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# If f(x) = x^2 - x - 6, is f(x) ≥ g(x)?

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If f(x) = x^2 - x - 6, is f(x) ≥ g(x)? [#permalink]

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04 Jan 2013, 01:21
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If f(x) = x^2 - x - 6, is f(x) ≥ g(x)?

(1) g(x) = x^2 - 2x - 8

(2) x < -2

I have tried this Question and got it wrong first time (Wrong approach). But I don't agree with OA.
[Reveal] Spoiler: OA

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Last edited by Bunuel on 04 Jan 2013, 01:54, edited 1 time in total.
Renamed the topic and edited the question.

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Re: If f(x) = x^2 - x - 6, is f(x) ≥ g(x)? [#permalink]

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04 Jan 2013, 01:59
If f(x) = x^2 - x - 6, is f(x) ≥ g(x)?

(1) g(x) = x^2 - 2x - 8. The question becomes: is $$x^2 - x - 6\geq{x^2 - 2x - 8}$$? --> is $$x\geq{-2}$$? We don't know that, hence this statement is insufficient.

(2) x < -2. We know noting about g(x). Not sufficient.

(1)+(2) From (1) the question became: is $$x\geq{-2}$$? and (2) answers this question with a NO. Sufficient.

Hope it's clear.
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Re: If f(x) = x^2 - x - 6, is f(x) ≥ g(x)? [#permalink]

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12 May 2013, 07:02
Bunuel wrote:
If f(x) = x^2 - x - 6, is f(x) ≥ g(x)?

(1) g(x) = x^2 - 2x - 8. The question becomes: is $$x^2 - x - 6\geq{x^2 - 2x - 8}$$? --> is $$x\geq{-2}$$? We don't know that, hence this statement is insufficient.

(2) x < -2. We know noting about g(x). Not sufficient.

(1)+(2) From (1) the question became: is $$x\geq{-2}$$? and (2) answers this question with a NO. Sufficient.

Hope it's clear.

hi banuel,

From statement 1 , how you conclude we hace solve for whether x<= -2 ?
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Re: If f(x) = x^2 - x - 6, is f(x) ≥ g(x)? [#permalink]

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12 May 2013, 09:55
kabilank87 wrote:
Bunuel wrote:
If f(x) = x^2 - x - 6, is f(x) ≥ g(x)?

(1) g(x) = x^2 - 2x - 8. The question becomes: is $$x^2 - x - 6\geq{x^2 - 2x - 8}$$? --> is $$x\geq{-2}$$? We don't know that, hence this statement is insufficient.

(2) x < -2. We know noting about g(x). Not sufficient.

(1)+(2) From (1) the question became: is $$x\geq{-2}$$? and (2) answers this question with a NO. Sufficient.

Hope it's clear.

hi banuel,

From statement 1 , how you conclude we hace solve for whether x<= -2 ?

$$x^2 - x - 6\geq{x^2 - 2x - 8}$$
Cancel x^2 on both sides and re-arrange $$- x +2x\geq{ - 8+6}$$ -->$$x\geq{-2}$$.

Hope it's clear.
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Re: If f(x) = x^2 - x - 6, is f(x) ≥ g(x)? [#permalink]

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12 May 2013, 10:33
Question: $$f(x) = x^2 - x - 6$$. Is $$f(x) >= g(x)$$?

1) $$g(x) = x^2 - 2x - 8$$
2) $$x < - 2$$

Consider statement (1):

For $$f(x)$$ to be greater than or equal to $$g(x)$$, $$x^2-x-6$$ should be greater than or equal to $$x^2-2x-8$$. This implies that: $$x^2-x-6>=x^2-2x-8$$ (or)
$$x>=-2$$.

We don't know for sure that this is true. So this is not sufficient.

Consider statement (2):

$$x < -2$$. However, we know nothing about the form of the function $$g(x)$$. So this is not sufficient, either.

Consider both together:

If $$g(x)$$ is given by the function in option (1), then we require $$x>=-2$$ in order for $$f(x)$$ to be greater than or equal to $$g(x)$$. From (2) we know that $$x < -2$$. Therefore, both statements together are sufficient to prove that $$f(x)$$ is NOT greater than or equal to $$g(x)$$.

The correct answer, therefore, is C.
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Re: If f(x) = x^2 - x - 6, is f(x) ≥ g(x)? [#permalink]

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12 May 2013, 22:42
mridulparashar1 wrote:
If f(x) = x^2 - x - 6, is f(x) ≥ g(x)?

(1) g(x) = x^2 - 2x - 8

(2) x < -2

I have tried this Question and got it wrong first time (Wrong approach). But I don't agree with OA.

where are the options??

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Re: If f(x) = x^2 - x - 6, is f(x) ≥ g(x)? [#permalink]

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13 May 2013, 02:14
shankg06 wrote:
mridulparashar1 wrote:
If f(x) = x^2 - x - 6, is f(x) ≥ g(x)?

(1) g(x) = x^2 - 2x - 8

(2) x < -2

I have tried this Question and got it wrong first time (Wrong approach). But I don't agree with OA.

where are the options??

This is a data sufficiency question.

