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:IPMAT Indore 2021If a function f(a)=max(a,0)f(a) = max (a, 0)f(a)=max(a,0) then the smallest integer value of xxx for which the equation f(x−3)+2f(x+1)=8f(x - 3) + 2f(x + 1) = 8f(x−3)+2f(x+1)=8 holds true is _______.IPMAT Indore 2022A 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 isIPMAT Indore 2022If 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 _________.