If f(x) = 100 - x^2, is f(a) > f(b)? (1) a^2 > b^2 (2) a/b > 1

The two cases youre assuming :
first when b<0 , it is not always that a-b will always be <0 . for eg. a could be -2 and b could be -3 . so a-b>0.
Am I making a mistake? please help.

[quote="Archit3110"]

does the statement 2 as explained by you holds true as "b" can be either positive or negative so considering a>b is correct ?? as a<b can also be correct .

Kindly shed some light.

and give kudos if i pointed out right.

regards,
Mahi..

mahipal
#2 given a/b > 1
so both a & b have to be of same sign and a >b then only #2 relation will stand true

mahipal
#2 given a/b > 1
so both a & b have to be of same sign and a >b then only #2 relation will stand true[/quote]


Hi,

I feel you didn't get me.

a/b>1 does not imply a>b

suppose a=-3, b=-2

a/b>1 holds true ,but does a>b holds true ? ==> No.

If a/b>1 ==> we can imply that a^2>b^2 after squaring both sides.

Kindly correct me ,if i am wrong.

Regrads,
Mahi.

(1) a^2 > b^2
(2) a/b > 1[/quote]

#1
a^2>b^2
a & b being + /-ve wont make any difference to say that f(a) <f(b)

sufficeint
#2
a/b>1
means a>b
if a >b then fa <fb
sufficient
IMO D[/quote]

[/quote]

mahipal
#2 given a/b > 1
so both a & b have to be of same sign and a >b then only #2 relation will stand true[/quote]


Hi,

I feel you didn't get me.

a/b>1 does not imply a>b

suppose a=-3, b=-2

a/b>1 holds true ,but does a>b holds true ? ==> No.

If a/b>1 ==> we can imply that a^2>b^2 after squaring both sides.

Kindly correct me ,if i am wrong.

Regrads,
Mahi.[/quote]
mahipal
please understand that in DS questions we need to use the given statements/ relations as the valid and true in #2 it says a/b > 1
so a/b>1 will be true in cases when a>b with both signs as same.. we cannot have a & b in opposite sign as it will break #2

Hi,

I feel you didn't get me.

a/b>1 does not imply a>b

suppose a=-3, b=-2

a/b>1 holds true ,but does a>b holds true ? ==> No.

If a/b>1 ==> we can imply that a^2>b^2 after squaring both sides.

Kindly correct me ,if i am wrong.

Regrads,
Mahi.[/quote]
mahipal
please understand that in DS questions we need to use the given statements/ relations as the valid and true in #2 it says a/b > 1
so a/b>1 will be true in cases when a>b with both signs as same.. we cannot have a & b in opposite sign as it will break #2[/quote]


Hi,

Yes i agree ,completely with you that the given statements are considered to be right and then in light of these statements we have to consider the original question.

But statement #2 says

a/b>1

which you have inferred as a>b.

but here i am not considering opposite signs of a and b infant i am saying they both have same signs but sign can be both negative as well as both positive.

if we consider the case of both positive then what you are saying holds true.i.e if a/b>1 ==> a>b
But

if we consider the case of both negative then a/b>1 does not imply a>b

suppose a=-3 b=-2

then a/b>1 will hold true , but a>b will not hold true.

This is what i am saying.

If i am not clear now also ,do let me know .
As it is important for me to know what i am missing.

Regards,
Mahi.

chetan2u Why aren't we considering the case of fractions in St2?

Hi Anubhav21

Try some values for fractions and you will see that for a/b>1, a needs to be a bigger value than b.
Say a = 1/2 and b = 1/3 so a square > b square

But if you try a = 1/3 and b = 1/2, a/b>1 will not hold true.

Hope this helps.

Also a/b > 1 means a and b should have the same sign and a should have a bigger absolute value than b. If a/b were more than zero we could consider fractions.

First, let's try to simplify the question stem:

Is f(a) > f(b) ?

Can be written as, is 100-a^2 > 100- b^2 ?

Also can be written as is a^2 < b^2?


Statement 1, tells us that a^2> b^2, that is a definite NO to the question asked.
Therefore statement 1 is sufficient.

Statement 2, tells us that a/b > 1
I don't know if a is greater than b or if a is less than b because I don't know the sign of b.
However, I do know that the absolute value of a is greater than the absolute value of b
Therefore, a^2 is greater than b^2. And that is a definite NO to the question asked.
Statement 2 is also sufficient.

Correct answer is D

Bunuel can you please check the reasoning above and let me know if it stands? Thank you