Let the domain of f(x)=_2_4_6(3+4x-x^2) be (a,b).

Question

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

Accepted Answer

Requiring $\log_6(3+4x-x^2)>1 \implies 3+4x-x^2>6 \implies x^2-4x+3<0 \implies (x-1)(x-3)<0$, so domain is $(1,3)$, giving $b-a=2$. $$\int_0^2[x^2]\,dx = \int_1^{\sqrt2}1\,dx+\int_{\sqrt2}^{\sqrt3}2\,dx+\int_{\sqrt3}^2 3\,dx = (\sqrt2-1)+2(\sqrt3-\sqrt2)+3(2-\sqrt3)$$ $$= 5-\sqrt2-\sqrt3$$ So $p=5,\,q=2,\,r=3$, and $p+q+r=10$.

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