IPMAT IndoreAlgebra > Easy16152526✅ Correct Option: 1Related questions:IPMAT Indore 2024Let fff and ggg be two functions defined by f(x)=∣x+∣x∣∣f(x) = |x + |x||f(x)=∣x+∣x∣∣ and g(x)=1xg(x) = \frac{1}{x}g(x)=x1 for x≠0x \neq 0x=0. If f(a)+g(f(a))=136f(a) + g(f(a)) = \frac{13}{6}f(a)+g(f(a))=613 for some real aaa, then the maximum possible value off(g(a)) f(g(a))f(g(a)) is:IPMAT Indore 2023If f(1)=1f(1) = 1f(1)=1 and f(n)=3n−f(n−1)f(n) = 3n - f(n - 1)f(n)=3n−f(n−1) for all integers n>1n > 1n>1 , then the value of f(2023)f(2023)f(2023) isIPMAT 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 is