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(D) The maximum value of the quadratic expression $1-2x-5x^2$ is found at $x = -\frac{b}{2a} = -\frac{-2}{2(-5)} = -\frac{1}{5}$.Substituting $x = -\f

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$(-\infty, \infty)$

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(D) The maximum value of the quadratic expression $1-2x-5x^2$ is found at $x = -\frac{b}{2a} = -\frac{-2}{2(-5)} = -\frac{1}{5}$.Substituting $x = -\frac{1}{5}$,we get $a = 1 - 2(-\frac{1}{5}) - 5(-\frac{1}{5})^2 = 1 + \frac{2}{5} - \frac{1}{5} = \frac{6}{5}$.The minimum value of the quadratic expression $x^2-2x+5$ is found at $x = -\frac{b}{2a} = -\frac{-2}{2(1)} = 1$.Substituting $x = 1$,we get $b = (1)^2 - 2(1) + 5 = 4$.Now,substitute $a = \frac{6}{5}$ and $b = 4$ into the expression $5ax^2+bx+7$:$5(\frac{6}{5})x^2 + 4x + 7 = 6x^2 + 4x + 7$.For the quadratic $6x^2 + 4x + 7 > 0$,we check the discriminant $D = b^2 - 4ac = (4)^2 - 4(6)(7) = 16 - 168 = -152$.Since $D  0$,the expression $6x^2 + 4x + 7$ is always positive for all real values of $x$.Thus,the set of all values of $x$ is $(-\infty, \infty)$.

If $a$ and $b$ are the maximum and minimum values of the quadratic expressions $1-2x-5x^2$ and $x^2-2x+5$ respectively,then the set of all values of $x$ for which the expression $5ax^2+bx+7$ is positive,is

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