Let the domain of the function f(x)= _2 _4 _6(3+4 x-x^2) be (a, b).

Question

Let the domain of the function $f(x)=\log _2 \log _4 \log _6\left(3+4 x-x^2\right)$ be $(a, b)$. If $\int_0^{b-a}\left[x^2\right] d x=p-\sqrt{q}-\sqrt{r}, p, q, r \in \mathbb{N}, \operatorname{gcd}(p, q, r)=1$, where $[\cdot]$ is the greatest integer function, then $p+q+r$ is equal to

Accepted Answer

Ensure the innermost function is greater than zero:

$    \log_6(3 + 4x - x^2) > 1    $

Simplify the inequality from  $    3 + 4x - x^2 > 6 \implies x^2 - 4x + 3

Solve the quadratic inequality:

$    (x - 1)(x - 3)

With the domain of $ f(x) $ identified as $ (a, b) = (1, 3) $, we calculate the definite integral over $[0, b-a]$:
Calculate the Integral:
Given:

$ \int_0^{b-a} [x^2] \, dx $

For $ b - a = 2 $, we compute:

$ \int_1^2 [x^2] \, dx = \int_1^{\sqrt{2}} 1 \, dx + \int_{\sqrt{2}}^{\sqrt{3}} 2 \, dx + \int_{\sqrt{3}}^2 3 \, dx  $
Computation of Each Integral Segment:

$\int_1^{\sqrt{2}} 1 \, dx = \sqrt{2} - 1$

$\int_{\sqrt{2}}^{\sqrt{3}} 2 \, dx = 2(\sqrt{3} - \sqrt{2})$

$\int_{\sqrt{3}}^2 3 \, dx = 3(2 - \sqrt{3})$

Summing these, we have:

$ 5 - \sqrt{2} - \sqrt{3} $
Conclusion:
The values for $ p, q, $ and $ r $ are $ p = 5 $, $ q = 2 $, and $ r = 3 $, with the greatest common divisor of these numbers being 1. Therefore, adding them together gives:

$  p + q + r = 5 + 2 + 3 = 10 $

Solutions · 1