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

Question

Let the domain of the function $f(x)={\log }_{2}{\log }_{4}{\log }_{6}(3+4x-{x}^{2})$ be $(a,b)$. If ${\int }_{b-a}^{0}[{x}^{2}]dx=p-\sqrt{q}-\sqrt{r},p,q,r\in N,gcd(p,q,r)=1$, 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+3Solve the quadratic inequality:

$(x-1)(x-3)This inequality indicates that $x$ must lie between the roots, giving the interval $x\in (1,3)$.

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:

${\int }_{b-a}^{0}[{x}^{2}]\, dx$

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

${\int }_{2}^{1}[{x}^{2}]\, dx={\int }_{\sqrt{2}}^{1}1\, dx+{\int }_{\sqrt{3}}^{\sqrt{2}}2\, dx+{\int }_{2}^{\sqrt{3}}3\, dx$

Computation of Each Integral Segment:

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

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

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

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