The parabola y = x^2 + px + q cuts the straig | vedclass

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(D) Given the line $y = 2x - 3$. At $x = 1$,$y = 2(1) - 3 = -1$. Since the parabola $y = x^2 + px + q$ passes through $(1, -1)$,we have $-1 = 1^2 + p(

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Both $(B)$ and $(C)$

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(D) Given the line $y = 2x - 3$. At $x = 1$,$y = 2(1) - 3 = -1$. Since the parabola $y = x^2 + px + q$ passes through $(1, -1)$,we have $-1 = 1^2 + p(1) + q$,which simplifies to $p + q = -2$,or $q = -2 - p$. The vertex of the parabola $y = x^2 + px + q$ is at $x = -p/2$. The $y$-coordinate of the vertex is $y_v = (-p/2)^2 + p(-p/2) + q = p^2/4 - p^2/2 + q = q - p^2/4$. The distance of the vertex from the $x$-axis is $|y_v| = |q - p^2/4|$. Substituting $q = -2 - p$,we get $d(p) = |(-2 - p) - p^2/4| = |- (p^2/4 + p + 2)|$. To minimize the distance,we analyze $f(p) = p^2/4 + p + 2$. Setting $f'(p) = p/2 + 1 = 0$,we get $p = -2$. For $p = -2$,$q = -2 - (-2) = 0$. The distance is $|0 - (-2)^2/4| = |-1| = 1$. Thus,both $(B)$ and $(C)$ are correct.

The parabola $y = x^2 + px + q$ cuts the straight line $y = 2x - 3$ at a point with abscissa $1$. If the distance between the vertex of the parabola and the $x$-axis is least,then:

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