JEE Main 2025 — 3 April, Morning Shift — Question 36

$\log _{4} \log _{6}\left(3+4 \mathrm{x}-\mathrm{x}^{2}\right)>0$

$\log _{6}\left(3+4 \mathrm{x}-\mathrm{x}^{2}\right)>1$

$3+4 x-x^{2}>6$

$\mathrm{x}^{2}-4 \mathrm{x}+3<0$

$(x-1)(x-3)<0$ $x \in(1,3)$

so $\mathrm{a}=1 \& \mathrm{~b}=3$

$\Rightarrow \int_{0}^{2}\left[\mathrm{x}^{2}\right] \mathrm{dx}=$ ?

$I=\int_{0}^{1}\left[x^{2}\right] d x+\int_{1}^{\sqrt{2}}\left[x^{2}\right] d x+\int_{\sqrt{2}}^{\sqrt{3}}\left[x^{2}\right] d x+\int_{\sqrt{3}}^{\sqrt{4}}\left[x^{2}\right] d x$

$=0+|x|_{1}^{\sqrt{2}}+2|x|_{\sqrt{2}}^{\sqrt{3}}+3|x|_{\sqrt{3}}^{\sqrt{4}}$

$=(\sqrt{2}-1)+2(\sqrt{3}-\sqrt{2})+3(2-\sqrt{3})$

$=5-\sqrt{2}-\sqrt{3} \Rightarrow \mathrm{p}+\mathrm{q}+\mathrm{r}=10$