Question

Let the function $f\left(x\right)$ and $g\left(x\right)$ be defined as
$f\left(x\right)=\{\begin{array}{cc}\sqrt{x}& 0\le x<1\\ 2-x& 1\le x<2\\ f(x+2)& \forall x\in R\end{array}$ and $g\left(x\right)=4f\left(3x\right)+1,\forall x\in R$.

Let $A$ denotes the sum of all the solutions of the equation $f\left(x\right)=0.6,3\le x\le 7$.
$B$ denotes the fundamental period of $g\left(x\right)$.
$C$ denotes the value of $g^{\prime }\left(6.75\right)$.
Then, the value of $\left[ABC\right]$ is
(where [.] represents greatest integer function)

Answer 1

As $f(x+2)=f\left(x\right)$, so $f\left(x\right)$ is a periodic function with period as $2$,
So, the period of $g\left(x\right)$ is $\frac{2}{3}$
Drawing the graph of $y=f\left(x\right)$, we get

Now,
$f\left(x\right)=0.6$
When $x\in [0,1)$
$\Rightarrow \sqrt{x}=0.6\Rightarrow x=0.36$
When $x\in [1,2)$
$\Rightarrow 2-x=0.6\Rightarrow x=1.4$
Now , the sum of solution of $f\left(x\right)=0.6$ for $x\in [3,7]$, we get
$A=4.36+6.36+3.4+5.4=19.52B=\frac{2}{3}$
Now,
$g\left(x\right)=4f\left(3x\right)+1\Rightarrow g^{\prime }\left(x\right)=12f^{\prime }\left(3x\right)\Rightarrow g^{\prime }\left(6.75\right)=12f^{\prime }(3\times 6.75)\Rightarrow g^{\prime }\left(6.75\right)=12f^{\prime }\left(20.25\right)\Rightarrow g^{\prime }\left(6.75\right)=12f^{\prime }(20+0.25)\Rightarrow g^{\prime }\left(6.75\right)=12f^{\prime }\left(0.25\right)$

When $x\in [0,1)$
$f\left(x\right)=\sqrt{x}\Rightarrow f^{\prime }\left(x\right)=\frac{1}{2\sqrt{x}}\Rightarrow f^{\prime }\left(0.25\right)=\frac{1}{2\sqrt{0.25}}\Rightarrow f^{\prime }\left(0.25\right)=1$
Therefore,
$g^{\prime }\left(6.75\right)=12\therefore \left[ABC\right]=[19.52\times \frac{2}{3}\times 12]=156$