Let $f: (0, \frac{6}{5}) \to \mathbb{R}$ & $g: (0, \frac{6}{5}) \to \mathbb{R}$ be functions defined by

$f(x) = [x^2]$ and $g(x) = (|x - 1| + |x - 2|)f(x)$

Here $[a] =$ the highest integer $\leq a$. Then

Looking at the domain $(0, \frac{6}{5}) = (0, 1.2)$.

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Analyzing $f(x) = [x^2]$

Since $x \in (0, 1.2)$, we have $x^2 \in (0, 1.44)$.

The floor function $[x^2]$ gives the greatest integer $\leq x^2$.

Range of $x$	Range of $x^2$	Value of $[x^2]$
$(0, 1)$	$(0, 1)$	$0$
$[1, 1.2)$	$[1, 1.44)$	$1$

The floor function jumps (and hence $f$ becomes discontinuous) when $x^2$ crosses an integer value. In our domain, this happens only when $x^2 = 1$, i.e., at $x = 1$.

Therefore, $f$ is discontinuous at exactly one point in $(0, \frac{6}{5})$.

Options 1 and 3 are incorrect.

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Analyzing $g(x) = (|x - 1| + |x - 2|)f(x)$

First, simplify $|x - 1| + |x - 2|$ for $x \in (0, 1.2)$.

Since all values in the domain are less than 2:

Range of $x$	$\|x - 1\|$	$\|x - 2\|$	$\|x - 1\| + \|x - 2\|$
$(0, 1)$	$1 - x$	$2 - x$	$3 - 2x$
$[1, 1.2)$	$x - 1$	$2 - x$	$1$

Now multiply by $f(x)$:

Range of $x$	$f(x)$	$\|x - 1\| + \|x - 2\|$	$g(x)$
$(0, 1)$	$0$	$3 - 2x$	$0$
$[1, 1.2)$	$1$	$1$	$1$

So $g(x) = \begin{cases} 0 & \text{if } 0 < x < 1 \\ 1 & \text{if } 1 \leq x < 1.2 \end{cases}$

This is a step function that jumps at $x = 1$. Since $g$ is not even continuous at $x = 1$, it cannot be differentiable there.

For all other points in the domain, $g$ is constant (and hence differentiable) on each piece.

Therefore, $g$ is not differentiable at exactly one point in $(0, \frac{6}{5})$.

Option 2 is correct.