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If $f( x)=\dfrac{5x+2}{ 3x-5}$ and $g( x )=x^{2}-2x-1,$ then the value of $g( f( f( 3 ) ) )$ is

1. $2$
2. $1/3$
3. $6$
4. $2/3$

Given that, $f(x) = \frac{5x+2}{3x-5}$

And, $g(x) = x^{2}-2x-1$

Now, $f(3) = \frac{5(3)+2}{3(3)-5} = \frac{17}{4}$

$\Rightarrow f(f(3)) = f\left(\frac{17}{4}\right)$

$\Rightarrow f\left(\frac{17}{4}\right) = \frac{5\left(\frac{17}{4}\right)+2}{{3\left(\frac{17}{4}\right)-5}}$

$\Rightarrow f\left(\frac{17}{4}\right) = \dfrac{\frac{85+8}{4}}{\frac{51-20}{4}}$

$\Rightarrow f\left(\frac{17}{4}\right) =\frac{93}{31} = 3$

$\Rightarrow g(f(f(3))) = g\left(f\left(\frac{17}{4}\right)\right) = g(3)$

$\Rightarrow g(3) = 3^{2} – 2(3)-1 = 9-6-1 = 9-7 = 2$

$\therefore$ The value of $g(f(f(3))) = 2.$

Correct Answer $:\text{A}$
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