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Question

Mathematics Question on Relations and functions

Let f:RRf :R→R be a function defined by f(x)=(2(1x252)(2+x25))150f(x)=(2(1−\frac {x^{25}}{2})(2+x^{25}))^{\frac {1}{50}}. If the function g(x)=f(f(f(x)))+f(f(x))g(x) = f (f (f (x))) + f (f (x)), then the greatest integer less than or equal to g(1)g(1) is _________.

Answer

f(x)=(2(1x252)(2+x25))150f(x)=(2(1−\frac {x^{25}}{2})(2+x^{25}))^{\frac {1}{50}}

f(x)=(2(2x252)(2+x25))150f(x)=(2(\frac {2−x^{25}}{{2}})(2+x^{25}))^{\frac {1}{50}}

=(4x50)150=(4−x^{50})^{\frac {1}{50}}
f(f(x))=(4((4x50)150)50)150=xf(f(x))=(4−((4−x^{50})^{\frac {1}{50}})^{50})^{\frac {1}{50}}=x
As f(f(x))=xf(f(x)) = x we have
g(x)=f(f(f(x)))+f(f(x))=f(x)+xg(x) = f(f(f(x))) + f(f(x)) = f(x) + x
g(x)=(4x50)150+xg(x) = (4 – x^{50})^{\frac {1}{50}} + x
g(1)=3150\+1g(1) = 3^{\frac {1}{50}} \+ 1
[g(1)]=2[g(1)] = 2

So, the answer is 22.