Is a positive?

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Is a positive?

by sachin_yadav » Sun Nov 15, 2015 12:00 pm
Is a positive?


(1) x^2 − 2x + a is positive for all x

(2) ax^2 + 1 is positive for all x

OA is A

Please give your explanation why correct answer is correct.

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by MartyMurray » Sun Nov 15, 2015 3:22 pm
Statement 1 can be broken into two parts, x² - 2x and a.

To prove that a must be positive, all we need is to create an x² - 2x that is not positive, because if x² - 2x is not positive, then a must be positive in order to generate a positive total.

Set x = 0.

x² - 2x = 0

In order for 0 + a to be positive, a must be positive.

So a has to be positive and Statement 1 is sufficient.

Statement 2 would be positive for all x if a were positive. It would also be positive for all x if a were 0. So a could be positive or 0, and so Statement 2 is insufficient.

Choose A.
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by Max@Math Revolution » Sat Nov 21, 2015 9:12 am
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

Is a positive?


(1) x^2 - 2x + a is positive for all x

(2) ax^2 + 1 is positive for all x

There is one variable (a) and 2 equations are given from the 2 conditions, making likely that (D) will be our answer.
For condition 1, from y=ax^2+bx+c D(disctriminant)=b^2-4ac=(-2)^2-4*1*a<0, it is always x^2-2x+a>0. Then 4-4a<0, or it is always sufficient when 1<a
For condition 2, it is 'yes' when a=1, and 'no' when a=0. This is insufficient and the answer becomes (A).

For cases where we need 1 more equation, such as original conditions with "1 variable", or "2 variables and 1 equation", or "3 variables and 2 equations", we have 1 equation each in both 1) and 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously there may be cases where the answer is A, B, C or E.

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by Matt@VeritasPrep » Fri Nov 27, 2015 12:50 am
S1::

We know

x² - 2x + a > 0

This is true for any value of x, so it must be true for x = 0. If x = 0, however, we have

a > 0

So a must be positive! Choosing a convenient value of x helped, and we thought of x = 0 because it eliminated the term we didn't care about (x) and left us with only the term we did (a).

S2::

ax² + 1 > 0

We'll use our convenient values of x again and suppose that x = 1. Then we have

a + 1 > 0

or

a > -1

Blast!

Another way to approach this would be to assume that a is 0. If it is, ax² + 1 will be > 0 for all x, since no matter what x is, 0*x² + 1 > 0 will be true. So a could be 0, and need not be positive.

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by sachin_yadav » Tue Jan 19, 2016 11:10 am
Matt@VeritasPrep wrote:S1::

We know

x² - 2x + a > 0

This is true for any value of x, so it must be true for x = 0. If x = 0, however, we have

a > 0

So a must be positive! Choosing a convenient value of x helped, and we thought of x = 0 because it eliminated the term we didn't care about (x) and left us with only the term we did (a).

S2::

ax² + 1 > 0

We'll use our convenient values of x again and suppose that x = 1. Then we have

a + 1 > 0

or

a > -1

Blast!

Another way to approach this would be to assume that a is 0. If it is, ax² + 1 will be > 0 for all x, since no matter what x is, 0*x² + 1 > 0 will be true. So a could be 0, and need not be positive.
How can A be the answer. If i put x = -2 in the equation i get 8 + a > 0
Now a can be 0 or any positive number.
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by DavidG@VeritasPrep » Tue Jan 19, 2016 12:06 pm
sachin_yadav wrote:
Matt@VeritasPrep wrote:S1::

We know

x² - 2x + a > 0

This is true for any value of x, so it must be true for x = 0. If x = 0, however, we have

a > 0

So a must be positive! Choosing a convenient value of x helped, and we thought of x = 0 because it eliminated the term we didn't care about (x) and left us with only the term we did (a).

S2::

ax² + 1 > 0

We'll use our convenient values of x again and suppose that x = 1. Then we have

a + 1 > 0

or

a > -1

Blast!

Another way to approach this would be to assume that a is 0. If it is, ax² + 1 will be > 0 for all x, since no matter what x is, 0*x² + 1 > 0 will be true. So a could be 0, and need not be positive.
How can A be the answer. If i put x = -2 in the equation i get 8 + a > 0
Now a can be 0 or any positive number.
We're told that x^2 - 2x + a is positive for all x, meaning that no matter what value we set x to, the entire expression has to be positive. If a = 0, then we have x^2 - 2x > 0, but this isn't true for all values of x. (If x = 1 or x = 0, for example.) Therefore, it cannot be the case that a = 0 because x^2 - 2x + a has to be positive for every single possible value of x.
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by GMATGuruNY » Tue Jan 19, 2016 12:46 pm
Is A positive?

1. x^2-2x+A is positive for all x
2. Ax^2+1 is positive for all x
Alternate approach:

Statement 1:
In other words, the graph of y = x² - 2x + A lies entirely above the x-axis, so that the value of y is always positive.

A parabola of the form y = ax² + bx + c, where a>0, opens UPWARD.
The result is a U-shaped graph that looks like this:
U.
The DISCRIMINANT of the parabola is equal to b² - 4ac.
The U-shaped graph will lie entirely above the x-axis -- and thus will yield only positive values for y -- if its discriminant is negative.
Implication:
Since y = x² - 2x + A must yield only positive values for y, its discriminant must be negative.

In y = x² - 2x + A, a=1, b=-2, and c=A.
Since b² - 4ac < 0, we get:
(-2)² - 4*1*A < 0
4 - 4A < 0
-4A < -4
A > 1.
SUFFICIENT.

Statement 2:
In other words, the graph of y = Ax² + 1 lies entirely above the x-axis, so that the value of y is always positive.

Case 1: A=0, so that y = Ax² + 1 becomes y=1.
Here, the graph is a horizontal line that lie entirely above the x-axis.

Case 2: A=1, so that y = Ax² + 1 becomes y = x² + 1
Here, since x² cannot be negative, every value for y will be positive, yielding a graph that lies entirely above the x-axis.

Since A=0 in Case 1 but A>0 in Case 2, INSUFFICIENT.

The correct answer is A.
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by Matt@VeritasPrep » Fri Jan 22, 2016 4:11 pm
sachin_yadav wrote: How can A be the answer. If i put x = -2 in the equation i get 8 + a > 0
Now a can be 0 or any positive number.
Remember that the equation must hold for ALL values of x. Since x = 0 is a possible value, whatever restrictions x = 0 gives you on a MUST HOLD for a in all cases. So if x = 0 forces a > 0, then a > 0, period, no matter what the other x values tell you.