JEE Advanced · 2015 · 24 May 2015 · Paper 1 · Definite Integrals

The inner function satisfies $f(v)=[v]$ for $v\le 2$ and $f(v)=0$ for $v>2$. In the numerator $f(x^2)$ is non-zero only where $x^2\le 2$, i.e. $|x|\le \sqrt{2}$, and there it equals $[x^2]$, which is $0$ for $x^2<1$ and $1$ for $1\le x^2<2$. In the denominator $f(x+1)$ becomes $0$ once $x+1>2$, i.e. $x>1$, so there the denominator is simply $2$. Combining these, the integrand vanishes on $[-1,1]$ because $[x^2]=0$, and it also vanishes on $(\sqrt{2},2]$ because $f(x^2)=0$, leaving only $1\le x\le \sqrt{2}$, where it reduces to $\frac{x}{2}$. Hence $I={\int }_1^{\sqrt{2}}\frac{x}{2}dx=\frac{1}{4}\left(x^2\right){|}_1^{\sqrt{2}}=\frac{1}{4}(2-1)=\frac{1}{4}$, so $4I-1=0$.