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

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

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Updated on: 27 Nov 2012, 03:10
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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)?

(1) a < b
(2) a + 4 < 0

Originally posted by darlingriyarai on 27 Nov 2012, 03:05.
Last edited by Bunuel on 27 Nov 2012, 03:10, edited 1 time in total.
Renamed the topic and edited the question.
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Re: A function f is defined by f(x,y) = x^2 - y. Is the value of  [#permalink]

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27 Nov 2012, 04:48
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.

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Re: A function f is defined by f(x,y) = x^2 - y. Is the value of  [#permalink]

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23 Aug 2014, 21:52
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darlingriyarai wrote:
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)?

(1) a < b
(2) a + 4 < 0

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.

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Re: A function f is defined by f(x,y) = x^2 - y. Is the value of  [#permalink]

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03 Jun 2016, 06:14
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Re: A function f is defined by f(x,y) = x^2 - y. Is the value of  [#permalink]

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03 Jun 2016, 10:04
darlingriyarai wrote:
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)?

(1) a < b
(2) a + 4 < 0

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.

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Re: A function f is defined by f(x,y) = x^2 - y. Is the value of  [#permalink]

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05 Jun 2016, 04:55
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.
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Re: A function f is defined by f(x,y) = x^2 - y. Is the value of  [#permalink]

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07 Sep 2018, 09:01
darlingriyarai wrote:
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)?

(1) a < b
(2) a + 4 < 0

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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Re: A function f is defined by f(x,y) = x^2 - y. Is the value of   [#permalink] 07 Sep 2018, 09:01
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