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p+q is always less than 1 and not greater than 1
Consider the graph of the function f(x) = x^2+px+q
Since the coeffiecient of x^2 is greater than zero the parabola would have the shape of a "bowl" and not that of a "mountain"
Now notice that when a bowl shaped parabola cuts the XAxis at two distinct points, it will take negative values for all values of x in between the roots.
[If you are having a hardtime visualizing it,just grab a pen and paper and see it for yourself!]
So f(1) < 0 or 1+p+q < 0 and hence the answer
You can also solve this problem by brute force.Not as elegant as the above one but yeah! it will get you the answer
Express the roots in terms of coefficients using the familiar formula for the roots of a quadratic equation
and then simplify the inequality
Alpha < 1 < Beta
You'll get the same answer.
One mate has already answered ... f(1) < 0
First of all lets write the coefficients ..!
a = 1 , b = p and c = q
SInce a > 0 the equation will open upward ...!
Hence for every x between root r1 and r2 ...the graph will have a negative value ..!
Given 1 lies between the root
So f(1) < 0
1 + p + q < 0
p+q < 1
And it will always be less than 1 ..never greater than 1 ..so please check out your question once again !
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question is wrong.
take p=4, q=2
1 lies between them but their sum is less than 1