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Is A positive? 1) x^2 -2x + A is positive for all x 2) Ax^2 [#permalink]

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20 Apr 2009, 05:14

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Is A positive?

1) x^2 -2x + A is positive for all x 2) Ax^2 + 1 is positive for all x

My answer is E = Both together are not sufficient. However, test answer say A - Statement 1 is sufficient.

Explanation for statement 1 not sufficient

Case 1: A>0 say 6 then equation becomes x^2 - 2x + 6. Now if x=6 then equation will be equal to 30 which is greater than 0 Case 2: A<0 say -6 then equation becomes x^2 - 2x -6. Now if x=6 then equation will be equal to 18 which is also greater than 0.

So, S1 is true for both A less than 0 and greater than 0. So, we cant definitely say if A > 0.

Can someone please explain if I am missing something here or the answer is incorrect.

1) x^2 -2x + A is positive for all x 2) Ax^2 + 1 is positive for all x

My answer is E = Both together are not sufficient. However, test answer say A - Statement 1 is sufficient.

Explanation for statement 1 not sufficient

Case 1: A>0 say 6 then equation becomes x^2 - 2x + 6. Now if x=6 then equation will be equal to 30 which is greater than 0 Case 2: A<0 say -6 then equation becomes x^2 - 2x -6. Now if x=6 then equation will be equal to 18 which is also greater than 0.

So, S1 is true for both A less than 0 and greater than 0. So, we cant definitely say if A > 0.

Can someone please explain if I am missing something here or the answer is incorrect.

Thanks Akshay

You haven't used a crucial piece of information in Statement 1:

x^2 -2x + A is positive for all x

That is, it must be positive for every value of x, not only for x = 6. In particular, x^2 - 2x + A must be positive when x = 0, so 0^2 - 2*0 + A > 0, or A > 0, and the statement is sufficient.

Statement 2 is not sufficient because A could be positive, or could be zero.
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For statement 1, simply plug in x=0 to find that A>0.

Ian, in statement 2, A can be positive, 0 or negative (Say x=1/2 and A=-1/2).

Also, Akshay, you cannot assume a value for A first and then go about proving that the inequality is positive. It says that the inequality is positive for all values of x NOT for all values of A.

Ian, in statement 2, A can be positive, 0 or negative (Say x=1/2 and A=-1/2).

No, if Statement 2 is true, A cannot be negative. Statement 2 tells us that: Ax^2 + 1 is positive for all x, and not only for x = 1/2. In your example, where A = -1/2, you can quickly see that Ax^2 + 1 will be negative for x = 10, for example, so A cannot be -1/2; that disagrees with the information in the statement. You can see that, if A is negative, Ax^2 + 1 will be zero (so certainly not positive) whenever \(x = \frac{1}{\sqrt{ |A| }\). Plugging that in for x:

and if \(|x| > \frac{1}{\sqrt{ |A| }\), then Ax^2 + 1 will be negative, if A is negative. So if Statement 2 is true, A cannot be negative. A can, however, be zero, which is why the statement is insufficient.
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I agree my approach for picking up value of A was not appropriate.

But what was confusing me was that range of A was becoming unpredictable for other values of x. Say x=4 then A>-8. From this its not possible to say whether A>0.

But yes, as it says FOR ALL X - we do have a value for x (i.e. 0) from where we can say A>0

What if we take the following values: x= 0 then A = +ve x= 1 then 1-2+A = +ve thus, -1 +A = +ve Thus A > 1 x= -1 then 1+2+A = +ve, CAN A here be "-2" then ?

1) x^2 -2x + A is positive for all x 2) Ax^2 + 1 is positive for all x

My answer is E = Both together are not sufficient. However, test answer say A - Statement 1 is sufficient.

Explanation for statement 1 not sufficient

Case 1: A>0 say 6 then equation becomes x^2 - 2x + 6. Now if x=6 then equation will be equal to 30 which is greater than 0 Case 2: A<0 say -6 then equation becomes x^2 - 2x -6. Now if x=6 then equation will be equal to 18 which is also greater than 0.

So, S1 is true for both A less than 0 and greater than 0. So, we cant definitely say if A > 0.

Can someone please explain if I am missing something here or the answer is incorrect.

Thanks Akshay

IMO A.

Since both statements > 0 for any value of x, then we should try to find such a value of x that combined with a<=0 would make the whole expression <=0. Plugging numbers for this one is not hard, and quick.

stmnt1 - whatever value you take for x, the expression will always be >0 only if a>0, try to pick numbers for x, e.g. x-2,0,5,etc. sufficient

stmnt2 - if x=1, and a=-1, then the expression becomes 0: -1*1^2+1=0 try other values for x and a, and you will either confirm the whole expression true or false insufficient