Dear Sir,
I would be most grateful if some friends can help to solve the attached question.
Thanks
Hello, kingman!
I'm coming up with a strange result . . .
A curve has the equation: . where is a constant.
By considering . . . . I don't see how
. . find the range of such that the equation: .
. . has exactly 2 real roots for all real values of
The equation is: .
. . which simplifies to: .
A quadratic has two real roots if its discriminant is positive.
The discriminant is: .
. . And we want: . .[1]
This is an up-opening parabola.
. . It will be positive if its vertex is above the -axis.
Its vertex is found at: .
. . Hence, the vertex is: .
The parabola is positive if: .
We have a constaint on
Yet we must find values of which satisfy [1] for all values of
Is there a flaw in my reasoning?
If eq 1 is written in powers of k instead of p we can get the answer but I still wonder why.But question required one to start from dy/dy with the hint that if dy/dy =0 has no turning point than the answer follows and again I wonder why this is so.
Thanks