question 4

[kutra]

Is x/3 + 3/x > 2?

1. x<3
2. x>1

[george_2010]

The Answer is B.

x/3 + 3/x = (x^2 + 3^2) /3x >2 ?

1. x<3 : Not Suff.
- For 0< x<3 ; the ecuation is >2
- For x<0 ; the ecuation is < 2

2. x> 1: Suff
- The ecuation is always > 2

[kutra]

Originally Posted by george_2010

The Answer is B.

x/3 + 3/x = (x^2 + 3^2) /3x >2 ?

1. x<3 : Not Suff.
- For 0< x<3 ; the ecuation is >2
- For x<0 ; the ecuation is < 2

2. x> 1: Suff
- The ecuation is always > 2

St 2 breaks for x=3

Wondering if there's any way to solve this algebraically

[finsisher]

there is a way...
sum of a +ve number and its reciprocal >= 2
and sum of a -ve number and its reciprocal <= -2

now the question says x is a +ve no.. the critical point is when x = 3 which is out of bound ( 2<x<3) so obviously the sum is >=2

hence C is the answer

[socker121314]

IMO A.

[gaurav211982]

Answer is C

First of all X is not defined as +ve or -ve integer or is floating value or not so X value can be -ve ,+ve or floating points values

1) X<3 disqualify because if X= -ve value than condition did not satisfy TO A and D
2) X>1 disqualify because at x=3 , it comes =2 but not >2 so it disqualify B and D
3) (1)+(2) = 1<x<3 then x =2 so 2/3+3/2 = 2.111 which is >2 so

C IS ANSWER

[socker121314]

I see my mstake..C .

[george_2010]

My mistake... C it is.

[Mits83]

IMO answer is A.

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

For this condition to occur, x can be anything BUT 3.

a) sufficient.
b) not sufficient as x could be 3.

[ramiy]

IMO A fisher and gaurav por favor, El Señor check your solution again.

I think Mits has solved it the right way