Let $$f(x) = \left\{\begin{array}{l l}\frac{ax^{2}+2ax+3}{4x^{2}+4x-3} ,& x\neq\quad -\frac{3}{2},\frac{1}{2}\\b, & \quad x=-\frac{3}{2},\frac{1}{2}\\\end{array}\right.$$

be continuous at $$x=-\frac{3}{2}$$. If $$fof(x)=\frac{7}{5}$$ then x is equal to:

The given function is
$$f(x)=\begin{cases}\dfrac{ax^{2}+2ax+3}{4x^{2}+4x-3}, & x\neq-\dfrac{3}{2},\,\dfrac{1}{2}\\[4pt] b, & x=-\dfrac{3}{2},\,\dfrac{1}{2}\end{cases}$$
For continuity, the limit of the first branch as $$x\rightarrow-\dfrac{3}{2}$$ must be finite, so both numerator and denominator must vanish there:

Denominator:
$$4x^{2}+4x-3=0$$ at $$x=-\dfrac{3}{2}$$ (verify: $$4(2.25)-6-3=0$$).

Numerator:
$$ax^{2}+2ax+3=0\quad\text{at}\;x=-\dfrac{3}{2}$$

Substitute $$x=-\dfrac{3}{2}$$:
$$a\left(\dfrac{9}{4}\right)+2a\left(-\dfrac{3}{2}\right)+3=0$$
$$a=4$$.
Numerator: $$4x^{2}+8x+3=(2x+3)(2x+1)$$
Denominator: $$4x^{2}+4x-3=(2x+3)(2x-1)$$

For $$x\neq-\dfrac{3}{2},\,\dfrac{1}{2}$$ the common factor $$2x+3$$ cancels, giving
$$f(x)=\dfrac{2x+1}{2x-1}\quad\text{(with a hole at }x=-\dfrac{3}{2}\text{)}.$$
The limit as $$x\rightarrow-\dfrac{3}{2}$$ of the simplified form is
$$b=f\!\Bigl(-\dfrac{3}{2}\Bigr)=\dfrac{2\left(-\dfrac{3}{2}\right)+1}{2\left(-\dfrac{3}{2}\right)-1}= \dfrac{-3+1}{-3-1}= \dfrac{-2}{-4}= \dfrac{1}{2}.$$
Let $$y=f(x)=\dfrac{2x+1}{2x-1}$$ (defined for $$x\neq\dfrac12,-\dfrac32$$).
$$f(f(x))=f(y)=\dfrac{2y+1}{2y-1}.$$
$$\dfrac{2y+1}{2y-1}=\dfrac{7}{5}$$
$$y=3.$$

$$\dfrac{2x+1}{2x-1}=3$$
$$x=1.$$

Check the domain: $$x=1$$ is not $$\dfrac12$$ or $$-\dfrac32$$, so it is valid.

Therefore, the required value of $$x$$ is $$1$$.
Option D.