Let the domain of the function f(x) = log_2 l | vedclass

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(A) For the function $f(x) = \log_2 \log_4 \log_6(3 + 4x - x^2)$ to be defined,we require:$\log_4 \log_6(3 + 4x - x^2) > 0 \implies \log_6(3 + 4x - x^

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$10$

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(A) For the function $f(x) = \log_2 \log_4 \log_6(3 + 4x - x^2)$ to be defined,we require:$\log_4 \log_6(3 + 4x - x^2) > 0 \implies \log_6(3 + 4x - x^2) > 1 \implies 3 + 4x - x^2 > 6$$x^2 - 4x + 3 Thus,$a = 1$ and $b = 3$,so $b - a = 2$.We need to evaluate $\int_0^2 [x^2] dx$.Since $[x^2] = k$ for $k \le x^2 $I = \int_0^1 [x^2] dx + \int_1^{\sqrt{2}} [x^2] dx + \int_{\sqrt{2}}^{\sqrt{3}} [x^2] dx + \int_{\sqrt{3}}^2 [x^2] dx$$I = 0 + 1(\sqrt{2} - 1) + 2(\sqrt{3} - \sqrt{2}) + 3(2 - \sqrt{3})$$I = \sqrt{2} - 1 + 2\sqrt{3} - 2\sqrt{2} + 6 - 3\sqrt{3} = 5 - \sqrt{2} - \sqrt{3}$.Comparing with $p - \sqrt{q} - \sqrt{r}$,we get $p = 5, q = 2, r = 3$.Thus,$p + q + r = 5 + 2 + 3 = 10$.

Let the domain of the function $f(x) = \log_2 \log_4 \log_6(3 + 4x - x^2)$ be $(a, b)$. If $\int_0^{b-a} [x^2] dx = p - \sqrt{q} - \sqrt{r}$,where $p, q, r \in \mathbb{N}$,$\gcd(p, q, r) = 1$,and $[\cdot]$ is the greatest integer function,then $p + q + r$ is equal to

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