M13-23

Official Solution:


Is \(a\) positive?

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

It's crucial to notice that we are given that the expression is positive not just for some value of \(x\) but for every value of \(x\).

Next, since the coefficient of \(x^2\) is positive, \(f(x)=x^2-2x+a\) represents an upward parabola. Since \(f(x)\) is positive for all \(x\), it means that the parabola lies entirely above the x-axis and has no intersections with it. Therefore, the equation \(x^2-2x+a=0\) has no real roots.

For a quadratic equation not to have real roots, its discriminant must be negative. Thus, we have \(discriminant = 2^2 - 4a = 4 - 4a \lt 0\), which simplifies to \(1-a \lt 0\) and finally to \(a \gt 1\). Sufficient.

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

Notice that, if \(a > 0\), then \(ax^2+1\) is positive for every \(x\). But the trick here is that \(ax^2+1\) is also positive for every \(x\) when \(a = 0\), giving a NO answer to the question! Not sufficient.


Answer: A
_________________

I used another approach for S1 and got inequality a>4. Is it correct?

x^2-2x+4-4+a>0
(x-2)^2-4+a>0 for all x. Min of (x-2)^2 if x=2.
Therefore a>4

Sufficient

I did it with another approach, question says that X^2+2x+a is positive for all values of X so we can take any value of X
if he take X=0 then X^2 and 2x will also be -ve and remaining portion should be positive which is a
1) sufficient

in (1) it is said that x^2 -2x+a>0 for all x
if we substitute x=10 then 100-20+a= 80+a>0
therefore a>-80
hence a insufficient

can someone please tell where i have gone wrong

Hi jacobneroth,

Statement 1 says the expression must be positive for 'all' X...x^2 -2x+a>0 ..So we want to make sure that the expression stays positive for any value of X..In your substitution, the expression is positive for 10, when a is negative; however, if we substitute 0 for X, the expression will be positive only if a is positive...for the expression to hold positive for all values of X, 'a' must be positive.

Hope this helps..

can we approach this with number picking?

(1) if we pick x=1 then the inequality would become (1)^2 - 2*(1) + a ---> 1 - 2 + a. for the statement 1 to hold true and remain positive a has to be positive - so to say a has to compensate for whatever negative result may come out of x^2-2x. sufficient

(2) statement can hold true with any value of a. consider 0 or -0.5 or 1.

ANSWER A

Hi Bunuel VeritasKarishma,

Can't figure what is wrong with my method. I rejected option A because of the following thought process :

Q: Is a positive?
(1) x^2−2x+a is positive for all x

Lets say x = 5
We get 25-10 + (a) > 0
Here 'a' can be 15 + (-3) > 0
This means A can be negative

Again x = 0
We get 0 - 0 + a > 0
Here A has to be positive

Since we get a Yes and No answer
I thought A is insufficient

Nums99
This is a problem of "what is given and what is asked"


GIVEN in stmnt 1:
(1) x^2−2x+a is positive for all x

So for every value of "x" that you put, "a" should be such that the entire expression is positive.

Say, I put x = 1. The expression becomes 1 - 2 + a.
This needs to be positive. So -1 + a > 0 or a > 1.

Say I put x = 0. The expression becomes a
This needs to be positive so a > 0

Say I put x = 3. The expression becomes 9 - 6 + a.
This needs to be positive. So 3 + a > 0 or a > -3

Now the point is that EACH of these cases needs to be satisfied since we know that the expression MUST be positive for EVERY value of x. So "a" must be greater than 0 and 1 and -3 and so on... Out of our examples, we have found that a must be at least greater than 1 so it must be positive.
_________________

I think this is a high-quality question and I agree with explanation.

Hello Bunuel KarishmaB . Requesting your response on this.

One of the usual way of solving any DS problem is to find the contradicting values and if we get two opposing values, then we can clearly say that the particular statement is insufficient in answering the question.

So if I consider statement 1 with this line of reasoning-

Let's say x = 10 and final value of expression should be positive
10^2 - 2 (10) + a > 0
100-20+a>0
80+a>0

So, to keep this expression true, I can say that a is -50 or 50, but I cannot say weather a is positive or negative.

Similarly, if I take x = 3
3^2 - 2 (3) + a > 0
9-6+a>0
3+a>0

Again, a can be -1 or +1, hence I cannot say weather a is positive or negative.

Moreover, I am able to substitute positive and negative values of "a" to keep the expression valid.

Can you please help in understanding why this is not correct?

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
_________________

I think this is a high-quality question and I agree with explanation.