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 $p,q,r \in \mathbb{N}$, $\text{gcd}(p,q,r)=1$, and $[\cdot]$ is the greatest integer function, then $p+q+r$ is equal to
- (1) 10
- (2) 8
- (3) 11
- (4) 9
Solution:
$$f(x) = \log_{2}(\log_{4}(\log_{6}(3+4x-x^2)))$$
$$\log_{4}(\log_{6}(3+4x-x^2)) > 0$$
$$\log_{6}(3+4x-x^2) > 4^0$$
$$\log_{6}(3+4x-x^2) > 1$$
$$3+4x-x^2 > 6^1$$
$$x^2 – 4x + 3 < 0$$
$$(x-1)(x-3) < 0$$
$$x \in (1, 3)$$
$$a = 1, \quad b = 3$$
$$b – a = 3 – 1 = 2$$
$$I = \int_{0}^{2} [x^2] dx$$
$$I = \int_{0}^{1} [x^2] dx + \int_{1}^{\sqrt{2}} [x^2] dx + \int_{\sqrt{2}}^{\sqrt{3}} [x^2] dx + \int_{\sqrt{3}}^{2} [x^2] dx$$
$$I = \int_{0}^{1} 0 \, dx + \int_{1}^{\sqrt{2}} 1 \, dx + \int_{\sqrt{2}}^{\sqrt{3}} 2 \, dx + \int_{\sqrt{3}}^{2} 3 \, dx$$
$$I = 0 + [x]_{1}^{\sqrt{2}} + [2x]_{\sqrt{2}}^{\sqrt{3}} + [3x]_{\sqrt{3}}^{2}$$
$$I = (\sqrt{2} – 1) + 2(\sqrt{3} – \sqrt{2}) + 3(2 – \sqrt{3})$$
$$I = \sqrt{2} – 1 + 2\sqrt{3} – 2\sqrt{2} + 6 – 3\sqrt{3}$$
$$I = 5 – \sqrt{2} – \sqrt{3}$$
$$p – \sqrt{q} – \sqrt{r} = 5 – \sqrt{2} – \sqrt{3}$$
$$p = 5, \quad q = 2, \quad r = 3$$
$$\text{gcd}(5, 2, 3) = 1$$
$$p + q + r = 5 + 2 + 3 = 10$$
Ans. (1)