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(B) The length of the intercept on the $X$-axis made by the curve $f(x) = a x^2 + b x + c$ is given by $A = |x_1 - x_2| = \frac{\sqrt{D}}{|a|}$,where

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$A_4 < A_2 < A_3$

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(B) The length of the intercept on the $X$-axis made by the curve $f(x) = a x^2 + b x + c$ is given by $A = |x_1 - x_2| = \frac{\sqrt{D}}{|a|}$,where $D = b^2 - 4 a c$.For $f_1(x) = 5 x^2 + 2 x + 1$,$D = 2^2 - 4(5)(1) = 4 - 20 = -16$. Since $D For $f_2(x) = 5 x^2 + 6 x + 1$,$D = 6^2 - 4(5)(1) = 36 - 20 = 16$. Thus,$A_2 = \frac{\sqrt{16}}{5} = \frac{4}{5} = 0.8$.For $f_3(x) = x^2 - 7 x + 6$,$D = (-7)^2 - 4(1)(6) = 49 - 24 = 25$. Thus,$A_3 = \frac{\sqrt{25}}{1} = 5$.For $f_4(x) = 64 x^2 + 48 x + 9$,$D = 48^2 - 4(64)(9) = 2304 - 2304 = 0$. Thus,$A_4 = \frac{\sqrt{0}}{64} = 0$.Comparing the values: $A_4 = 0$,$A_2 = 0.8$,$A_3 = 5$.Therefore,$A_4 < A_2 < A_3$ is true.

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$f_1(x) = 5 x^2 + 2 x + 1$	$f_2(x) = 5 x^2 + 6 x + 1$
$f_3(x) = x^2 - 7 x + 6$	$f_4(x) = 64 x^2 + 48 x + 9$