CATAlgebra > EasyEntered answer:✅ Correct Answer: 10Related questions:CAT 2017 Slot 2Let f(x)=x2f(x) = x^2f(x)=x2 and g(x)=2xg(x) = 2^xg(x)=2x, for all real xxx. Then the value of f(f(g(x))+g(f(x)))f(f(g(x)) + g(f(x)))f(f(g(x))+g(f(x))) at x=1x = 1x=1 isCAT 2022 Slot 2Suppose for all integers x, there are two function f and g such that f(x)+f(x−1)−1=0f(x) + f(x-1) - 1 = 0f(x)+f(x−1)−1=0 and g(x)=x2g(x) = x^2g(x)=x2. If f(x2−x)=5f(x^2 - x) = 5f(x2−x)=5, then the value of the sum f(g(5))+g(f(5))f(g(5)) + g(f(5))f(g(5))+g(f(5)) isCAT 2019 Slot 1Consider a function fff satisfying f(x+y)=f(x)f(y)f(\mathrm{x}+\mathrm{y})=f(\mathrm{x}) \mathrm{f}(\mathrm{y})f(x+y)=f(x)f(y) where x,yx , yx,y are positive integers, and f(1)=2f(1)=2f(1)=2. If f(a+1)+f(a+2)+……+f(a+n)=f(\mathrm{a}+1)+f(\mathrm{a}+2)+\ldots \ldots+f(\mathrm{a}+\mathrm{n})=f(a+1)+f(a+2)+……+f(a+n)= 16(2n−1)16\left(2^{n}-1\right)16(2n−1) then a is equal to