Is x a positive number? (1) (x – 2)^2 2 (2) 2^x 3^x

Question

Is x a positive number?
 
(1) (x – 2)^2 > 2
(2) 2^x > 3^x

Nikkb · Answer

x a positive number?

(1) (x – 2)^2 > 2
Assume x=1 (1-2)^2 > 2 No...
but x=0.1 positive (0.1-2)^2 = (1.9)^2 > 2... True..... 
Not sufficient

(2) 2^x > 3^x 
Assume x=1 => 2>3.. False
x=0.1 => 2^0.1 > 3^0.1 => False.. All positive powers will give answer "False"
so for 2^x > 3^x ... x is not positive 
Sufficient

Answer: B

KS15 · Answer

Bunuel,

I have a doubt here.

For statement 1,
 I took ((x – 2)^2- (\sqrt{2}) ^2> 0

Now, can we not take (a-b)(a+b)>0 seperately to find out the solution?

Thanks

Bunuel · Answer

Well yes but it will give the same answer.

Probably the easiest way to deal with the first statement would be to tests numbers or to take the square root from both sides:

|x-2|>\sqrt{2} ;

x-2>\sqrt{2}  or  -(x-2)>\sqrt{2}

x>2+\sqrt{2}  or  x<2-\sqrt{2}

KS15 · Answer

Can you show this using the formula a^2-b^2 i.e (a-b)(a+b) and then solving? I know it is easy to solve using numbers but not able to get using this formula

Thanks

Nikkb · Answer

KS15 (x – 2)^2 > 2 => (x – 2)^2 - 2 >0 => (x – 2)^2 - \sqrt{2} >0 => (x-2+\sqrt{2}) (x-2-\sqrt{2}) >0 This will term either both term are positive or both terms are negative both positive : x-2+\sqrt{2} > 0 and x-2-\sqrt{2} > 0 x>2-\sqrt{2} and x >2+\sqrt{2} OR both negative : x-2+\sqrt{2} < 0 and x-2-\sqrt{2} < 0 x<2-\sqrt{2} and x <2+\sqrt{2} Combining all 4 equations we can write : x >2+\sqrt{2} OR x<2-\sqrt{2}

Bunuel · Answer

(x – 2)^2 > 2 (x – 2)^2 -2> 0 (x – 2-\sqrt{2})(x – 2+\sqrt{2})> 0 . For the product to be positive both multiples must have the same sign. When both are positive: x – 2-\sqrt{2}> 0 and x – 2+\sqrt{2}> 0 . x > 2+\sqrt{2} and x > 2-\sqrt{2} . For both above to be true simultaneously, x > 2+\sqrt{2} should hold (so x must be more than the larger number). When both are negative: x – 2-\sqrt{2}< 0 and x – 2+\sqrt{2}< 0 . x < 2+\sqrt{2} and x < 2-\sqrt{2} . For both above to be true simultaneously, x < 2-\sqrt{2} should hold (so x must be less than the smaller number). So, x>2+\sqrt{2} or x<2-\sqrt{2} . Hope it's clear.

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Is x a positive number? (1) (x – 2)^2 > 2 (2) 2^x > 3^x

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