Let $$f:\left(0,\frac{7}{5}\right)\longrightarrow$$ $$\mathbb{R}$$ and $$g:\left(0,\frac{7}{5}\right)\longrightarrow$$ $$\mathbb{R}$$ be functions defined by $$f(x)=2[x^2]$$ and $$g(x)=(2|x-1|+3|x-2|)f(x)$$ (where $$[x]$$ is the greatest integer less than or equal to $$x$$). Let 

$$a=$$ number of points of discontinuity of $$f$$,

$$b=$$ number of points of non-differentiability of $$f$$,

$$c=$$ number of points of discontinuity of $$g$$, and

$$d=$$ number of points of non-differentiability of $$g$$.

What is the value of $$a+b+c+d$$?

The domain is given by $$x \in \left(0, \frac{7}{5}\right)$$, which means $$0 < x < 1.4$$.

Let us first analyze the function $$f(x) = 2[x^2]$$.

As $$x$$ ranges from $$0$$ to $$1.4$$, $$x^2$$ ranges from $$0$$ to $$1.96$$.

The greatest integer function $$[x^2]$$ changes its value wherever $$x^2$$ is an integer ($$1$$).This happens at $$x = 1$$.

Breaking down $$f(x)$$ for intervals:

• For $$x \in (0, 1)$$, $$x^2 \in (0, 1)$$, so $$[x^2] = 0 \implies f(x) = 0$$.
  
• For $$x \in \left[1, \frac{7}{5}\right)$$, $$x^2 \in [1, 1.96)$$, so $$[x^2] = 1 \implies f(x) = 2$$.
  

Thus, $$f(x)$$ has a point of discontinuity at $$x = 1$$, making $$a = 1$$. Since $$f(x)$$ is a step function constant on intervals, its derivative is zero everywhere except at $$x = 1$$ where it is non-differentiable, making $$b = 1$$.

Now, let us consider $$g(x) = (2\vert{}x - 1\vert{} + 3\vert{}x - 2\vert{})f(x)$$.

Since we are restricted to the domain $$x \in \left(0, \frac{7}{5}\right)$$, the term $$\vert{}x - 2\vert{}$$ is always equal to $$-(x - 2) = 2 - x$$ because $$x < 1.4 < 2$$.

Therefore, within our domain, $$g(x)$$ simplifies to:

$$g(x) = (2\vert{}x - 1\vert{} + 3(2 - x))f(x) = (2\vert{}x - 1\vert{} + 6 - 3x)f(x)$$

Let's look at the behaviour of $$g(x)$$ over the two sub-intervals:

• For $$x \in (0, 1)$$, $$f(x) = 0$$, which implies $$g(x) = 0$$ identically on $$(0, 1)$$. Thus, $$g(x)$$ is continuous and differentiable on $$(0, 1)$$.
  
• For $$x \in \left[1, \frac{7}{5}\right)$$, $$f(x) = 2$$, so $$g(x) = 2(2(x - 1) + 6 - 3x) = 2(2x - 2 + 6 - 3x) = 2(4 - x) = 8 - 2x$$.
  

Let us check continuity and differentiability of $$g(x)$$ at $$x = 1$$:

• Left-hand limit as $$x \to 1^-$$ is $$\lim g(x) = 0$$.
  
• Right-hand value at $$x = 1$$ is $$g(1) = 8 - 2(1) = 6$$.
  

Since the left-hand limit ($$0$$) does not equal the function value ($$6$$), $$g(x)$$ is discontinuous at $$x = 1$$. Thus, $$c = 1$$.

Since $$g(x)$$ is discontinuous at $$x = 1$$, it is automatically non-differentiable at $$x = 1$$.

Are there any other points of non-differentiability for $$g(x)$$?

Inside the interval $$\left[1, \frac{7}{5}\right)$$, $$g(x) = 8 - 2x$$, which is a straight line and is completely differentiable.

Hence, the only point of non-differentiability for $$g(x)$$ is $$x = 1$$, so $$d = 1$$.

Summing up all the values:

$$a + b + c + d = 1 + 1 + 1 + 1 = 4$$

The correct option is A.