Let $$f : \left(0, \frac{6}{5}\right) \rightarrow R$$ & $$g : \left(0, \frac{6}{5}\right) \rightarrow R$$ be functions defined by $$f\left(x\right) = \left[x^{2}\right] \text{and} g\left(x\right) = \left(|x-1| + |x-2|\right)f\left(x\right)$$ Here $$[a] = \text{the highest integer} \leq a$$. Then

We need to check the continuity of the function $$f(x) = [x]$$ in the interval $$\left(0,\ \frac{\ 6}{5}\right)$$
$$[x]$$ = 0 when $$x\in\ \left[0,\ \ 1\right)$$
$$[x]$$ = 1 when $$x\in\ \left[1,\ \frac{\ 6}{5}\right)$$
Left hand limit of $$f\left(x\right)$$ $$\ne\ $$ Right hand limit of $$f\left(x\right)$$ at $$x=1$$
The function $$f\left(x\right)$$ is continuous elsewhere in the given domain of values. So it is discontinuous at 1 point.

Now we need to check for the differentiability of  $$g\left(x\right)$$ in $$\left(0,\ \frac{\ 6}{5}\right)$$
$$|x - 1| + |x - 2|$$ = $$1-x+2-x$$ = $$3-2x$$ when $$x\in\ \left[0,\ \ 1\right)$$
$$|x - 1| + |x - 2|$$ = 1 when $$x\in\ \left[1,\ \frac{\ 6}{5}\right]$$
Left hand limit of $$f\left(x\right)$$ = Right hand limit of $$f\left(x\right)$$ at $$x=1$$
Now, differentiating the function $$g\left(x\right)$$ wrt $$x$$ we get,
$$g'\left(x\right)$$ = -2 when $$x\in\ \left[0,\ \ 1\right)$$
$$g'\left(x\right)$$ = 0 when $$x\in\ \left[1,\ \frac{\ 6}{5}\right)$$
$$g'\left(x\right)$$ is not continuous and $$ x=1$$, and continuous elsewhere in the given domain.
Thus, $$g\left(x\right)$$ is not differentiable at 1 point in $$\left(0,\ \frac{\ 6}{5}\right)$$