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Question

Let the function f(x) and g(x) be defined as
f(x)=x0x<12x1x<2f(x+2) xR and g(x)=4f(3x)+1, xR.

Let A denotes the sum of all the solutions of the equation f(x)=0.6, 3x7.
B denotes the fundamental period of g(x).
C denotes the value of g(6.75).
Then, the value of [ABC] is
(where [.] represents greatest integer function)

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Solution

As f(x+2)=f(x), so f(x) is a periodic function with period as 2,
So, the period of g(x) is 23
Drawing the graph of y=f(x), we get

Now,
f(x)=0.6
When x[0,1)
x=0.6x=0.36
When x[1,2)
2x=0.6x=1.4
Now , the sum of solution of f(x)=0.6 for x[3,7], we get
A=4.36+6.36+3.4+5.4=19.52B=23
Now,
g(x)=4f(3x)+1g(x)=12f(3x)g(6.75)=12f(3×6.75)g(6.75)=12f(20.25)g(6.75)=12f(20+0.25)g(6.75)=12f(0.25)

When x[0,1)
f(x)=xf(x)=12xf(0.25)=120.25f(0.25)=1
Therefore,
g(6.75)=12[ABC]=[19.52×23×12]=156

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