JEE Advanced 2021 Paper 2 · Q19 · Definite Integrals

Question

For any real number $x$, let $[x]$ denote the largest integer less than or equal to $x$. If $$I = \int_0^{10}\left[\sqrt{\dfrac{10x}{x+1}}\right]dx,$$ then the value of $9I$ is _____.

SolveKar AI · Accepted Answer

Let $f(x) = \sqrt{\dfrac{10x}{x+1}}$. $\lfloor f(x)floor = n$ iff $n^2\le \dfrac{10x}{x+1} < (n+1)^2$, i.e. $\dfrac{n^2}{10-n^2}\le x < \dfrac{(n+1)^2}{10-(n+1)^2}$ (for $n^2<10$). For $n=0$: $0\le x < 1/9$ (length $1/9$). For $n=1$: $1/9\le x < 4/6 = 2/3$ (length $2/3 - 1/9 = 5/9$). For $n=2$: $2/3\le x < 9$ (length $9 - 2/3 = 25/3$). For $n=3$: $x\ge 9$ to $10$ (length 1, since $f(10)=\sqrt{100/11}<\sqrt{10}\approx 3.16$ but $\le \sqrt{10}<4$). $I = 0\cdot	frac{1}{9} + 1\cdot	frac{5}{9} + 2\cdot	frac{25}{3} + 3\cdot 1 = 	frac{5}{9} + 	frac{50}{3} + 3 = 	frac{5 + 150 + 27}{9} = 	frac{182}{9}$. Hence $9I = 182$.

JEE Advanced 2021 Paper 2 · Q19 · Definite Integrals

Solution