JEE Main 2025 — Definite Integration Question with Solution

Step 1: Ensure the innermost function is greater than zero:

${\log }_6(3+4x-x^2)>1$

Step 2: Simplify the inequality from Step 1:

$3+4x-x^2>6⟹x^2-4x+3<0$

Step 3: Solve the quadratic inequality:

$(x-1)(x-3)<0$

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:

${\int }_0^{b-a}[x^2]dx$

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

${\int }_1^2[x^2]dx={\int }_1^{\sqrt{2}}1dx+{\int }_{\sqrt{2}}^{\sqrt{3}}2dx+{\int }_{\sqrt{3}}^{2}3dx$

Computation of Each Integral Segment:

${\int }_1^{\sqrt{2}}1dx=\sqrt{2}-1$

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

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

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:

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