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Is x/3 + 3/x > 2

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Re: Is \frac{x}{3} + \frac{3}{x} > 2 ? (1) x < 3 [#permalink] New post 14 Aug 2012, 23:16
suryadasika wrote:
My answer is B.

For any positive number 'a', (a+ 1/a )>=2
So , from the given options, all we need to do is ensure that x/3 is a positive quantity.

Hence B.

Please correct me if i am wrong.


OA for this question is given in the first post under the spoiler and it's C, not A.

To see that the second statement is not sufficient, consider x=3 for a NO answer (if x=3 then 3/3+3/3=2) and x=10 for an YES answer. Please refer to the solutions on the first page.

Hope it helps.
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Re: Tricky inequality problem [#permalink] New post 15 Aug 2012, 02:47
shrouded1 wrote:
hirendhanak wrote:
thanks for your explanation... can you explain the below mentioned rule with an example and its reasoning.

Never multiply or reduce inequality by an unknown (a variable) unless you are sure of its sign since you do not know whether you must flip the sign of the inequalit


Consider the inequality \frac{x^2-12}{x}>-1

Lets say I multiply both sides by 7x without considering the signs of the variable, what happens ?

x^2-12>-x
x^2+x-12>0
(x+4)(x-3)>0

Which is true whenever x>3 (both terms positive) or when x<-4 (both terms negative)

But since we haven't kept the Sign of x in mind when we multiplied in step 1, the solution is wrong.

For eg. Take x=-1 which according to us is not a solution. It is easy to see ((-1)^2-12)/(-1)=11>-1. So it should be a solution
Similarly take x=-6 which according to us is a solution, but ((-6)^2-12)/-6=-4<-1. So it should not be a solution


Very well illustrated why we shouldn't multiply an inequality with an expression of unknown sign.
Just a few more things:

The given inequality \frac{x^2-12}{x}>-1 is equivalent to \frac{x^2+x-12}{x}>0. The sign of the ratio of two numbers is the same as the sign of their product. So, the previous inequality is equivalent to x(x^2+x-12)>0, which in fact can be obtained from the original inequality by multiplying by x^2, which we know for sure that it is positive, because x being in the denominator, x\neq0.

When given A/B>0, it means:
1) A and B have the same sign (either both positive or both negative)
2) Neither A nor B can be 0, A is in the numerator and the fraction is greater than 0, B is in the denominator.
It is obvious that multiplying by B the given inequality leads to A>0, incorrect.
A/B>0 is equivalent to AB>0,\, B\neq0.

When given A/B\geq0, it means:
1) A and B have the same sign (either both positive or both negative). A can be 0.
2) B cannot be 0, being in the denominator.
It is obvious that multiplying by B the given inequality leads to A\geq0, incorrect.
A/B\geq0 is equivalent to AB\geq0,\, B\neq0.


So, when we have to compare a fraction to 0, we can just compare the product of all the factors involved to 0. We should check carefully for values of the unknown if equality to 0 is allowed and we should not forget the values for which the denominator becomes 0.
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Re: Is x/3 + 3/x > 2 [#permalink] New post 16 Jan 2013, 03:48
\frac{x}{3}+3/x > 2
\frac{{9+x^2}}{3x}>2
\frac{{9+x^2}}{3x}-2>0
\frac{x^2-6x+9}{3x}>0
\frac{(x-3)^2}{3x}>0

x is not equal 3 since it will make the value 0
x is not negative since it will make the value negative



(1) x < 3
x may be negative or positive. THUS INSUFFICIENT!

(2) x > 1[/quote]
x may be equal to 3 or not. THUS INSUFFICIENT.

Together (1) and (2): x is positive and not equal to 3

Answer: C
Re: Is x/3 + 3/x > 2   [#permalink] 16 Jan 2013, 03:48
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