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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