Dear Sir,

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

Thanks

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

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

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

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

Quote:

A curve has the equation: .$\displaystyle y \:=\:\dfrac{2x^2+px+2}{x+1}$ where $\displaystyle p$ is a constant.

By considering $\displaystyle \dfrac{dy}{dx}$ . . . . I don't see how

. . find the range of $\displaystyle p$ such that the equation: .$\displaystyle \dfrac{2x^2+px+2}{x+1} \:=\:k$

. . has exactly 2 real roots for all real values of $\displaystyle k.$

The equation is: .$\displaystyle 2x^2 + px + 2 \:=\:k(x+1)$

. . which simplifies to: .$\displaystyle 2x^2 + (p-k)x + (2-k) \:=\:0$

A quadratic has two real roots if its discriminant is positive.

The discriminant is: .$\displaystyle b^2 - 4ac \;=\;(p-k)^2 - (4)(2)(2-k)$

. . And we want: .$\displaystyle p^2 - 2kp + k^2 + 8x - 16 \;>\;0$ .[1]

This is an up-opening parabola.

. . It will be positive if its vertex is above the $\displaystyle p$-axis.

Its vertex is found at: .$\displaystyle p \:=\:\dfrac{\text{-}b}{2a} \:=\:\dfrac{\text{-}(\text{-}2k)}{2(1)} \:=\:k$

. . Hence, the vertex is: .$\displaystyle (k,\;8k-16)$

The parabola is positive if: .$\displaystyle 8k-16 \:>\:0 \quad\Rightarrow\quad k \:>\:2$

We have a constaint on $\displaystyle k.$

Yet we must find values of $\displaystyle p$ which satisfy [1] forvalues of $\displaystyle k.$*all*

Is there a flaw in my reasoning?

- Jul 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