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(C) Let $f(x) = -8x^2 + 6x - 1$. We need to find the intervals where $f(x)$ takes integer values.$f(x) = 0 \implies -8x^2 + 6x - 1 = 0 \implies 8x^2 -

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$\frac{\sqrt{17}-13}{8}$

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(C) Let $f(x) = -8x^2 + 6x - 1$. We need to find the intervals where $f(x)$ takes integer values.$f(x) = 0 \implies -8x^2 + 6x - 1 = 0 \implies 8x^2 - 6x + 1 = 0 \implies (2x-1)(4x-1) = 0$. So,$x = 1/4$ and $x = 1/2$.$f(x) = -1 \implies -8x^2 + 6x - 1 = -1 \implies -8x^2 + 6x = 0 \implies -2x(4x-3) = 0$. So,$x = 0$ and $x = 3/4$.$f(x) = -2 \implies -8x^2 + 6x - 1 = -2 \implies 8x^2 - 6x - 1 = 0$. Using the quadratic formula,$x = \frac{6 \pm \sqrt{36 - 4(8)(-1)}}{16} = \frac{6 \pm \sqrt{68}}{16} = \frac{3 \pm \sqrt{17}}{8}$. Since $x \in [0, 1]$,we take $x = \frac{3+\sqrt{17}}{8} \approx 0.89$.$f(x) = -3 \implies -8x^2 + 6x - 1 = -3 \implies 8x^2 - 6x - 2 = 0 \implies 4x^2 - 3x - 1 = 0 \implies (4x+1)(x-1) = 0$. So,$x = 1$.Now,we split the integral based on the values of $[f(x)]$:For $x \in [0, 1/4)$,$-1 For $x \in [1/4, 1/2]$,$0 \le f(x) \le 1/8$,so $[f(x)] = 0$.For $x \in (1/2, 3/4)$,$-1 For $x \in [3/4, \frac{3+\sqrt{17}}{8})$,$-2 \le f(x) For $x \in [\frac{3+\sqrt{17}}{8}, 1]$,$-3 \le f(x) Integral $I = \int_{0}^{1/4} (-1) dx + \int_{1/4}^{1/2} (0) dx + \int_{1/2}^{3/4} (-1) dx + \int_{3/4}^{\frac{3+\sqrt{17}}{8}} (-2) dx + \int_{\frac{3+\sqrt{17}}{8}}^{1} (-3) dx$$I = -[x]_{0}^{1/4} + 0 - [x]_{1/2}^{3/4} - 2[x]_{3/4}^{\frac{3+\sqrt{17}}{8}} - 3[x]_{\frac{3+\sqrt{17}}{8}}^{1}$$I = -\frac{1}{4} - (\frac{3}{4} - \frac{1}{2}) - 2(\frac{3+\sqrt{17}}{8} - \frac{3}{4}) - 3(1 - \frac{3+\sqrt{17}}{8})$$I = -\frac{1}{4} - \frac{1}{4} - 2(\frac{3+\sqrt{17}-6}{8}) - 3(\frac{8-3-\sqrt{17}}{8})$$I = -\frac{1}{2} - \frac{\sqrt{17}-3}{4} - \frac{15-3\sqrt{17}}{8} = \frac{-4 - 2\sqrt{17} + 6 - 15 + 3\sqrt{17}}{8} = \frac{\sqrt{17}-13}{8}$

Let $[t]$ denote the greatest integer less than or equal to $t.$ Then,the value of the integral $\int\limits_{0}^{1}\left[-8 x^{2}+6 x-1\right] d x$ is equal to

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