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m13/q23....wrong ans

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m13/q23....wrong ans [#permalink] New post 03 Jan 2010, 09:27
Is A positive?

i)x^2-2x+A is positive for all x
ii)Ax^2+1 is positive for all x
reason given < Statement (1) by itself is sufficient. x^2-2x+A = (x-1)^2+(A-1) . For this expression to be always positive (A-1) has to be more than 0.so A has to be more than 1>
A can be any -ive int till (x-1)^2 is greater than (A-1)... ex x=3... 4+A-1>0.. or A>-3
ans shud be E...pl confirm
i think where we have gone wrong is taking + in(x-1)^2+(A-1) as *....
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Re: m13/q23....wrong ans [#permalink] New post 03 Jan 2010, 18:13
chetan2u wrote:
Is A positive?

i)x^2-2x+A is positive for all x
ii)Ax^2+1 is positive for all x
reason given < Statement (1) by itself is sufficient. x^2-2x+A = (x-1)^2+(A-1) . For this expression to be always positive (A-1) has to be more than 0.so A has to be more than 1>
A can be any -ive int till (x-1)^2 is greater than (A-1)... ex x=3... 4+A-1>0.. or A>-3
ans shud be E...pl confirm
i think where we have gone wrong is taking + in(x-1)^2+(A-1) as *....


Step by step:

Question: is A>0?

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

f(x)=x^2-2x+A is a function of of upward parabola. We are told that it's positive for all x --> f(x)=x^2-2x+A>0, which means that this function is "above" X-axis OR in other words parabola has no intersections with X-axis OR equation x^2-2x+A=0 has no real roots.

Quadratic equation to has no real roots discriminant must be negative --> D=2^2-4A=4-4A<0 --> 1-A<0 --> A>1.

Sufficient.

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

Ax^2+1>0 --> when A\geq0 this expression is positive for all x. So A can be zero too.

Not sufficient.

So I think the answer (A) is correct.
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Re: m13/q23....wrong ans [#permalink] New post 04 Jan 2010, 04:43
hi .....
i do understand that as an eq of parabola c would be y-intercept and parabola being limited to i and ii quad , c would be always +ive...... but why cant we prove it taking it as a normal eq (as it wud seem to any layperson)... is there any way to prove it
Re: m13/q23....wrong ans   [#permalink] 04 Jan 2010, 04:43
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