A function f is defined by f(x,y) = x^2 - y. Is the value of

A function f is defined by f(x,y) = x^2 - y. Is the value of f(3,b) less than the value of f(a,3)?

Since \(f(x,y) = x^2 - y\), then \(f(3,b)=9-b\) and \(f(a,3)=a^2-3\). The question asks whether \(9-b<a^2-3\) --> is \(a^2+b>12\)?

(1) a < b. If \(a=1\) and \(b=2\), then the answer is NO but \(a=3\) and \(b=4\), then the answer is YES. Not sufficient.

(2) a + 4 < 0 --> \(a<-4\). If \(a=-5\) and \(b=0\), then the answer is YES but \(a=-5\) and \(b=-15\), then the answer is NO. Not sufficient.

(1)+(2) From (1) we have that \(a<b\). Add \(a^2\) to both parts: \(a^2+a<a^2+b\). Now, since \(a<-4\), then \(a^2+a>12\) (if a=-4 then a^2+a=12 and if we decrease a, then we increase a^2+a). Thus we have that \(12<a^2+a<a^2+b\). Sufficient.

Answer: C.
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The question is asking is 9 - b less than a(a) - 3

Is 9 - b < a(a) - 3?
Is 12 < a(a) + b

Statement I is insufficient:
a = 2, b = 3 and a = 4, b = 15 will give us a Yes and a No

Statement II is insufficient:
a< -4, a = -5, b = 1 and a = -5 and b = -16 will give us a Yes and a No

Now if a < -4 and b>a we know that b will have to be greater than -4 hence it a(a) + b will always be greater than 12.

Hence the answer is C

3^2 -b< a^2-3?

12<a^2+b

(1) a < b

We don't know that values of a and b. Hence, we can't figure out if a^2+b>12. Not sufficient.

(2) a + 4 < 0

a<-4
a^2> 16 but we don't know the value of b and hence can't get the value of a^2+b>12. Not sufficient

Combining both statements:-

a^2 > 16 and b> a
a^2+b>12 will always be true.

C is the answer
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If we modify the original condition and the question, we get the question as f(3,b)<f(a,3)?, 3^2-b<a^2-3?. In other words, it becomes a^2+b-12>0?. There are 2 variables (a and b) in the original condition. In order to match the number of variables to the number of equations, we need 2 equations. Since the condition 1) and the condition 2) each has 1 equation, there is high chance that C is the correct answer.
Using both the condition 1) and the condition 2), the correct answer is C.

OA: C

\(f(x,y) = x^2 - y\)
\(f(3,b) = 9 -b\)
\(f(a,3) = a^2-3\)

Question Stem : \(Is \quad f(a,3)>f(3,b)?\)

\(Is \quad a^2-3>9 -b? \quad or \quad Is \quad a^2 +b>12?\)

\((1) \quad a < b\)

Taking \(a = 1\) and \(b = 2\) , \(1<2\)

\(Is \quad a^2 +b>12?\quad No\) , as \(1^2+2 =3\) is less than \(12\).

Taking \(a = 2\) and \(b = 10\) , \(2<10\)

\(Is \quad a^2 +b>12?\quad Yes\) , as \(2^2+10 =14\) is greater than \(12\).

Statement \(1\) alone is not sufficient

\((2) \quad a + 4 < 0 \quad i.e \quad a<-4\)

Value of \(b\) can be anything so Statement \(2\) alone is not sufficient.

Combining \((1)\) and \((2)\) , we get \(a<-4\) and \(b≥-4\)

The minimum value of \(a^2\) would be slightly more than \(16\).

The minimum value of \(b\) would be \(-4\)

So Minimum value of \(a^2+b\) would be \(16.X-4 =12.X\) where \(X\) is greater than \(0\),So \(a^2+b\) would be always greater than \(12\).

Combining \((1)\) and \((2)\) \(Is \quad a^2 +b>12?\quad Yes\)

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