Question

Show that the expression\[(px^{2}+3x-4) / (p+3x-4x^{2})\]
will be capable of all values when x is real, provided that p has any value between 1 and 7.

Answer 1

@nitz

Answer 2

i've tried it by taking the expression equal to p,then i made a quadratic eq. in terms of y.

Answer 3

As x is capable of all values, i applied the quadratic eq. in D<0.

Answer 4

Then i am stuck.

Answer 5

\[y = \frac{px^{2}+3x-4}{-4x^{2}+3x+p}\]
\[\Rightarrow -4yx^{2} + 3xy + py = px^{2}+3x-4\]
\[\Rightarrow (p+4y)x^{2}+(3-3y)x+(-4-py) = 0\]we know that \(x\) can take all real values, so, the equation above in terms of \(x\) must have real solutions.
\(\Rightarrow D > 0\) i.e. \((3-3y)^{2} - 4(p+4y)(-4-py) > 0\)
\[(9+16p)y^{2}+(4p^{2}+46)y+(9+16p) > 0\] since this must be true for all values of y, there is no solution for the equation set to 0, i.e. \(D < 0\) i.e. \((4p^{2}+46)^{2} - 4(9+16p)^{2} < 0\) or \((2p^{2}+23)^{2} -(9+16p)^{2}<0\)
\[\Rightarrow (2p^{2}+16p+32)(2p^{2}-16p+14)< 0\] note that \(2p^{2} + 16p +32 \ge 0\) for all values of p. So we have \(2p^{2}-16p+14 < 0\) i.e. \(p^{2}-8p+7 < 0\)
or \((p-1)(p-7) < 0\) So, \(p\) lies between \((1,7)\)

Answer 6

thank you.