IPMAT Indore 2019Algebra > Hardpositive and monotonically increasing for x ∈(−∞,5−572)\in (-\infty, \frac{5-\sqrt{57}}{2})∈(−∞,25−57) and x ∈(5+572,+∞)\in (\frac{5+\sqrt{57}}{2}, +\infty)∈(25+57,+∞)negative and monotonically decreasing for x ∈(−∞,5−572)\in (-\infty, \frac{5-\sqrt{57}}{2})∈(−∞,25−57) and x ∈(5+572,+∞)\in (\frac{5+\sqrt{57}}{2},+\infty)∈(25+57,+∞)negative and monotonically increasing for x ∈(−∞,5−572)\in (-\infty, \frac{5-\sqrt{57}}{2})∈(−∞,25−57) and positive and monotonically increasing for x ∈(5+572,+∞)\in (\frac{5+\sqrt{57}}{2},+\infty)∈(25+57,+∞)positive and monotonically increasing for x ∈(−∞,5−572)\in (-\infty, \frac{5-\sqrt{57}}{2})∈(−∞,25−57) and negative and monotonically decreasing for x ∈(5+572,+∞)\in (\frac{5+\sqrt{57}}{2},+\infty)∈(25+57,+∞)✅ Correct Option: 3Related questions:Suppose 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).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 isA real-valued function fff satisfies the relation f(x)f(y)=f(2xy+3)+3f(x+y)−3f(y)+6yf(x)f(y) = f(2xy + 3) + 3f(x + y) - 3f(y) + 6yf(x)f(y)=f(2xy+3)+3f(x+y)−3f(y)+6y, for all real numbers xxx and yyy, then the value of f(8)f(8)f(8) is