Dear Sir,

I would be most grateful if some friends can help to solve the attached question.

Thanks

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- July 24th 2010, 12:27 AMkingmanTurning point
Dear Sir,

I would be most grateful if some friends can help to solve the attached question.

Thanks - July 24th 2010, 06:32 AMSoroban
Hello, kingman!

I'm coming up with a strange result . . .

Quote:

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] forvalues of*all*

Is there a flaw in my reasoning?

- July 24th 2010, 07:24 AMkingman
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