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so we have x/3 + 3/x > 2...assume x > 0 (as when x < 0, equation fails, thus statement 1 alone is not true), then we have x^2/3 + 3 - 2x > 0 => x^2 - 6x + 9 > 0 => (x-3)^2 > 0 so for all positive x except for 3 does the equation stand. thus statement 2 alone doesn't stand either...we need both. 1 < x < 3 definitely works.

I think better way to look at this as an eqn a+1/a ... see the behaviour of this eqn....first image is for eqn (a+1/a) and the later is the actual behaviour of (x/3+3/x) value of this equation cannot go below 2 (except in fractional values less than 1)..thats the property...hence OA C

If you like my post ....consider it for Kudos

Attachments

File comment: behaviour of a+1/a

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File comment: behaviour of x/3+3/x

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_________________

Bhushan S. If you like my post....Consider it for Kudos

Q: is x/3+3/x>2? Let's check when this statement holds true.

x/3+3/x>2 --> (x^2-6x+9)/3x>0 --> (x-3)^2/3x>0.

Nominator (x-3)^2 is NEVER negative, so expression is negative only when denominator 3x is negative, or simply when x is negative. Hence when x>0 our expression (x-3)^2/3x>0 is always positive EXCEPT when x=3, because at that case nominator (x-3)^2 becomes 0, thus expression becomes 0 and we need expression to be more then 0.

So we get that (x-3)^2/3x>0 holds true when x>0 and x#3.

Let's move to the statements: (1) x<3, tells us that x#3, but it's not enough, we need x to be positive besides that. Not sufficient.

(2) x>1, tell us that x positive, but it's not enough we need x not to be equal to 3. Not sufficient.

(1) + (2) 1<x<3 in this range x is positive and not equal to 3. Sufficient.

Bunuel can you please elaborate how did you arrive at this step: x/3+3/x>2 --> (x^2-6x+9)/3x>0 --> (x-3)^2/3x>0.

Just noticed that you addressed the question. I simply reduced to a common denominator x/3, 3/x and -2: x/3+3/x>2 --> x/3+3/x-2>0 --> (x^2+9-6x)/3>0 --> (x-3)^2/3>0 _________________

Q: is x/3+3/x>2? Let's check when this statement holds true.

x/3+3/x>2 --> (x^2-6x+9)/3x>0 --> (x-3)^2/3x>0.

Nominator (x-3)^2 NEVER negative, so expression is negative only when denominator 3x is negative, or simply when x is negative. Hence when x>0 our expression (x-3)^2/3x>0 is always positive EXCEPT when x=3, because at that case nominator (x-3)^2 becomes 0, thus expression becomes 0 and we need expression to be more then 0.

So we get that (x-3)^2/3x>0 holds true when x>0 and x#3.

Let's move to the statements: (1) x<3, tells us that x#3, but it's not enough, we need x to be positive besides that. Not sufficient.

(2) x>1, tell us that x positive, but it's not enough we need x not to be equal to 3. Not sufficient.

(1) + (2) 1<x<3 in this range x is positive and not equal to 3. Sufficient.

Answer: C.

Hi Bunuel,

Why do we have to keep denominator here?

x/3+3/x>2 --> (x^2-6x+9)/3x>0 --> (x-3)^2/3x>0.

Can't i reduce this equation to (x^2-6x+9) ---> (x-3)^2 > 0.

Then with Stm1: we can prove the Stem..... _________________

"Where are my Kudos" ............ Good Question = kudos

Q: is x/3+3/x>2? Let's check when this statement holds true.

x/3+3/x>2 --> (x^2-6x+9)/3x>0 --> (x-3)^2/3x>0.

Nominator (x-3)^2 NEVER negative, so expression is negative only when denominator 3x is negative, or simply when x is negative. Hence when x>0 our expression (x-3)^2/3x>0 is always positive EXCEPT when x=3, because at that case nominator (x-3)^2 becomes 0, thus expression becomes 0 and we need expression to be more then 0.

So we get that (x-3)^2/3x>0 holds true when x>0 and x#3.

Let's move to the statements: (1) x<3, tells us that x#3, but it's not enough, we need x to be positive besides that. Not sufficient.

(2) x>1, tell us that x positive, but it's not enough we need x not to be equal to 3. Not sufficient.

(1) + (2) 1<x<3 in this range x is positive and not equal to 3. Sufficient.

Answer: C.

Hi Bunuel,

Why do we have to keep denominator here?

x/3+3/x>2 --> (x^2-6x+9)/3x>0 --> (x-3)^2/3x>0.

Can't i reduce this equation to (x^2-6x+9) ---> (x-3)^2 > 0.

Then with Stm1: we can prove the Stem.....

Never multiply or reduce inequality by an unknown (a variable) unless you are sure of its sign./b]

So you CANNOT multiply \(\frac{(x-3)^2}{x}>0\) by \(3x\) since you don't know the sign of \(x\): if x>0, then you'll have \((x-3)^2>0\) but if x<0, then you'll have \((x-3)^2<0\) (flip the sign when multiplying by negative value).

Hope it's clear.

[b]OPEN DISCUSSION OF THIS QUESTION IS HERE: is-x-3-3-x-97331.html _________________

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