IPMAT Indore 2021Algebra > MediumEntered answer:✅ Correct Answer: 3Related questions:A set of all possible values the function f(x)=x∣x∣f(x)=\dfrac{x}{|x|}f(x)=∣x∣x, where x≠0x \neq 0x=0, takes isSuppose that a real-valued function f(x)f(x)f(x) of real numbers satisfies f(x+xy)=f(x)+f(xyf(x + xy) = f(x) + f(xyf(x+xy)=f(x)+f(xy) for all real x,y,x, y,x,y, and that f(2020)=1f(2020) = 1f(2020)=1. Compute f(2021)f(2021)f(2021).If f(x2+f(y))=xf(x)+yf\left(x^{2}+f(y)\right)=x f(x)+yf(x2+f(y))=xf(x)+y for all non-negative integers xxx and yyy, then the value of [f(0)]2+f(0)[f(0)]^{2}+f(0)[f(0)]2+f(0) equals _________.