefore the answer is a confirm no...

actually from y < 1 it is proved that y is not negative not zero but between 0 and 1 only


Hence D....



If you see any mistake then please point out my mistake....

[email protected] wrote:If y is an integer and y = |x| + x, is y=0?
(1) x < 0
(2) y < 1

This is a yes or no question and statement 1 everybody got as sufficient.

There is a strong contradiction for statement 2 whether it is sufficient or not.

According to me, statement 2 is sufficient but because of a different reason. According to me y can never be equal to zero from statement 2 and that is what I got...

statement 2 says y < 1.

hence can we say that |x| + x < 1, from the equation that we are given...

Now try putting all the values of x.

if x = o then the inequality does not stand. x < 0 is also not possible as it does not stand in the inequality...

According to the prephrase that we received from the question, that y = 0 only if x < 0 or x = 0...

Both the conditions do not meet. statement 2 gives us an inequality and we have to assume that it is true and from that see if our condition is met.

It does not meet and therefore the answer is a confirm no...

actually from y < 1 it is proved that y is not negative not zero but between 0 and 1 only


Hence D....



If you see any mistake then please point out my mistake....

Acc to the stem, we infer the following-

If x>0,
y = 2x

If x<= 0,
y = 0

St(1)
x<0
y = 0

Suff


St(2)-
y < 1

For x<=0,
y =0

Lets check other possible values of y
For x= 1/2, y = 1
for x= 1, y=2
for x=3/2 , y = 3

If we make a set of all possible values of y
{0, 1, 2, 3, 4, 5, 6, 7.....}

Now acc to st(2)
The only value of y less than 1 is 0

Hence, suff

Option D


Amit -
How does x <= 0 not stand in the inequality, x+|x|<1?

For x = 0
0<1

For x= -1
-1+1<0

Yes I got my mistake brother Rijul... Thank You very much...

I am convinced that y = 0 satisfies the equation...