Define a function f(x) = frac{16x^2 - 96x + 1 | vedclass

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(D) Given,$f(x) = \frac{16x^2 - 96x + 153}{x - 3}$.Let $f(x) = y$.Then,$y = \frac{16x^2 - 96x + 153}{x - 3}$.$y(x - 3) = 16x^2 - 96x + 153$.$16x^2 - (

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$24$

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(D) Given,$f(x) = \frac{16x^2 - 96x + 153}{x - 3}$.Let $f(x) = y$.Then,$y = \frac{16x^2 - 96x + 153}{x - 3}$.$y(x - 3) = 16x^2 - 96x + 153$.$16x^2 - (96 + y)x + (153 + 3y) = 0$.Since $x$ is a real number,the discriminant $D \geq 0$.$D = (96 + y)^2 - 4(16)(153 + 3y) \geq 0$.$9216 + 192y + y^2 - 64(153 + 3y) \geq 0$.$9216 + 192y + y^2 - 9792 - 192y \geq 0$.$y^2 - 576 \geq 0$.$y^2 \geq 576$.This implies $y \in (-\infty, -24] \cup [24, \infty)$.Thus,the least positive value of $f(x)$ is $24$.

Define a function $f(x) = \frac{16x^2 - 96x + 153}{x - 3}$ for all real $x \neq 3$. The least positive value of $f(x)$ is

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