The data sufficiency problem consists of a question and two statements, labeled (1) and (2), in which certain data are given. You have to decide whether the data given in the statements are sufficient for answering the question. Using the data given in the statements, plus your knowledge of mathematics and everyday facts (such as the number of days in July or the meaning of the word counterclockwise), you must indicate whether—

A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient to answer the question asked.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient to answer the question asked.
C. BOTH statements (1) and (2) TOGETHER are sufficient to answer the question asked, but NEITHER statement ALONE is sufficient to answer the question asked.
E. Statements (1) and (2) TOGETHER are NOT sufficient to answer the question asked, and additional data specific to the problem are needed.

Hope it's clear.
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Re: If f(x) = x^2 - x - 6, is f(x) ≥ g(x)? [#permalink]

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14 May 2013, 01:13
Bunuel wrote:
If f(x) = x^2 - x - 6, is f(x) ≥ g(x)?

(1) g(x) = x^2 - 2x - 8. The question becomes: is $$x^2 - x - 6\geq{x^2 - 2x - 8}$$? --> is $$x\geq{-2}$$? We don't know that, hence this statement is insufficient.

(2) x < -2. We know noting about g(x). Not sufficient.

(1)+(2) From (1) the question became: is $$x\geq{-2}$$? and (2) answers this question with a NO. Sufficient.

Hope it's clear.

Hi Bunuel.. I understood Your solution but I have a doubt.

$$x^2 -x -6 = (x-3)*(x+2)$$
$$x^2-2x - 8 = (x-4)*(x+2)$$

now the question becomes is $$(x-3)*(x+2) >= (x-4)*(x+2)$$

if x is not equal to -2 then it will become is x-3 > x-4 so YES
if x is equal to -2 then both the sides are equal so YES

So isn't A sufficient ??
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Re: If f(x) = x^2 - x - 6, is f(x) ≥ g(x)? [#permalink]

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15 May 2013, 01:44
SrinathVangala wrote:
Bunuel wrote:
If f(x) = x^2 - x - 6, is f(x) ≥ g(x)?

(1) g(x) = x^2 - 2x - 8. The question becomes: is $$x^2 - x - 6\geq{x^2 - 2x - 8}$$? --> is $$x\geq{-2}$$? We don't know that, hence this statement is insufficient.

(2) x < -2. We know noting about g(x). Not sufficient.

(1)+(2) From (1) the question became: is $$x\geq{-2}$$? and (2) answers this question with a NO. Sufficient.

Hope it's clear.

Hi Bunuel.. I understood Your solution but I have a doubt.

$$x^2 -x -6 = (x-3)*(x+2)$$
$$x^2-2x - 8 = (x-4)*(x+2)$$

now the question becomes is $$(x-3)*(x+2) >= (x-4)*(x+2)$$

if x is not equal to -2 then it will become is x-3 > x-4 so YES
if x is equal to -2 then both the sides are equal so YES

So isn't A sufficient ??

The red part is not correct.

If x is not equal to -2, then x+2 could be more (for example, if x=0) as well as less than zero (for example if x=-3).

If x+2 is less than zero, then when reducing $$(x-3)*(x+2)\geq{ (x-4)*(x+2)}$$ by negative x+2 we should flip the sign and we'll get $$x-3\leq{{x-4}$$ --> so the answer is NO.

Hope it's clear.
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Re: If f(x) = x^2 - x - 6, is f(x) ≥ g(x)? [#permalink]

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15 May 2013, 02:06
Bunuel wrote:
SrinathVangala wrote:
Bunuel wrote:
If f(x) = x^2 - x - 6, is f(x) ≥ g(x)?

(1) g(x) = x^2 - 2x - 8. The question becomes: is $$x^2 - x - 6\geq{x^2 - 2x - 8}$$? --> is $$x\geq{-2}$$? We don't know that, hence this statement is insufficient.

(2) x < -2. We know noting about g(x). Not sufficient.

(1)+(2) From (1) the question became: is $$x\geq{-2}$$? and (2) answers this question with a NO. Sufficient.

Hope it's clear.

Hi Bunuel.. I understood Your solution but I have a doubt.

$$x^2 -x -6 = (x-3)*(x+2)$$
$$x^2-2x - 8 = (x-4)*(x+2)$$

now the question becomes is $$(x-3)*(x+2) >= (x-4)*(x+2)$$

if x is not equal to -2 then it will become is x-3 > x-4 so YES
if x is equal to -2 then both the sides are equal so YES

So isn't A sufficient ??

The red part is not correct.

If x is not equal to -2, then x+2 could be more (for example, if x=0) as well as less than zero (for example if x=-3).

If x+2 is less than zero, then when reducing $$(x-3)*(x+2)\geq{ (x-4)*(x+2)}$$ by negative x+2 we should flip the sign and we'll get $$x-3\leq{{x-4}$$ --> so the answer is NO.

Hope it's clear.

Yes!!!!!!! Awesome!!!! I didn't take into account that we cannot cancel negative numbers on the both sides of a inequality without changing the sign.

All I was thinking about was to avoid the condition where x = -2 so that we cancel the terms.

Thanks a lot!!!! Will remember this!!!!
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Re: If f(x) = x^2 - x - 6, is f(x) ≥ g(x)? [#permalink]

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Re: If f(x) = x^2 - x - 6, is f(x) ≥ g(x)?   [#permalink] 30 Aug 2016, 16:01
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